TWO-DIMENSIONAL MOTION

 

 

 

 

 

 

First Notation:

 

 = 65 m [120 N of E] = 65 m [780 E of N]

= 34 m [270 W of N] = 34 m [630 N of W]

= 47 m [700 S of W] = 47 m [200 W of S]

= 28 m [500 S of E] = 28 m [400 E of S]

 

 

Other Notation:

 

 = 65 m [E 120 N] = 65 m [N 780 E] (Point north and turn 78 degrees east)

= 34 m [N 270 W] = 34 m [W 630 N]

= 47 m [W 700 S] = 47 m [S 200 W]

= 28 m [E 500 S] = 28 m [S 400 E]

 

 

VECTOR ADDITION USING SINE AND COSINE LAW:

 

 

 

 

Cosine Law:  c2 = a2 + b2 -2ab. cos C

 

ΔdR2 = Δd12 + Δd22 – 2* Δd1* Δd2*cos 108

ΔdR2 = 42 + 82 – 2* 4* 8*cos 108

ΔdR = 10 m

 

Sine Law:   

 

   

 

                

 

B = 50o

C = 22o

90 - (50 + 30) = 10

ΔdR = 10 m [E10S]

                              

 

VECTOR ADDITION USING COMPONENT METHOD:

 

 

 

 

 

The vector sum of x components:

 

 

 

= 10 m

 

 

A = 10o

ΔdR = 10 m [E10S]

 

 

 

VECTOR SUBTRACTION:

 

Find 

 

 

 

 

 

AVERAGE VELOCITY IN TWO DIMENSIONS:

 

 

 

RELATIVE MOTION:

 

 

Sample Vector Problem:

 

Kevin walked the following distances in the given directions:

 

80 m [N 400 W],

100 m [W 550 S],

150 m [N 600 E],

50 m [W 200 N]

 

Find the Kevin’s final displacement.

 

∆dx = - 80*cos 50 – 100*cos 55 + 150*cos30 - 50*cos20 = - 25.9

 

∆dy = 80*sin 50 – 100*sin 55 + 150*sin30 - 50*sin20 = 71.47

 

∆d = (∆dx2 + ∆dy2)1/2

 

∆d = [(-25.9)2 + 71.472]1/2

 

∆d = 76 m

 

tan θ = 71.47: 25.9

 

θ = 700

 

∆d = 76 m [W 700 N]