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

Δd_R² = Δd₁² + Δd₂² – 2* Δd₁* Δd₂*cos 108

Δd_R² = 4² + 8² – 2* 4* 8*cos 108

Δd_R = 10 m

Sine Law:   

B = 50^o

C = 22^o

90 - (50 + 30) = 10

ΔdR = 10 m [E10S]

VECTOR ADDITION USING COMPONENT METHOD:

The vector sum of x components:

= 10 m

A = 10^o

Δd_R = 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]