Vectors 1w4f4

VECTORS Level 1 Physics

Objectives and Essential Questions  Objectives  Distinguish between basic trigonometric functions (SOH CAH TOA)

 Distinguish between vector and scalar quantities

 Add vectors using graphical and analytical methods

 Essential Questions  What is a vector quantity? What is a scalar quantity? Give examples of each.

SCALAR A SCALAR quantity is any quantity in physics that has MAGNITUDE ONLY

Number value with units

Scalar Example

Magnitude

Speed

35 m/s

Distance

25 meters

Age

16 years

VECTOR A VECTOR quantity is any quantity in physics that has BOTH MAGNITUDE and DIRECTION

r r r r x, v , a, F

Vector Example

Magnitude and Direction

Velocity

35 m/s, North

Acceleration

10 m/s2, South

Force

20 N, East

An arrow above the symbol illustrates a vector quantity. It indicates MAGNITUDE and DIRECTION

VECTOR APPLICATION ADDITION: When two (2) vectors point in the SAME direction, simply add them together. EXAMPLE: A man walks 46.5 m east, then another 20 m east. Calculate his displacement relative to where he started. 46.5 m, E

+

66.5 m, E

20 m, E

MAGNITUDE relates to the size of the arrow and DIRECTION relates to the way the arrow is drawn

VECTOR APPLICATION SUBTRACTION: When two (2) vectors point in the OPPOSITE direction, simply subtract them. EXAMPLE: A man walks 46.5 m east, then another 20 m west. Calculate his displacement relative to where he started. 46.5 m, E

20 m, W

26.5 m, E

NON-COLLINEAR VECTORS When two (2) vectors are PERPENDICULAR to each other, you must use the PYTHAGOREAN THEOREM FINISH

Example: A man travels 120 km east then 160 km north. Calculate his resultant displacement.

the hypotenuse is called the RESULTANT 160 km, N

c 2  a2  b2  c  a2  b2 c  resul tan t  c  200km

VERTICAL COMPONENT

120  160  2

2

S R T T A

120 km, E

HORIZONTAL COMPONENT

WHAT ABOUT DIRECTION? In the example, DISPLACEMENT asked for and since it is a VECTOR quantity, we need to report its direction.

N W of N

E of N N of E

N of E

N of W E

W S of W

NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE.

S of E

W of S

E of S S

NEED A VALUE – ANGLE! Just putting N of E is not good enough (how far north of east ?). We need to find a numeric value for the direction. To find the value of the angle we use a Trig function called TANGENT.

200 km 160 km, N

Tan 

 N of E

opposite side 160   1.333 adjacent side 120

  Tan1 (1.333)  53.1o

120 km, E

So the COMPLETE final answer is : 200 km, 53.1 degrees North of East

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What are your missing components? Suppose a person walked 65 m, 25 degrees East of North. What were his horizontal and vertical components? H.C. = ?

V.C = ?

25

65 m

The goal: ALWAYS MAKE A RIGHT TRIANGLE! To solve for components, we often use the trig functions since and cosine.

adjacent side opposite side sine  hypotenuse hypotenuse adj  hyp cos  opp  hyp sin 

cosine 

adj  V .C.  65 cos 25  58.91m, N opp  H .C.  65 sin 25  27.47m, E

Example A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement.

-

12 m, W

-

=

6 m, S 20 m, N

35 m, E

14 m, N

R 

23 m, E =

14 m, N

R  142  232  26.93m 14 Tan   .6087 23   Tan 1 (0.6087)  31.3

23 m, E The Final Answer: 26.93 m, 31.3 degrees North of East

Example A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

Rv  82  152  17 m / s 8.0 m/s, W 15 m/s, N

Rv



8 Tan   0.5333 15   Tan 1 (0.5333)  28.1

The Final Answer : 17 m/s, @ 28.1 degrees West of North

Example A plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate the plane's horizontal and vertical velocity components.

adjacent side opposite side cosine  sine  hypotenuse hypotenuse adj  hyp cos  opp  hyp sin 

H.C. =? 32 63.5 m/s

V.C. = ?

adj  H .C.  63.5 cos 32  53.85 m / s, E opp  V .C.  63.5 sin 32  33.64 m / s, S

Example A storm system moves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement. 1500 km

adjacent side opposite side sine  hypotenuse hypotenuse V.C. adj  hyp cos  opp  hyp sin  cosine 

40

5000 km, E

H.C.

5000 km + 1149.1 km = 6149.1 km

R

964.2 km



adj  H .C.  1500 cos 40  1149.1 km, E opp  V .C.  1500 sin 40  964.2 km, N

R  6149.12  964.2 2  6224.2 km 964.2 Tan   0.157 6149.1   Tan1 (0.157)  8.92 o The Final Answer: 6224.2 km @ 8.92 degrees, North of East

6149.1 km

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