Rectiliner Motion – Jee (main+advanced) w2n14

RECTILINEAR MOTION  MECHANICS Mechanics is the branch of physics which deals with the cause and effects of motion of a particle, rigid objects and deformable bodies etc. Mechanics is classified under two streams namely Statics and Dynamics. Dynamics is further divided into Kinematics and Kinetics.

1.

MOTION AND REST Motion is a combined property of the object and the observer. There is no meaning of rest or motion without the observer. Nothing is in absolute rest or in absolute motion. An object is said to be in motion with respect to a observer, if its position

changes with

respect to that observer. It may happen by both ways either observer moves or object moves.

2.

RE CTI L IN E AR M OTI ON Rectilinear motion is motion, along a straight line or in one dimension. It deals with the kinematics of a particle in one dimension.

2.1

Position The position of a particle refers to its location in the space at a certain moment of time. It is concerned with the question  “where is the particle at a particular moment of time?”

2.2

Displacement The change in the position of a moving object is known as 

displacement. It is the vector ing the initial position ( r1 ) of

r2



the particle to its final position ( r2 ) during an interval of time. Displacement can be negative positive or zero.

2.3

D i s ta n c e The length of the actual path travelled by a particle during a given time interval is called as distance. The distance travelled is a scalar quantity which is quite different from displacement. In general, the distance travelled between two points may not be equal to the magnitude of the displacement between the same points.

RESONANCE

RECTILINEAR MOTION - 1

Example 1. 2

Ram takes path 1 (straight line) to go from P to Q and Shyam takes path 2 (semicircle). (a)

Find the distance travelled by Ram and Shyam?

(b)

Find the displacement of Ram and Shyam?

(a)

Distance travelled by Ram = 100 m

P

1 100 m

Q

Solution : Distance travelled by Shyam = (50 m) = 50 m (b)

Displacement of Ram = 100 m Displacement of Shyam = 100 m

 2.4

Average Velocity (in an interval) : The average velocity of a moving particle over a certain time interval is defined as the displacement divided by the lapsed time. Average Velocity =

displaceme nt time int erval

for straight line motion, along xaxis, we have v av = v = =

xf  xi x = tf  ti t

The dimension of velocity is [LT -1 ] and its SI unit is m/s. The average velocity is a vector in the direction of displacement. For motion in a straight line, directional aspect of a vector can be taken care of by +ve and -ve sign of the quantity.

2.5

Instantaneous Velocity (at an instant) : The velocity at a particular instant of time is known as instantaneous velocity. The term “velocity” usually means instantaneous velocity.

dx  x    = V inst. = lim t  0  t  dt In other words, the instantaneous velocity at a given moment (say , t) is the limiting value of the average velocity as we let t approach zero. The limit as t  0 is written in calculus notation as dx/dt and is called the derivative of x with respect to t.

Note : 

The magnitude of instantaneous velocity and instantaneous speed are equal.



The determination of instantaneous velocity by using the definition usually involves calculation of

derivative. W e can find v =

dx by using the standard results from dt

differential calculus. 

Instantaneous velocity is always tangential to the path.

RESONANCE

RECTILINEAR MOTION - 2

Example 2.

A particle starts from a point A and travels along the solid curve shown in figure. Find approximately the position B of the particle such that the average velocity between the positions A and B has the same direction as the instantaneous velocity at B.

Answer : Solution :

x = 5m, y = 3m The given curve shows the path of the particle starting at y = 4 m. Average velocity =

displacement time taken ; where displacement is straight line distance between points

Instantaneous velocity at any point is the tangent drawn to the curve at that point.

Now, as shown in the graph, line AB shows displacement as well as the tangent to the given curve. Hence, point B is the point at which direction of AB shows average as well as instantaneous velocity.

2.6

Averag e Sp eed (in an interva l) Average speed is defined as the total path length travelled divided by the total time interval during which the motion has taken place. It helps in describing the motion along the actual path. Average Speed =

distance travelled time interval

The dimension of velocity is [LT -1 ] and its SI unit is m/s.

Note : 

Average speed is always positive in contrast to average velocity which being a vector, can be positive or negative.



If the motion of a particle is along a straight line and in same direction then, average velocity = average speed.



Average speed is, in general, greater than the magnitude of average velocity.

RESONANCE

RECTILINEAR MOTION - 3

Example 3.

In the example 1, if Ram takes 4 seconds and Shyam takes 5 seconds to go from P to Q, find (a) Average speed of Ram and Shyam? (b) Average velocity of Ram and Shyam?

Solution :

(a)

Average speed of Ram =

100 m/s = 25 m/s 4

50 5 m/s = 10 m/s

Average speed of Shyam =

(b)

100 m/s = 25 m/s (From P to Q) 4

Average velocity of Ram =

Average velocity of Shyam = Example 4. Solution :

100 m/s = 20 m/s (From P to Q) 5

A particle travels half of total distance with speed v 1 and next half with speed v 2 along a straight line. Find out the average speed of the particle? Let total distance travelled by the particle be 2s.

s v1 s Time taken to travel next half = v2 Time taken to travel first half =

Average speed =

Total distance covered = Total time taken

2s s s  v1 v 2

=

2v1v2 (harmonic progression) v1  v2

Example 5.

A person travelling on a straight line moves with a uniform velocity v1 for some time and with uniform velocity v2 for the next equal time. The average velocity v is given by

Answer :

v

v1 v 2 (Arithmetic progression) 2

Solution : As shown, the person travels from A to B through a distance S, where first part S1 is travelled in time t/2 and next S2 also in time t/2. So, according to the condition : v1 =

S1 S2 and v2 = t/2 t/2

Total displacement S1  S2 Average velocity = = = Total time taken t

v1 t v 2 t  2 2 = v1  v 2 2 t

 2.6

Averag e a cc eleration (in a n interva l): The average acceleration for a finite time interval is defined as : Average acceleration =

change in velocity time interval

Average acceleration is a vector quantity whose direction is same as that of the change in velocity.

 v  a av = = t

RESONANCE

  vf  vi t RECTILINEAR MOTION - 4

Since for a straight line motion the velocities are along a line, therefore a av =

vf  vi v = tf  ti t

(where one has to substitute v f and v i with proper signs in one dimensional motion)

2.8

I nsta nta neou s Ac c elera tion (a t an instant): The instantaneous acceleration of a particle is its acceleration at a particular instant of time. It is defined as the derivative (rate of change) of velocity with respect to time. W e usually mean instantaneous acceleration when we say “ acceleration”. For straight motion we define instantaneous acceleration as :

 v  dv  a = = lim  t 0  t  dt

and in general

   v  dv  lim a = dt = t0  t   

The dimension of acceleration is [LT -2 ] and its SI unit is m/s 2.

3.

GRAP HI CAL I N TE RPRE TATI ON OF SOM E QUAN TI TIE S 3 . 1 Avera ge V elo c ity If a particle es a point P (x i) at time t = t i and reaches Q (x f) at a later time instant t = t f , its average velocity in the interval PQ is V av =

xf  xi x = tf  ti t x xf

This expression suggests that the average velocity is equal to t he s lo pe o f th e li ne ( c h or d) j oi ni ng t he p oi nt s corresponding to P and Q on the xt graph.

3.2

Q

xi

P

O

I ns ta nta neo u s V elo c ity

ti

tf

t

Consider the motion of the particle between the two points P and Q on the xt graph shown. As the point Q is brought closer and closer to the point P, the time interval between PQ (t, t , t,......) get progressively smaller. The average velocity for each time interval is the slope of the appropriate dotted line (PQ, PQ, PQ......). As the point Q approaches P, the time interval approaches zero, but at the same time the slope of the dotted line approaches that of the tangent to the curve at the point P. As t  0, V av (=x/t)  V inst. Geometrically, as t  0, chord PQ  tangent at P. Hence the instantaneous velocity at P is the slope of the tangent at P in the x  t graph. W hen the slope of the x  t graph is positive, v is positive (as at the point A in figure). At C, v is negative becaus e the tangent has negative slope. The ins tantaneous velocity at point B (turning point) is zero as the slope is zero.

3.3

v

I nsta nta neo u s Ac c eleratio n : The derivative of velocity with respect to time is the slope of the tangent in velocity time (vt) graph.

a=0 a>0

a<0

a=0 0

RESONANCE

t a=0

RECTILINEAR MOTION - 5

Example 6.

Position of a particle as a function of time is given as x = 5t 2 + 4t + 3. Find the velocity and acceleration of the particle at t = 2 s?

Solution :

Velocity;

v =

At

t = 2 s v = 10(2) + 4 v = 24 m/s

dx = 10t + 4 dt

d 2x Acceleration; a = = 10 dt 2 a = 10 m/s 2

Acceleration is constant, so at t = 2 s

 4.

MOTION WITH UNI FORM V ELOCI T Y Consider a particle moving along xaxis with uniform velocity u starting from the point x = xi at t = 0. Equations of x, v, a are : x (t) = x i + ut ; v (t) = u ; a (t) = 0   

x t graph is a straight line of slope u through x i. as velocity is constant, v  t graph is a horizontal line. at graph coincides with time axis because a = 0 at all time instants. x xi

slo pe =

u

u is negative t

O v u

v positive velocity

t

O

5.

t

O u

negative velocity

UN IFORMLY AC C E LE RATE D MOTI ON : If a particle is accelerated with constant acceleration in an interval of time, then the motion is termed as uniformly accelerated motion in that interval of time. For uniformly accelerated motion along a straight line (xaxis) during a time interval of t seconds, the following important results can be used.

v u t

(a)

a

(b)

Vav 

(c)

S = (v av )t

(d)

v u S t  2 

(e)

v = u + at

v u 2

RESONANCE

RECTILINEAR MOTION - 6

6.

(f)

s = ut + 1/2 at 2 s = vt  1/2 at 2 x f = x i + ut + 1/2 at 2

(g)

v 2 = u 2 + 2as

(h)

s n = u + a/2 (2n  1) u = initial velocity (at the beginning of interval) a = acceleration v = final velocity (at the end of interval) s = displacement (x f  x i) x f = final coordinate (position) x i = initial coordinate (position) s n = displacement during the n th sec

DIRECTIONS OF VECTORS IN STRAIGHT LINE MOTION In straight line motion, all the vectors (position, displacement, velocity & acceleration) will have only one component (along the line of motion) and there will be only two possible directions for each vector. 

For example, if a particle is moving in a horizontal line (xaxis), the two directions are right and left. Any vector directed towards right can be represented by a positive number and towards left can be represented by a negative number. -

+

line of motion

For objects moving vertically near the surface of the earth, the only force acting on the particle is its weight (mg) i.e. the gravitational pull of the earth. Hence acceleration for this type of motion will always be a = g i.e. a =  9.8 m/s 2 (ve sign, because the force and acceleration are directed downwards, If we select upward direction as positive).

-

line o

-

+ fm oti on



line of motion

For vertical or inclined motion, upward direction can be taken +ve and downward as ve

+



Note :  

Example 7.

Solution :

If acceleration is in same direction as velocity, then speed of the particle increases. If acceleration is in opposite direction to the velocity then speed decreases i.e. the particle slows down. This situation is known as retardation.

A particle moving rectilinearly with constant acceleration is having initial velocity of 10 m/s. After some time, its velocity becomes 30 m/s. Find out velocity of the particle at the mid point of its path? Let the total distance be 2x.  distance upto midpoint = x Let the velocity at the mid point be v and acceleration be a. From equations of motion v 2 = 10 2 + 2ax ____ (1) ____ (2) 30 2 = v 2 + 2ax

RESONANCE

RECTILINEAR MOTION - 7

(2) - (1) gives v 2 - 30 2 = 10 2 - v 2  Example 8.

Solution :

v 2 = 500



v = 10 5 m/s

Mr. Sharma brakes his car with constant acceleration from a velocity of 25 m/s to 15 m/s over a distance of 200 m. (a) How much time elapses during this interval? (b) W hat is the acceleration? (c) If he has to continue braking with the same constant acceleration, how much longer would it take for him to stop and how much additional distance would he cover? (a) W e select positive direction for our coordinate system to be the direction of the velocity and choose the origin so that x i = 0 when the braking begins. Then the initial velocity is u x = +25 m/s at t = 0, and the final velocity and position are v x = +15 m/s and x = 200 m at time t. Since the acceleration is constant, the average velocity in the interval can be found from the average of the initial and final velocities. 

v av, x =

1 1 (u x + v x) = (15 + 25) = 20 m/s. 2 2

The average velocity can also be expressed as vav, x = and

t = t  0, we can solve for t: t =

(b)

x . W ith x = 200 m t

Äx 200 v av,x = 20 = 10 s.

W e can now find the acceleration using v x = u x + a xt ax =

vx  u x 15  25 = t 10

=  1 m/s 2.

The acceleration is negative, which means that the positive velocity is becoming smaller as brakes are applied (as expected). (c)

Now with known acceleration, we can find the total time for the car to go from velocity u x = 25 m/s to v x = 0. Solving for t, we find t =

vx  u x ax

=

0  25 = 25 s. 1

The total distance covered is x = x i + u xt +

1 2 at 2 x

= 0 + (25)(25) +

1 (1)(25) 2 2

= 625 – 312.5 = 312.5 m. Additional distance covered = 312.5 – 200 = 112.5 m.

RESONANCE

RECTILINEAR MOTION - 8

Example 9.

A police inspector in a jeep is chasing a pickpocket an a straight road. The jeep is going at its maximum speed v (assumed uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance d away, and the motorcycle starts with a constant acceleration a. Show that the pick pocket will be caught if

v  2ad . Solution :

Suppose the pickpocket is caught at a time t after motorcycle starts. The distance travelled by the motorcycle during this interval is

1 s  at2 2

____ (1)

During this interval the jeep travels a distance ____ (2)

s  d  vt By (1) and (2),

1 2 at  d  vt 2 t

or,

v  v 2  2ad a

The pickpocket will be caught if t is real and positive. This will be possible if

v 2  2ad

or,

v

2ad

Example 10. A man is standing 40 m behind the bus. Bus starts with 1 m/sec 2 constant acceleration and also at the same instant the man starts moving with constant speed 9 m/s. Find the time taken by man to catch the bus.

1m/sec2

x=0 t=0 Solution :

40 m

t=0 x = 40

Let after time ‘t’ man will catch the bus For bus x = x 0 + ut +

x = 40 +

1 at 2 , 2

t2 2

For man, x = 9t From (i) & (ii) 40 +

t2 = 9t 2

RESONANCE

x = 40 + 0(t) +

1 (1) t 2 2

............. (i) ............. (ii)

or t = 8 s or t = 10s.

RECTILINEAR MOTION - 9

Example 11. A particle is dropped from height 100 m and another particle is projected vertically up with velocity 50 m/s from the ground along the same line. Find out the position where two particle will meet? (take g = 10 m/s 2) Solution :

Let the upward direction is positive. Let the particles meet at a distance y from the ground. For particle A, y0 = + 100 m u = 0 m/s a =  10 m/s 2 y = 100 + 0(t) 

1  10  t 2 2

= 100 - 5t 2 For particle B, y0 = 0 m u = + 50 m/s a =  10 m/s 2 y = 50(t) 

[y = y0 + ut +

y=100m A

1 2 at ] 2

y=0m

u=0 m/s

u=50 m/s

B

---- (1)

1  10  t 2 2

= 50t  5t 2 ---- (2) According to the problem; 50t  5t 2 = 100  5t 2 t = 2 sec Putting t = 2 sec in eqn. (1), y = 100  20 = 80 m Hence, the particles will meet at a height 80 m above the ground. Example 12. A particle is dropped from a tower. It is found that it travels 45 m in the last second of its journey. Find out the height of the tower? (take g = 10 m/s 2) Solution : Let the total time of journey be n seconds. Using;

sn  u 

a ( 2n  1) 2



45 = 0 +

1  10  5 2 2

= 125 m

10 ( 2n  1) 2

n = 5 sec Height of tower; h =

1 2 gt 2

=

 7.

RE AC T I ON TI M E W hen a situation demands our immediate action. It takes some time before we really respond. Reaction time is the time a person takes to observe, think and act.

Example 13. A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If the balloon was 60 m high when the stone was dropped, find its height when the stone hits the ground. Take g = 10 m/s 2.

RESONANCE

RECTILINEAR MOTION - 10

Solution :

S = ut +

1 at 2 2

– 60 = 5(t) +

1 (–10) t 2 2

– 60 = 5t – 5t 2 5t 2 – 5t – 60 = 0 t 2– t – 12 = 0 t 2 – 4t + 3t – 12 = 0 (t – 4) (t + 3) = 0  t = 4 Height of balloon from ground at this instant = 60 + 4 × 5 = 80 m

+ve 60m

–ve

Example 14. A balloon is rising with constant acceleration 2 m/sec 2. Two stones are released from the balloon at the interval of 2 sec. Find out the distance between the two stones 1 sec. after the release of second stone. Solution : Acceleration of balloon = 2 m/sec 2 Let at t = 0, y = 0 when the first stone is released. By the question, y1 = 0 t 1 +

1 gt 12 2

(taking vertical upward as – ve and downward as + ve) 

Position of  st stone =

9 g 2

(1 second after release of second stone will be the 3 rd second for the 1 st stone) For second stone y2 = ut 2 +

1 gt 22 2

u = 0 + at = – 2 × 2 = – 4m/s (taking vertical upward as – ve and downward as + ve) 

y2 = – 4 × 1 +

1 g × (1) 2 2

(t 2 = 1 second)

The 2 nd stone is released after 2 second

1 1 at 2 = – × 2 × 4 = – 4 2 2



y = –

So,

Position of second stone from the origin = – 4 +

Distance between two stones =

1 g – 4 2

1 1 g × 9 – g × 1 + 8 = 48 m. 2 2

Note : 

As the particle is detached from the balloon it is having the same velocity as that of balloon, but its acceleration is only due to gravity and is equal to g.

 8.

STRAIGHT LIN E-EQUATION , GRAPH, SLOPE (+ve, –ve, zero slo pe). If  is the angle at which a straight line is inclined to the positive direction of xaxis, & 0°  < 180°,  90°, then the slope of the line, denoted by m, is defined by m = tan . If  is 90°, m does not exist, but the line is parallel to the yaxis. If  = 0, then m = 0 & the line is parallel to the x-axis.

RESONANCE

RECTILINEAR MOTION - 11

Slope  intercept form : y = mx + c is the equation of a straight line whose slope is m & which makes an intercept c on the yaxis. dy dx

m = slope = tan =

y

y

y

–ve slope

+ve slope 

9.

C

C

C

slope = 0

x

 x

x

PARABOLIC CURVE-EQUATION, GRAPH (VARIOUS SITUATIONS UP, DOWN, LEFT, RIGHT WITH CONDITIONS) y

y

y2 = kx

x

y2 = –kx

x

y

x2 = ky

y

x

x2 = –ky

x

W here k is a positive constant.

y

E qu a tio n o f p ara bo la : y = ax 2 + bx + c For a > 0 The nature of the parabola will be like that of the of nature x 2 = ky Minimum value of y exists at the vertex of the parabola. D ymin = where D = b 2 – 4ac 4a Case (i) :

x

 b D  ,  Coordinates of vertex =   2a 4a 

Case (ii) : a < 0 The nature of the parabola will be like that of the nature of x 2 = –ky Maximum value of y exists at the vertex of parabola. ymax =

RESONANCE

y

D where D = b 2 – 4ac 4a

x

RECTILINEAR MOTION - 12

10.

GRAPHS IN UNIFORMLY ACCELERATED MOTION (a  0)  x is a quadratic polynomial in of t. Hence x  t graph is a parabola. x

x a<0

xi

xi

a>0 t

0

t

0

x-t graph  v is a linear polynomial in of t. Hence vt graph is a straight line of slope a. v pe slo

=

v

a

u

slo

pe =

a

u a is positive 0

a is negative

t

0

t

v-t graph  at graph is a horizontal line because a is constant. a a

a

positive acceleration

0

0 a

t

negative acceleration

a-t graph

11.

IN TERPRETATI ON OF SOME MORE GRAP HS 1 1. 1 Position vs Time g raph (i) Zero V eloc ity As position of particle is fixed at all the time, so the body is at rest. Slope;

dx = tan = tan 0º = 0 dt

Velocity of particle is zero

(ii)

Unif o rm V elo c ity Here tanis constant tan =

(iii)

dx dt



dx is constant. dt



velocity of particle is constant.

Non uniform velocity (increasing with time) In this case; As time is increasing,  is also increasing. 

dx = tan is also increasing dt

Hence, velocity of particle is increasing.

RESONANCE

RECTILINEAR MOTION - 13

(iv)

Non uniform velocity (decreasing with time) In this case; As time increases,  decreases. 

dx = tan also decreases. dt

Hence, velocity of particle is decreasing.

1 1. 2 Velocity vs time graph (i) Zero acceleration Velocity is constant. tan = 0 

dv = 0 dt

Hence, acceleration is zero.

(ii)

Unif o rm a c c eler a tio n tan is constant. 

dv = constant dt

Hence, it shows constant acceleration.

(iii)

Un if o rm re ta rd a ti o n Since  > 90º  tan is constant and negative. 

dv = negative constant dt

Hence, it shows constant retardation.

11.3 Acceleration vs time graph (i) Co nsta nt a cceleration tan = 0 

da = 0 dt Hence, acceleration is constant.

(ii)

Unifo rm ly inc rea sing a c c elera tio n  is constant. 0º <  < 90º  tan > 0 

da = tan = constant > 0 dt

Hence, acceleration is uniformly increasing with time.

RESONANCE

RECTILINEAR MOTION - 14

(iii)

Unif o rmly d ec reasing a c c elera tio n Since  > 90º  tan is constant and negative. 

da = negative constant dt

Hence, acceleration is uniformly decreasing with time

Example 15.

Solution :

The displacement vs time graph of a particle moving along a straight line is shown in the figure. Draw velocity vs time and acceleration vs time graph. x = 4t 2 v =

dx = 8t dt

Hence, velocity-time graph is a straight line having slope i.e. tan = 8. a =

dv = 8 dt

Hence, acceleration is constant throughout and is equal to 8.

A –v2

Solution :

For particle A : (i) Displacement vs time graph is 1 2 y = ut + at 2 u = + 10 m/sec 2 1 y = 10t – ×10t 2 2 = 10 t – 5t 2 dy v = = 10 – 10 t = 0 dt t = 1 ; hence, velocity is zero at t = 1 10 t – 5 t 2 = – 100 t 2 – 2t – 20 = 0 t = 5.5 sec. i.e. particle travels up till 5.5 seconds.

RESONANCE

Displacement

B 10 m/s

100 m

10 m/s

Example 16. At the height of 100 m, a particle A is thrown up with V = 10 m/s, B particle is thrown down with V =10m/s and C particle released with V = 0 m/s. Draw graphs of each particle. (i) Displacement–time (ii) Speed–time (iii) Velocity–time (iv) Acceleration–time

0 m/sec –v

2

C

25 20 15 10 5

–2 –1 1 –5 –10 –15 –20 –25 –100

2 3 4 T

5 5.5 time

RECTILINEAR MOTION - 15

(ii)

Speed vs time graph : Particle has constant acceleration = g  throughout the motion, so v-t curve will be straight line. when moving up, v = u + at 0 = 10 – 10 t or t = 1 is the time at which speed is zero. there after speed increases at constant rate of 10 m/s 2 . Resulting Graph is : (speed is always positive). This shows that particle travels till a time of 21 seconds

(iii)

Velocity vs time graph : V = u + at V = 10 – 10t ; this shows that velocity becomes zero at t = 1 sec and thereafter the velocity is negative with slope g.

(iv)

Acceleration vs time graph : throughout the motion, particle has constant acceleration = –10 m/s 2.

Acceleration

1 +

Time

–10

For particle B : u = – 10 m/s. y = – 10t – (i)

1 (10) t 2 = – 10t – 5t 2 2

Displacement time graph : y = 10t – 5t 2 dy = – 10t – 5t 2 = – 10 – 10t dt

this shows that slope becomes more negative with time.

(ii)

Speed time graph : dy = – 10t – 5t 2 = – 10 – 10t dt

hence, speed is directly proportional to time with slope of 10. initial speed = 10 m/s

RESONANCE

RECTILINEAR MOTION - 16

Velocity time graph : Velocity

(iii)

dy = – 10t – 5t 2 = – 10 – 10t dt

hence, velocity is directly proportional to time with slope of –10. Initial velocity = –10 m/s

(iv)

Time t –10 m/s

Acceleration vs time graph : throughout the motion, particle has constant acceleration = –10 m/s 2. dv a = = – 10 dt

For Particle C : (i) Displacement time graph : u = 0 , y = –

1 × 10t 2 = – 5t 2 2

(ii)

Speed vs time graph : dy v = = – 10 t dt hence, speed is directly proportional to time with slope of 10.

(iii)

Velocity time graph : V = u + at V = – 10t ; hence, velocity is directly proportional to time with slope of –10.

Velocity

this shows that slope becomes more negative with time.

Time 20

–10 20

RESONANCE

RECTILINEAR MOTION - 17

Acceleration vs time graph : throughout the motion, particle has constant acceleration = –10 m/s 2.

Acceleration

(iv)

Time

–10 m/s

 12.

DISPLACEMENT FROM v - t GRAPH & CHANGE IN VELOCITY FROM a -t GRAPH Displacement = x = area under vt graph. Since a negative velocity causes a negative displacement, areas below the time axis are taken negative. In similar way, can see that  v = a t leads to the conclusion that area under a  t graph gives the change in velocity v during that interval.

Example 17. Describe the motion shown by the following velocity-time graphs.

(a)

Solution :

(a)

(b)

(b)

During interval AB: velocity is +ve so the particle is moving in +ve direction, but it is slowing down as acceleration (slope of v-t curve) is negative. During interval BC: particle remains at rest as velocity is zero. Acceleration is also zero. During interval CD: velocity is -ve so the particle is moving in -ve direction and is speeding up as acceleration is also negative. During interval AB: particle is moving in +ve direction with constant velocity and acceleration is zero. During interval BC: particle is moving in +ve direction as velocity is +ve, but it slows down until it comes to rest as acceleration is negative. During interval CD: velocity is -ve so the particle is moving in -ve direction and is speeding up as acceleration is also negative.

Imp ortant Poin ts to 

For uniformly accelerated motion (a  0), xt graph is a parabola (opening upwards if a > 0 and opening downwards if a < 0). The slope of tangent at any point of the parabola gives the velocity at that instant.



For uniformly accelerated motion (a  0), vt graph is a straight line whose slope gives the acceleration of the particle.



In general, the slope of tangent in xt graph is velocity and the slope of tangent in vt graph is the acceleration.



The area under at graph gives the change in velocity.



The area between the vt graph gives the distance travelled by the particle, if we take all areas as positive.



Area under vt graph gives displacement, if areas below the taxis are taken negative.

RESONANCE

RECTILINEAR MOTION - 18

Example 18. For a particle moving along x-axis, velocity-time graph is as shown in figure. Find the distance travelled and displacement of the particle? Solution : Distance travelled = Area under v-t graph (taking all areas as +ve.) Distance travelled = Area of trapezium + Area of triangle =

1 2  6 8 + 1  4  5 2 2

= 32 + 10 = 42 m Displacement = Area under v-t graph (taking areas below time axis as –ve.) Displacement = Area of trapezium  Area of triangle

1 2  6 8  1  4  5 2 2

=

= 32  10 = 22 m Hence, distance travelled = 42 m and displacement = 22 m.

 13.

MOTION WITH NON-UNI FORM ACCE LERATI ON (USE OF DEF IN I TE I NTEGRALS) tf

x =

 v( t )dt

(displacement in time interval t = t i to t f)

ti

The expression on the right hand side is called the definite integral of v(t) between t = t i and t = t f. . Similarly change in velocity tf

v = v f  v i =

 a(t )dt ti

1 3. 1 Solving Prob lems which Involves Non uniform Accelera tion (i)

Acceleration depending on velocity v or time t By definition of acceleration, we have a =

v

 dv

v

t

=

v0

dv . dt

 a ( t ) dt .

If a is in of v,

0



v0

If a is in of t,

dv  a (v )

t

 dt . 0

On integrating, we get a relation between v and t, and then x

using

 dx

x0

RESONANCE

t

=

 v (t ) dt ,

x and t can also be related.

0

RECTILINEAR MOTION - 19

(ii)

Acceleration depending on velocity v or position x a =

dv dt

dv dx dx dt



a =



a= v



a =

dx dv dt dx

dv dx

This is another important expression for acceleration. If a is in of x, x

v

 a ( x ) dx

 v dv = v

If a is in of v,



v0

.

x0

v0

v dv  a( v )

x

 dx

x0

On integrating, we get a relation between x and v. x



Using

x0

Example 19. An (a) (b) Solution : As

t

=

 dt , we can relate x and t. 0

object starts from rest at t = 0 and accelerates at a rate given by a = 6t. W hat is its velocity and its displacement at any time t? acceleration is given as a function of time, v(t)



dx v (x)

t

 dv   a(t)dt

v(t 0 )

t0

Here t 0 = 0 and v(t 0) = 0 t



 6tdt

v(t) =

0

So,

 t2  t = 6   2 0

= 6 (

 t3  t 3 =    3 0

 t3  3 =  0 = t 3 3 

t2 - 0) = 3t 2 2

v(t) = 3t 2 t

As

x   v(t)dt t0

t



x   3t 2dt 0

Hence, velocity v(t) = 3t 2 and displacement x  t 3 .

RESONANCE

RECTILINEAR MOTION - 20

Example 20. For a particle moving alongv + x-axis, acceleration is given as a = x. Find the position as a function of time? Given that at t = 0 , x = 1 v = 1. Solution :

a = x

vdv = x dx





 t = 0, x = 1 and v = 1  C = 0  v2 = x2 v = ± x 

v = x

nx = t + C x = et

v2 x2 = + C 2 2

but given that x = 1 when v = 1



dx = x dt



dx = dt x



0 = 0 + C



nx = t

Example 21. For a particle moving along x-axis, acceleration is given as a = v. Find the position as a function of time ? Given that at t = 0 , x = 0 v = 1. Solution :

a = v



dv =v dt

nv = t + c



0 = 0 + c

v = et



dx = et dt

Problem 1.

A particle covers

 dt





 dx =  e dt



0 = 1 + c

t

x = et + c x = et – 1



dv = v



3 1 of total distance with speed v 1 and next with v 2. Find the average 4 4

speed of the particle? Answer :

4v 1v 2 v 1  3v 2

Solution :

Let the total distance be s

A

3s/4

B s/4

C

Total dis tance average speed (< v >) = Total time taken

< v > =

Problem 2.

Answer :

s 3s s  4v1 4v 2

=

1 3 1  4v1 4v 2

=

4v 1v 2 v 1  3v 2

A car is moving with speed 60 Km/h and a bird is moving with speed 90 km/h along the same direction as shown in figure. Find the distance travelled by the bird till the time car reaches the. tree? 360 m

RESONANCE

240 m

RECTILINEAR MOTION - 21

Solution :

Time taken by a car to reaches the tree (t) =

240 m 0.24 hr = 60 km / hr 60

Now, the distance travelled by the bird during this time interval (s) = 90  Problem 3

Answer : Solution :

0.24 = 0.12 × 3 km = 360 m. 60

The position of a particle moving on X-axis is given by x = At 3 + Bt 2 + Ct + D. The numerical values of A, B, C, D are 1, 4, -2 and 5 respectively and SI units are used. Find (a) the dimensions of A, B, C and D, (b) the velocity of the particle at t = 4 s, (c) the acceleration of the particle at t = 4s, (d) the average velocity during the interval t =0 to t = 4s, (e) the average acceleration during the interval t = 0 to t = 4 s. (a) [A] = [LT -3], [B] = [LT -2], [C] = [LT -1] and [D] = [L] ; (b) 78 m/s ; (c) 32 m/s 2 ; (d) 30 m/s ; (e) 20 m/s 2 As x = At 3 + Dt 2 + Ct + D (a) Dimensions of A, B, C and D , [At 3] = [x] (by principle of homogeneity) [A] = [LT –3 ] similarly, [B] = [LT -2], [C] = [LT -1] and [D] = [L] ; dx = 3At 2 + 2Bt + C dt

(b) As v =

velocity at t = 4 sec. v = 3(1) (4) 2 + 2(4) (4) – 2 = 78 m/s. (c) Acceleration (a) =

dv = 6At + 2B ; a = 32 m/s 2 dt

(d) average velocity as x = At 3 + Bt 2 + Ct = D position at t = 0, is x = D = 5m. Position at t = 4 sec is (1)(64) + (4)(16) – (2) (4) + 5 = 125 m Thus the displacement during 0 to 4 sec. is 125 – 5 = 120 m        < v > = 120 / 4 = 30 m/s (e) v = 3At 2 + 20 t + C , velocity at t = 0 is c = – 2 m/s velocity at t = 4 sec is 78 m/s

 < a > =

v 2  v1 = 20 m/s 2 t 2  t1

Problem 4.

For a particle moving along x-axis, velocity is given as a function of time as v = 2t2 + sin t. At t = 0, particle is at origin. Find the position as a function of time?

Solution :

v = 2t 2 + sin t x

t

 dx =

 ( 2t

0

Problem 5. Solution :

0

2



dx = 2t 2 + sin t dt

 sin t ) dt = x  2 t 3  cos t  1 3

Ans.

A car decelerates from a speed of 20 m/s to rest in a distance of 100 m. W hat was its acceleration, assumed constant? v = 0 u = 20 m/s s = 100 m  as v 2 = u 2 + 2 as 0 = 400 + 2a × 100  a = – 2 m/s 2  acceleration = 2 m/s Ans.

RESONANCE

RECTILINEAR MOTION - 22

Problem 6.

Solution :

A 150 m long train accelerates uniformly from rest. If the front of the train es a railway worker 50 m away from the station at a speed of 25 m/s, what will be the speed of the back part of the train as it es the worker? v 2 = u 2 + 2as 25 × 25 = 0 + 100 a a =

25 m/s 2 4

Now, for time taken by the back end of the train to the worker v´ 2 = v 2 + 2al = (25) 2 + 2 ×

we have

25 × 150 48

v´ 2 = 25 × 25 × 4 v´ = 50 m/s. Ans. Problem 7. Answer : Solution :

Problem 8.

Solution :

A particle is thrown vertically with velocity 20 m/s . Find (a) the distance travelled by the particle in first 3 seconds, (b) displacement of the particle in 3 seconds. 25m, 15m Highest point say B VB = 0 v = u + gt 0 = 20 – 10 t t = 2 sec.  distance travel in first 2 seconds. s = s(t =0 to 2sec) + s (2sec. to 3sec.) s = [ut + 1/2 at 2] t = 0 to t = 2s + [ut +1/2at 2] t = 2 to t = 3s s = 20 × 2 – 1/2 × 10 × 4 + 1/2 × 10 × 1 2 = (40 – 20) + 5 = 25 m. and displacement = 20 – 5 = 15 m. A car accelerates from rest at a constant rate  for some time after which it decelerates at a constant rate  to come to rest. If the total time elapsed is t. Find the maximum velocity acquired by the car. V t = t1 + t2 A vmax v max slope of OA curve = tan =  = t 1   v max slope of AB curve =  = t t1 t2 2 O B t t = t1 + t2 t =



v max v max +  



    t v max =   

Problem 9.

In the above question find total distance travelled by the car in time ‘t’ .

Solution :

v max =

 t (  )



t1 =

t v max = (   ) 



t2 =

t v max = (   ) 

 Total distance travelled by the car in time ‘t’ =

1 2 1 2 1  2 t 2  2 t 2 1  2 t 2 t 1 + v max t 2 –  t 2 = + – 2 2 2 (   ) 2 (   ) 2 2 (  ) 2

Area under graph (directly) =

RESONANCE

1  t 2 = 2 (   )

 t 2 Ans. 2(   )

RECTILINEAR MOTION - 23

Problem 10. The displacement vs time graph of a particle moving along a straight line is shown in the figure. Draw velocity vs time and acceleration vs time graph.

Solution :

Upwards direction is taken as positive, downwards direction is taken as negative. (a) The equation of motion is : x = –8t 2  v =

dx = – 16 t ; dt

this shows that velocity is directly proportional to time and

slope of velocity-time curve is negative i.e. – 16. Hence, resulting graph is (i) (b) Acceleration of particle is : a =

dv = –16. dt

This shows that acceleration is constant but negative. Resulting graph is (ii)

Problem 11. Draw displacement–time and acceleration–time graph for the given velocity–time graph.

Solution :

Part AB : v-t curve shows constant slope i.e. constant acceleration or Velocity increases at constant rate with time. Hence, s-t curve will show constant increase in slope and a-t curve will be a straight line. Part BC : v-t curve shows zero slope i.e. constant velocity. So, s-t curve will show constant slope and acceleration will be zero. Part CD : v-t curve shows negative slope i.e. velocity is decreasing with time or acceleration is negative. Hence , s-t curve will show decrease in slope becoming zero in the end. and a-t curve will be a straight line with negative intercept on y-axis.

RESONANCE

RECTILINEAR MOTION - 24

RESULTING GRAPHS ARE:

Problem 12. For a particle moving along x-axis, following graphs are given. Find the distance travelled by the particle in 10 s in each case?

Solution :

(a)

Distance area under the v - t curve  distance = 10 × 10 = 100 m

(b)

Ans.

Area under v – t curve  distance =

1 × 10 × 10 = 50 m 2

Ans.

2

Problem 13. For a particle moving along x-axis, acceleration is given as a  2v . If the speed of the particle is v0 at x = 0, find speed as a function of x. Solution :

a = 2v 2

v v



v0

dv = 2v 2 dx

dv v =

n

dv = 2v 2 dt



or



dv = 2v dx



nv vv



v = v0e2x

or

dv dx × = 2v 2 dx dt

x

 2 dx 0

v v 0 = 2x

RESONANCE

0

= 2 x 0x

Ans.

RECTILINEAR MOTION - 25

Type (I) : Very Short Answer Type Questions :

[01 Mark Each]

1.

Are rest and motion absolute or relative ?

2.

Under what condition will the distance and displacement of a moving object will have the same magnitude?

3.

What will be nature of x–t graph for a uniform motion?

4.

Can x-t graph be a straight line parallel to position axis ?

5.

Can x-t graph be a straight line parallel to time-axis for an object which is moving ?

6.

what does slope of v-t graph represent ?

7.

Two balls of different masses (one lighter and other heavier) are thrown vertically upward with same initial speed. Which one will rise to the grater height ?

8.

Is it possible that velocity of an object is zero but its acceleration is none zero ? If yes, then give an example.

9.

What does speedometer of a car measure ?

Type (II) : Short Answer Type Questions :

[02 Marks Each]

10.

Given below are some examples of motion. State in each case, if the motion is one, two or three dimensional : (a) a kite flying on a windy day (b) a speeding car on a long straight highway (c) an insect crawling on a globe (d) a carrom coin rebounding from the side of the board (e) a planet revolving around its star.

11.

A car travelling with a speed of 90 kmh–1 on a straight road is ahead of scooter travelling with a speed of 60 kmh–1,calculate velocity of car wrt to scooter. solve above problem if scooter is ahead of car.

12.

Show that slope of displacement-time graph is equal to the velocity of uniform motion.

13.

Delhi is at a distance of 250 km from Chandigarh. A sets out from Chandigarh at a speed of 80 kmh–1 and B sets out at the same time from Delhi at a speed of 45 kmh–1. When will they meet each other.

14.

Derive the equation x(t) = x(0) + v(0) t +

1 2 at 2

15.

‘The direction in which an object moves is given by the direction of velocity of the object and not by the direction of acceleration.’ Explain the above statement with some suitable example.

16.

Fig. below gives x-t plot of a particle in one dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest and in which is it the least ? Give the sign of average velocity for each interval.

RESONANCE

RECTILINEAR MOTION - 26

17.

Prove that v2 – u2 = 2aS, where symbols have their usual meanings.

18.

Prove that the distance travelled by a body in nth second is given by

a (2n – 1) 2 where symbols have their usual meanings. Snth = u +

Type (III) : Long Answer Type Questions:

[04 Mark Each]

19.

Derive an expression for the distance travelled by uniformly accelerated body in t seconds. Also derive an expression for the distance travelled by the body in the nth second.

20.

Deduce the following relations analytically for a uniform motion along a straight time, where the have their usual meanings : (i) v = u + at

(ii) S = ut +

1 2 at 2

(iii) v2 – u2 = 2aS

21.

The driver of a car A going at 25 ms–1 applies the brakes, decelerates uniformly, and stops in 10s. The driver of another car B going at 15 ms–1 puts less pressure on this brakes and stops in 20 s. On the same graph, plot speed-time for each of the two cars. (a) Which of the two cars travelled farther after the brakes were applied ? (b) Add a line to the graph, which shows the car B decelerating at the same rate as the car A. How long does it take the car B to stop at this rate of deceleration ?

22.

Draw velocity-time graph of uniform motion and prove that the displacement of an object in a time interval is equal to the area under velocity-time graph in that time interval.

RESONANCE

RECTILINEAR MOTION - 27

PART - I : SUBJECTIVE QUESTIONS SECTION (A) : DISTANCE AND DISPLACEMENT A-1.

A car starts from P and follows the path as shown in figure. Finally car stops at R. Find the distance travelled and displacement of the car if a = 7 m, b = 8 m and r =

A-2.

11 22 m? [Take   ]  7

A man moves to go 50 m due south, 40 m due west and 20 m due north to reach a field. (a) What distance does be have to walk to reach the field ? (b) What is his displacement from his house to the field?

SECTION (B) : AVERAGE SPEED AND AVERAGE VELOCITY B-1.

When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hours the reading is 12416 km. During journey he stay for 15 minute at midway. (a) What is the average speed of the car during this period ? (b) What is the average velocity?

B-2.

A particle covers each

1 of the total distance with speed v1, v2 and v3 respectively. Find the average speed of 3

the particle ?

SECTION (C) : VELOCITY, ACCELERATION, AVERAGE ACCELERATION C-1.

The position of a body is given by x = At + 4Bt3, where A and B are constants, x is position and t is time. Find (a) acceleration as a function of time, (b) velocity and acceleration at t = 5 s.

C-2.

Find the velocity as a function of time if x = At + Bt–3 , where A and B are constants, x is position and t is time.

C-3.

An athelete takes 2s to reach his maximum speed of 18 km/h after starting from rest. What is the magnitude of his average accleration?

SECTION (D) : EQUATIONS OF MOTION AND MOTION UNDER GRAVITY D-1.

A car accelerates from 36 km/h to 90 km/h in 5 s on a straight rod. What was its acceleration in m/s2 and how far did it travel in this time? Assume constant acceleration and direction of motion remains constant.

D-2.

A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute after applying breaks. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed. (Assume retardation to be constant)

D-3.

A particle moving along a straight line with constant acceleration is having initial and final velocity as 5 m/s and 15 m/s respectively in a time interval of 5 s. Find the distance travelled by the particle and the acceleration of the particle. If the particle continues with same acceleration, find the distance covered by the particle in the 8th second of its motion. (direction of motion remains same)

D-4.

A car travelling at 72 km/h decelerates uniformly at 2 m/s2. Calculate (a) the distance it goes before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and third seconds.

RESONANCE

RECTILINEAR MOTION - 28

D-5.

A ball is dropped from a tower. In the last second of its motion it travels a distance of 15 m. Find the height of the tower. [take g = 10m/sec2]

D-6.

A toy plane P starts flying from point A along a straight horizontal line 20 m above ground level starting with zero initial velocity and acceleration 2 m/s 2 as shown. At the same instant, a man P throws a ball vertically upwards with initial velocity 'u'. Ball touches (coming to rest) the base of the plane at point B of plane's journey when it is vertically above the man. 's' is the distance of point B from point A. Just after the of ball with the plane, acceleration of plane increases to 4 m/s 2. Find: (i) (ii) (iii)

Initial velocity 'u' of ball. Distance 's'. Distance between man and plane when the man catches the ball back. (g = 10 m/s 2)

SECTION (E) : GRAPH RELATED QUESTIONS E-1.

For a particle moving along x-axis, velocity-time graph is as shown in figure. Find the distance travelled and displacement of the particle? Also find the average velocity of the particle in intervel 0 to 5 second.

E-2.

A cart started at t = 0, its acceleration varies with time as shown in figure. Find the distance travelled in 30 seconds and draw the position-time graph.

E-3.

Two particles A and B start from rest and move for equal time on a straight line. The particle A has an acceleration a for the first half of the total time and 2a for the second half. The particle B has an acceleration 2a for the first half and a for the second half. Which particle has covered larger distance?

PART - II : OBJECTIVE QUESTIONS * Marked Questions are having more than one correct option. SECTION (A) : DISTANCE AND DISPLACEMENT A-1.

A hall has the dimensions 10 m × 10 m × 10 m. A fly starting at one corner ends up at a farthest corner. The magnitude of its displacement is: (A) 5 3 m

(B) 10 3 m

(C) 20 3 m

(D) 30 3 m

SECTION (B) : AVERAGE SPEED AND AVERAGE VELOCITY B-1.

A car travels from A to B at a speed of 20 km h–1 and returns at a speed of 30 km h–1. The average speed of the car for the whole journey is : (A) 5 km h–1 (B) 24 km h–1 (C) 25 km h–1 (D) 50 km h–1

RESONANCE

RECTILINEAR MOTION - 29

B-2.*

A person travelling on a straight line without changing direction moves with a uniform speed v1 for half distance and next half distance he covers with uniform speed v2 . The average speed v is given by

2v 1v 2 (A) v  v  v 1 2

B 3.

2 1 1 (C) v  v  v 1 2

1 1 1 (D) v  v  v 1 2

1 1 part of its journey with a velocity of 2 m/s, next part with a velocity of 3 m/s 3 3 and rest of the journey with a velocity 6m/s. The average velocity of the body will be

A body covers first

(A) 3 m/s B 4.

(B) v v 1 v 2

(B)

11 m/s 3

(C)

8 m/s 3

(D)

4 m/s 3

A car runs at constant speed on a circular track of radius 100 m taking 62.8 s on each lap. What is the average speed and average velocity on each complete lap? ( = 3.14) (A) velocity 10m/s, speed 10 m/s (B) velocity zero, speed 10 m/s (C) velocity zero, speed zero (D) velocity 10 m/s, speed zero

SECTION (C) : VELOCITY, ACCELERATION AND AVERAGE ACCELERATION C 1.

The displacement of a body is given by 2s = gt2 where g is a constant. The velocity of the body at any time t is: (A) gt (B) gt/2 (C) gt2/2 (D) gt3/6

SECTION (D) : EQUATIONS OF MOTION AND MOTION UNDER GRAVITY D-1.

A particle performs rectilinear motion in such a way that its initial velocity has opposite direction with its uniform acceleration. Let xA and xB be the magnitude of displacements in the first 10 seconds and the next 10 seconds, then: (A) xA < xB (B) xA = xB (C) xA > xB (D) the information is insufficient to decide the relation of xA with xB.

D-2.

A body starts from rest and is uniformly acclerated for 30 s. The distance travelled in the first 10 s is x1, next 10 s is x2 and the last 10 s is x3. Then x1 : x2 : x3 is the same as (A) 1 : 2 : 4 (B) 1 : 2 : 5 (C) 1 : 3 : 5 (D) 1 : 3 : 9

D-3.

A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the 3 m height of a window some distance from the top of the building. If the speed of the ball at the top and at the bottom of the window are vT and vB respectively, then (g = 9.8 m/sec2) (A) vT + vB = 12 ms–1

(B) vT – vB = 4.9 m s–1

(C) vBvT = 1 ms–1

vB (D) v = 1 ms–1 T

D-4.

A stone is released from an elevator going up with an acceleration a and speed u. The acceleration and speed of the stone just after the release is (A) a upward, zero (B) (g-a) upward, u (C) (g-a) downward, zero (D) g downward, u

D-5.

The initial velocity of a particle is given by u (at t = 0) and the acceleration by f, where f = at (here t is time and a is constant). Which of the following relation is valid? (A) v = u + at2

D-6.

(B) v = u +

at 2 2

(C) v = u + at

(D) v = u

A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T after dropping the stone at which the splash is heard is given by (A) T = 2h/v

RESONANCE

(B) T 

2h h  g v

(C) T 

2h h  g 2v

(D) T 

h 2h  2g v

RECTILINEAR MOTION - 30

D-7.

A student determined to test the law of gravity for himself walks off a sky scraper 320 m high with a stopwatch in hand and starts his free fall (zero initial velocity). 5 second later, superman arrives at the scene and dives off the roof to save the student. What must be superman's initial velocity in order that he catches the student just before reaching the ground ? [Assume that the superman's acceleration is that of any freely falling body.] (g = 10 m/s 2) (A) 98 m/s

D 8.

(B)

275 m/s 3

(C)

187 m/s 2

(D) It is not possible

In the above question, what must be the maximum height of the skyscraper so that even superman cannot save him. (A) 65 m (B) 85 m (C) 125 m (D) 145 m

SECTION (E) : GRAPH RELATED QUESTIONS E-1*.

Figure shows position-time graph of two cars A and B. Lines are parallel. x(m) A B 5 0

(A) Car A is faster than car B. (C) Both cars are moving with same velocity.

(B) Car A always leads Car B. (D) Both cars have positive acceleration.

E-2.

Figure shows the position time graph of a particle moving on the X-axis. (A) the particle is continuously going in positive x direction (B) area under x–t curve shows the displacement of particle (C) the velocity increases up to a time to, and then becomes constant. (D) the particle moves at a constant velocity up to a time to, and then stops.

E-3.

In the displacement–time graph of a moving particle is shown, the instantaneous velocity of the particle is negative at the point :

E-4.

E-5.

(A) C

(B) D

(C) E

(D) F

x D E

F

C t

The variation of velocity of a particle moving along a straight line is shown in the figure. The distance travelled by the particle in 4 s is :

(A) 25 m (B) 30 m (C) 55 m (D) 60 m A particle starts from rest and moves along a straight line with constant acceleration. The variation of velocity v with displacement S is : v

v

(A)

(B) S

RESONANCE

v

v

C) S

(D) S

S RECTILINEAR MOTION - 31

E-6.

The displacement time graphs of two particles A and B are straight lines making angles of respectively 300

vA and 600 with the time axis. If the velocity of A is vA and that of B is vB, then the value of v is B (A)

1

1 2

(B)

3

(C)

3

(D)

1 3

PART - III : ASSERTION / REASON 1. 1.

STATEMENT-1: A particle having negative acceleration will slow down. STATEMENT-2 : Direction of the acceleration is not dependent upon direction of the velocity. (A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1 (B) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-1 (C) STATEMENT-1 is true, STATEMENT-2 is false (D) STATEMENT-1 is false, STATEMENT-2 is true (E) Both STATEMENTS are false

2.

STATEMENT-1 : Magnitude of average velocity is equal to average speed. STATEMENT-2 : Magnitude of instantaneous velocity is equal to instantaneous speed. (A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1 (B) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-1 (C) STATEMENT-1 is true, STATEMENT-2 is false (D) STATEMENT-1 is false, STATEMENT-2 is true (E) Both STATEMENTS are false

3.

STATEMENT-1 : When velocity of a particle is zero then acceleration of particle must be zero at that instant.  dv  STATEMENT-2 : Acceleration is equal to a  v   , where v is the velocity at that instant..  dx 

(A) (B) (C) (D) (E) 4.

STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1 STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-1 STATEMENT-1 is true, STATEMENT-2 is false STATEMENT-1 is false, STATEMENT-2 is true Both STATEMENTS are false

STATEMENT-1 : A particle moves in a straight line with constant accleration. The average velocity of this particle cannot be zero in any time interval STATEMENT-2 : For a particle moving in straight line with constant acceleration, the average velocity uv in a time interval is , where u and v are initial and final velocity of the particle in the given time 2 interval. (A) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1 (B) STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-1 (C) STATEMENT-1 is true, STATEMENT-2 is false (D) STATEMENT-1 is false, STATEMENT-2 is true (E) Both STATEMENTS are false

RESONANCE

RECTILINEAR MOTION - 32

PART - I : SUBJECTIVE QUESTIONS 1.

Figure shows four paths along which objects move from a starting point to a final point (particle is moving along the same straight line), all in the same time. The paths over a grid of equally spaced straight lines. Rank the paths according to

(a) the magnitude of average velocity of the objects (b) the average speed of the objects, greatest first. 2.

A particle moving in straight line, traversed half the distance with a velocity v0. The remaining part of the distance was covered with velocity v1 for half the time and with velocity v2 for the other half of the time. Find the mean velocity of the particle averaged over the whole time of motion.

3.

The displacement of a particle moving on a straight line is given by x = 16t – 2t2. Find out (a) Displacement upto 2 and 6 s. (b) Distance travelled upto 2 and 6 s.

4.

A man walking with a speed ' v ' constant in magnitude and direction es under a lantern hanging at a height H above the ground (consider lantern as a point source). Find the velocity with which the edge of the shadow of the man's head moves over the ground, if his height is ' h '.

5.

A police jeep is chasing a culprit going on a moter bike. The motor bike crosses a turn at a speed of 72 km/ h. The jeep follows it at a speed of 108 km/h, crossing the turn 10 seconds later than bike (keeping constant speed). After crossing the turn, jeep acclerates with constant accleration 2 m/s2. Assuming bike travels at constant speed, how far from the turn will the jeep catch the bike?

6.

A healthy youngman standing at a distance of 6 m from a 11.5 m high building sees a kid slipping from the top floor. With what uniform acceleration (starting from rest) should he run to catch the kid at the arms height (1.5 m)? Take g = 10 m/s2.

7.

A lift is descending with uniform acceleration. To measure the acceleration , a person in the lift drops a coin at the moment when lift was descending with speed 6 ft/s. The coin is 5 ft above the floor of the lift at time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data, the acceleration of the lift. [Take g = 32 ft/s2]

8.

A body starts with an initial velocity of 10 m/s and moves along a straight line with a constant acceleration. When the velocity of the particle becomes 50 m/s the acceleration is reversed in direction without changing magnitude. Find the velocity of the particle when it reaches the starting point.

9.

The accompanying figure shows the velocity v of a particle moving on a coordinate line. (m/s)

2

-4

(a) When does the particle move forward? move backward? Speed up? slow down? (b) When is the particle's acceleration positive? Negative ? zero? (c) When does the particle move at its greatest speed ? (d) When does the particle stand still for more than an instant?

RESONANCE

RECTILINEAR MOTION - 33

10.

A point moves rectilinearly in one direction. Fig. shows the displacement s traversed by the point as a function of the time t. Using the plot find: (a) the average velocity of the point during the time of motion; (b) the maximum velocity; (c) the time t0 at which the instantaneous velocity is equal to the mean velocity averaged over the first t0 seconds.

11.

A lift starts from the top of a mine shaft and descends with a constant speed of 10 m/s. 4 s later a boy throws a stone vertically upwards from the top of the shaft with a speed of 30 m/s. Find when and where stone hits the lift.[ Take: g = 10 m/s² ]

PART - II : OBJECTIVE QUESTIONS Single choice type 1.

Two balls of equal masses are thrown upward, along the same vertical line at an interval of 2 seconds, with the same initial velocity of 40 m/s. Then these collide at a height of (Take g = 10 m/s 2) (A) 120 m (B) 75 m (C) 200 m (D) 45 m

2.

A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3 – 2t), where t is in second and velocity is in m/s. What is acceleration of the particle, when it is (A) 28 m/s 2 (B) 22 m/s 2 (C) 12 m/s 2 (D) 10 m/s 2

3.

The displacement (from origin) of a body in motion is given by x = a sin (t + ). The time at which the displacement is maximum is ( a, and  are constants) (A)

 

     (B)   2  

 2     (D)    

(C)  / 2

4.

A body is released from the top of a tower of height h metre. It takes T seconds to reach the ground. Where is the ball at the time T/2 seconds ? (A) at h/4 metre from the ground (B) at h/2 metre from the ground (C) at 3h/4 metre from the ground (D) depends upon the mass of the ball

5.

A stone is thrown vertically upward with an initial speed u from the top of a tower, reaches the ground with a speed 3u. The height of the tower is:

3u 2 (A) g 6.

6u 2 (C) g

9u 2 (D) g

A particle starts from rest with uniform acceleration a. Its velocity after n seconds is v. The displacement of the particle in the last two seconds is : (A)

7.

4u 2 (B) g

2v(n - 1) n

(B)

v(n - 1) n

(C)

v(n  1) n

(D)

2v(2n  1) n

A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after time.

v 2gh  (A) g 1  1  2  v  

RESONANCE

v 2gh  (B) g 1  1  2  v  

1/ 2

(C)

v 2gh  1  2  g v 

(D)

v 2gh  1  2  g v 

1/ 2

RECTILINEAR MOTION - 34

8.

A balloon is moving upwards with velocity 10 ms–1. It releases a stone which comes down to the ground in 11 s. The height of the balloon from the ground at the moment when the stone was dropped is : (A) 495 m (B) 592 m (C) 460 m (D) 500 m

9.

Water drops fall at regular intervals from a tap which is 5m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant ? (Take g = 10 ms –2) (A)

10.

5 m 4

(C)

5 m 2

(D)

15 m 4

Two particles held at different heights a and b above the ground are allowed to fall from rest. The ratio of their velocities on reaching the ground is : (A) a : b

11.

(B) 4 m

(B)

(C) a2 : b2

a: b

(D) a3 : b3

In the one-dimensional motion of a particle, the relation between position x and time t is given by x 2 + 2x = t (here x > 0). Choose the correct statement :

1 (A) The retardation of the particle is

4( x  1)3 1

(B) The uniform acceleration of the particle is

( x  1)3

1 (C) The uniform velocity of the particle is

( x  1)3

(D) The particle has a variable acceleration of 4t + 6. 12.

Mark the correct statement : (A) Ideal speedometer shows the magnitude of the instantanteous velocity. (B) The magnitude of displacement in an interval is equal to distance covered in that time interval. (C) It is possible to have a situation in which the speed of a particle is some time zero and some time nonzero but the average speed is zero. (D) It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.

13.

Figure shows the position of a particle moving on X-axis as function of time. (A) The particle has come to rest 5 times (B) Initial speed of particle was zero (C) The velocity remains positive for t = 0 to t = 6 s (D) The average velocity for the total period shown is negative.

14.

A body freely falling from rest has a velocity v after it falls through distance h. The distance it has to fall down further for its velocity to become double is : (A) h (B) 2h (C) 3h (D) 4h

More than one choice type 15.

The acceleration time plot for a particle (starting from rest) moving on a straight line is shown in figure. For given time interval, (A) The particle has zero average acceleration (B) The particle has never turned around. (C) The particle has zero displacement (D) The average speed in the interval 0 to 10s is the same as the average speed in the interval 10s to 20s.

RESONANCE

RECTILINEAR MOTION - 35

16.

The acceleration of a particle is zero at t = 0 (A) Its velocity must be constant. (B) The speed at t = 0 may be zero. (C) If the acceleration is zero from t = 0 to t = 5 s, the speed is constant in this interval. (D) If the speed is zero from t = 0 to t = 5 s the acceleration is also zero in the interval.

17.

Consider the motion of the tip of the minute hand of a clock. In one hour (A) the displacement is zero (B) the distance covered is zero (C) the average speed is zero (D) the average velocity is zero

18.

Mark the correct statements for a particle going on a straight line (x–position coordinate, v–velocity, a– acceleration) : (A) If the v and a have opposite sign, the object is slowing down. (B) If the x and v have opposite sign, the particle is moving towards the origin. (C) If the v is zero at an instant, the a should also be zero at that instant. (D) If the v is zero for a time interval, then a is zero at every instant within the time interval.

PART - III : MATCH THE COLUMN 1.

Column  gives some graphs for a particle moving along x-axis in positive x–direction. The variables v, x and t represent velocity of particle, x–coordinate of particle and time respectively. Column  gives certain resulting interpretation. Match the graphs in Column  with the statements in Column . Column  Column  v

(A)

(p) Acceleration of particle is uniform

x v - x graph 2

v

(B)

(q) Acceleration of particle is nonuniform

x 2

v - x graph

v

(C)

(r) Acceleration of particle is directly proportional to ‘t’

t v - t graph

v

(D)

(s) Acceleration of particle is directly proportional to ‘x’.

t

2

2

v - t graph

RESONANCE

RECTILINEAR MOTION - 36

2.

Match the following : Column  (A) Rate of change of displacement (B) Average speed is always greater than or equal to (C) Displacement has the same direction as that of (D) Motion under gravity is considered as the case of

Column  (p) Magnitude of average velocity (q) Initial to final position (r) Velocity (s) Uniform acceleration

PART - IV : COMPREHENSION Comprehension # 1 Read the following write up and answer the questions based on that. The graph below gives the coordinate of a particle travelling along the X-axis as a function of time. AM is the tangent to the curve at the starting moment and BN is tangent at the end moment (1 = 2 =120°).

1. 2.

The average velocity during the first 20 seconds is (A) – 10 m/s (B) 10 m/s (C) zero

(D) 20 m/s

The average acceleration during the first 20 seconds is (A) – 1 m/s2 (B) 1 m/s2 (C) zero

(D) 2 m/s2

3.

The direction ( ˆi or – ˆi ) of acceleration during the first 10 seconds is _____________ .

4.

Time interval during which the motion is retarded. (A) 0 to 20sec. (B) 10 to 20sec. (C) 0 to 10sec.

(D) None of these

Comprehension # 2 The position of a particle is given by x = 2 (t – t2) where t is expressed in seconds and x is in meter. Possitive direction is twords right. 5.

6.

The acceleration of the particle is (A) 0 (B) 4 m/s2

(C) –4 m/s2

(D) None of these.

The maximum value of position co-ordinate of particle on positive x-axis is (A) 1 m

(B) 2 m

(C)

1 m 2

(D) 4 m

7.

The particle (A) never goes to negative x-axis (B) never goes to positive x-axis (C) starts from the origin then goes upto x = 1/2 in the positive x-axis then goes to left (D) final velocity of the particle is zero

8.

The total distance travelled by the particle between t = 0 to t = 1 s is : (A) 0 m

RESONANCE

(B) 1 m

(C) 2 m

(D)

1 m 2

RECTILINEAR MOTION - 37

PART - I : IIT-JEE PROBLEMS (LAST 10 YEARS) * Marked Questions are having more than one correct option. 1.

A block is moving down a smooth inclined plane starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval t = n – 1 to t = n. The ratio

(A)

2.

2n  1 2n

(B)

(C)

2n  1 2n  1

[JEE (Scr.), 2004, 3]

(D)

2n 2n  1

A particle is initially at rest, It is subjected to a linear acceleration a , as shown in the figure. The maximum speed attained by the particle is [JEE Scr. 2004; 3]

(A) 605 m/s 3.

2n  1 2n  1

Sn is S n1

(B) 110 m/s

(C) 55 m/s

(D) 550 m/s

The velocity displacement graph of a particle moving along a straight line is shown.

The most suitable acceleration-displacement graph will be

(A)

(B)

(C)

[JEE Scr. 2005; 3]

(D)

PART - II : AIEEE PROBLEMS (LAST 10 YEARS) 1.

If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest? [AIEEE - 2002, 4/300] (1) 1 cm (2) 2 cm (3) 3 cm (4) 4 cm

2.

From a building two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically with same speed). If VA and VB are their respective velocities on reaching the ground, then [AIEEE - 2002, 4/300] (1) vB> vA (2) vA =vB (3) vA> vB (4) their velocities depends on their masses

3.

Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances at which the two cars are stopped at the same instant is : [AIEEE - 2002, 4/300] (1) 1 : 1

RESONANCE

(2) 1 : 4

(3) 1 : 8

(4) 1 : 16

RECTILINEAR MOTION - 38

4.

The coordinates of a moving particle at any time t are given by x = t3 and y = t3. The speed of the particle at time t is given by : [AIEEE - 2003, 4/300] (1)

 2  2

(2) 3t2

 2  2

(3) t2

 2  2

(4)

 2  2

5.

A car moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. if the same car is moving at a speed of 100 km/hr, the minimum stopping distance is : [AIEEE - 2003, 4/300] (1) 12 m (2) 18 m (3) 24 m (4) 6 m

6.

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds? [AIEEE - 2004, 4/300] (1) h/9 metre from the ground (2) 7h/9 metre from the ground (3) 8h/9 metre from the ground (3) 17h/9 metre from the ground

7.

An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, ie. 120 km/h, the stopping distance will be [AIEEE - 2004, 4/300] (1) 20 m (2) 40 m (3) 60 m (4) 80 m

8.

The relation between time t and distance x is t = ax2 + bx, where a and b are constants. The acceleration is: [AIEEE 2005, 4.300] (1) –2abv2 (2) 2bv2 (3) –2av3 (4) 2av3

9.

A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate

(1) S = ft 10.

(2) S =

1 2 ft 6

f to come to rest. If the total distance travelled is 15 S, then : 2 [AIEEE 2005, 4/300] (3) S =

1 2 ft 72

(4) S =

1 2 ft 4

A particle is moving eastwards with a velocity of 5 ms–1. In 10 second the velocity changes to 5 ms–1 northwards. The average acceleration in this time is : [AIEEE 2005, 4/300]

1 (1)

2

ms–1 towards north-west

(3) zero

(2)

1 ms–2 towards north 2

(4)

1 ms–2 towards north-west. 2

11.

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height approximately, did he bail out? [AIEEE 2005, 4/300] (1) 91 m (2) 182 m (3) 293 m (4) 111 m

12.

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity v that varies as v =  x The displacement of the particle varies with time as (1) t1/2 (2) t3 (3) t2

13.

(4) t

The velocity of a particle is v = v0 + gt + ft2 . If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is [AIEEE 2007, 3/120] f g g + (3) v0 + g + f (4) v0 + +f 3 2 2 An object moving with a speed of 6.25 m/s, is decelerated at a rate given by :

(1) v0 + 2g + 3f 14.

[AIEEE-2006, 3/180]

(2) v0 +

d   2. 5  dt where  is the instantaneous speed. The time taken by the object, to come to rest, would be : [AIEEE - 2011, 4/120, –1] (1) 1 s (2) 2 s (3) 4 s (4) 8 s

RESONANCE

RECTILINEAR MOTION - 39

15.

A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves ? [AIEEE 2012 ; 4/120, –1] Hint : Acceleration =

Force Mass v(t) F0b mb

(1)

(2) t

(3)

(4)

The relative velocity of the car w.r.t. the scooter,    or VCS = VC – VS VCS  VC  VS

BOARD LEVEL EXERCISE : ANSWER & SOLUTIONS 1.

Both rest and motion are relative .

2.

When the object moves along a straight line and

When the car is ahead of the scooter :

always in the same direction. 3.

It will be a straight line, inclined with time-axis.

4.

No. It is because, the x-t graph parallel to position

VCS = 90 – 60 = 30 km h–1 (away from the scooter) When the scooter is ahead of the car : vCS = 90 – 60 = 30 km h–1 (towards the scooter)

axis indicates that the position of the object is 5.

changing at a given instant of time.

13.

two hours

No. It is because, the x-t graph parallel to time-

15.

When an object is thrown up, the direction of mo-

axis indicates that the object is at rest.

tion of the object and its velocity are along verticaly

6.

Acceleration .

downward and upward direction but acceleration

7.

Both the balls will rise to the same height. It is

is along verticaly downward direction. Therefore,

because, for a body moving with given initial veloc-

the direction of motion of the object is that of the

ity and acceleration, the distance covered by the

velocity and not that of acceleration.

body does not depend on the mass of the body. 8.

The average speed in a small time interval is equal

if an object is thrown verticle upward than at top

to the slope of the x-t graph in that interval. Since

point velocity will be zero but acceleration will be

the slope of the graph is maximum in the time interval 3 and least in the interval 2; the average speed

non zero.

is greatest in interval 3 and is least in interval 2.

9.

Instantaneous speed.

10.

(a) Three dimensional

(b) One dimensional

(c) Three dimensional

(d) Two dimensional.

11.

16.

(e) Two dimensional   Let VC and VS be the velocities of the car and the scooter respectively.   Here VC = 90 kmh–1 and VS = 60 kmh–1

RESONANCE

RECTILINEAR MOTION - 40

EXERCISE # 1

Further, the slope of x-t graph is positive in the time intervals 1 and 2 and it is negative in the inter-

PART - I

val 3. Therefore, in interval 1 and 2, the average velocity is positive; and in the interval 3, the average velocity is negative. 18.

sn  u.n 

1 .a.n2 2

sn 1  u(n  1) 

1 2 .an  1 2

SECTION (A) : A-1.

A-2.

Distance travelled by the car = 48 m, Displacement of the car = 36 m 4 (a) 110 m (b) 50 m,tan–1 west of south 3

SECTION (B) : a  sn  sn 1  u  2n  1 2

snth

(a) 32 km/h (b) zero B-2.

For a uniformly accelerated motion, the speed-time graph is a straight line, whose slope is equal to the acceleration of the motion. Since both the cars start with a given speed at zero time and have zero speed at a later time, the v-t graph in each case can be obtained by ing these points with a straight line. Thus, AB represents v-t graph for the

A

25 20 –1

CarA

15

C

10

CarB

3v 1v 2 v 3 v 1v 2  v 2 v 3  v 1v 3

SECTION (C) : C-1. C-2.

(a) 24 Bt ; (b) A + 300 B, 120 B 5 A – 3 Bt–4 C-3. = 2.5 m/s2 2

SECTION (D) : 175 = 87.5 m 2

D-1.

a = 3 m/s2 ;

D-2.

(b) 60 m/s,

D-3. D-4. D-5.

(a) 2700 m = 2.7 km , (c) 225 m and 2.25 km 50m ; 2m/s2 ; 20 m (a) 100 m ; (b) 10 s ; 20m

D-6.

(i) 20 m/s

(iii)

car A and CD represents that for the car B.

Speed(ms )

21.

B-1.

(ii) 4 m

(c) 19 m, 15 m

656 m.

SECTION (E) :

5

O

2

4

D' B 6 8 10 12 Time (s)

14

16

18

D 20

E-1.

distance travelled = 10 m; displacement = 6 m; average velocity =

(a) The distance travelled by each car is equal to the area under its v-t graph. Therefore,

E-2.

6 = 1.2 m/s 5

2000 m. ,

distance travelled by the car A, x1 = area of OAB =

1 1 AO × OB = × 25 × 10 = 125 m 2 2

,

distance travelled by the car B, x2 = area of COD =

1 1 CO × OD = × 15 × 20 = 150 m 2 2

x(m) parabolic curve 2000

Since x2 > x1, the car B travels farther. 1500

(b) If the car B decelerates at the same rate as the

straight line

car A, then its v-t graph should be a straight line drawn from the point C and parallel to AB. There500

fore, dotted line CD’ represents the v-t graph of the

parabolic curve

car B decelerating at the same rate as the car A. Since corresponding to point D’, t = 6s, the car B will then stop in 6s. E-3.

RESONANCE

Particle B RECTILINEAR MOTION - 41

(b)

PART - II

(3, 6) s (0, 2) s & (6, 7) s

SECTION (A) A-1.

(2, 3) s & (7, 9) s

(B)

SECTION (B) B-1.

(B)

B-2.

(A,C) B 3.

(A)

B 4.

(B)

SECTION (C) C 1.

(c)

0 s & (2, 3) s

(d)

[7, 9] s

10.

(a) 10 cm/s; (b) 25 cm/s; (c) t0 = 16s

11.

(8  2 6 ) sec after the lift starts descending,

(A)

129m below the top of the shaft

SECTION (D) : D-1.

(D)

D-2.

(C)

D-3.

(A)

D-4.

(D)

D-5.

(B)

D-6.

(B)

D-7.

(B)

D 8.

(C)

SECTION (E) : E-1.

(B,C) E-2.

(D)

E-5.

(B)

(D)

E-6.

E-3.

(C)

E-4.

(C)

PART - III 1.

1.

(D)

2.

(D)

3.

(D)

4.

PART - II 1.

(B)

2.

(B)

3.

(B)

4.

(C)

5.

(B)

6.

(A)

7.

(A)

8.

(A)

9.

(D)

10.

(B)

11.

(A)

12.

(A)

13.

(A)

14.

(C)

15.

(A,B,D)

16.

(B,C,D)

17.

(A,D)

EXERCISE # 2

1.

(A) q, s (B) p (C) p (D) q, r

PART - I

2.

(A) r, (B) p, (C) q, (D) s

     v1  =  v 2  =  v 3  =  v 4 

(b)

v4, avg > v1, avg = v2, avg > v3, avg

2.

  2v 0  v1  v 2  2 v  v  v 1 2  0

3.

(a)

4.

 H   H h

6.

6 m/s2 7.

9.

(a)

24 m, 24 m (b)

  v 5. 

22 ft/s2

PART - IV : 1.

(A)

2.

(C)

3.

ˆi

4.

(C)

5.

(C)

6.

(C)

7.

(C)

8.

(B)

EXERCISE # 3 PART - I

24 m, 40 m 1.

400 m 8.

(A,B,D)

PART - III :

(D)

(a)

18.

(B)

2.

70 m/s

(C)

3.

(B)

PART - II

[0,1) s & (5,7)s

1.

(1)

2.

(2)

3.

(2)

4.

(2)

(1, 5)s

5.

(3)

6.

(3)

7.

(4)

8.

(3)

(1, 2) s & (5, 6) s

9.

(3)

10.

(1)

11.

(3)

12.

(3)

(0, 1) s & (3, 5) s & (6, 7) s

13.

(2)

14.

(2)

15.

(3)

RESONANCE

RECTILINEAR MOTION - 42