Motion in 2D - An Algebraic Approach

Section 2.1 shows how to solve motion problems in 2D with vector scale diagrams, which have limitations, as it's not precise and scale diagrams can be cumbersome when 2 vectors are needed.

Pic 1 shows many legs of a trip, each being someone's displacement. Without the map and the scale adding these by scale diagram is hard. In many ways an algebraic way is a better to add vectors.

This method again uses trigonometry.

Right way to write bearing and distance notation 

Distance with unit of measurement (e.g. 3 km) [1st: Direction angle is from (the direction it's sticked on), 2nd: angle, 3rd: 2nd direction of the angle] 

E.g. Joe walks 12 km at 34° east of north would be written as: 

12 km [N 34° E]

The vector arrows aren't shown above the two displacements on the sketch because only their are shown.

To find the magnitude and direction of dT. 

As it's a right triangle, use Pythagorean theorem:

Problem 2

Joe walks 20 m [W], then turns 10.0 m [S 40° E]. 

What's his total displacement?

Solution:
ΔdT = Δd1 + Δd2

ΔdT = 20 m [W] + 10 m [S 40° E]

Notice Δd1 has only 1 component - the x-component and no y-component,

1. Find the total x component of ΔdT:

ΔdTx = 20 m [W] + 10 m cos 50° [E]

ΔdTx = 20 m [W] + 10 m x 0.64 [E]

ΔdTx = 20 m [W] + 6.4 [E]

ΔdTx = 20 m [W] - 6.4 [W] (change from east to west from positive to negative)

ΔdTx = 13.6 m [W] is the total x displacement.

2. Find the total y component of ΔdT:

ΔdTy = 0 + 10 m x sin 50° [S]

ΔdTy = 10 m x 0.77 [S]

ΔdTy = 7.7 m [S] is the total x displacement

3. Find the total displacement:

Use the x and y components with the Pythagorean theorem to find the total displacement (bigger triangle's hypothenus).

ΔdT =13.6² + 7.7²

ΔdT = √244.25 

ΔdT = 15.63 m is the total displacement

4. Find the total displacement's angle:

tan Angle = 7.7/13.6 

tan Angle = 0.57

Angle = cos⁻¹ 0.57

Angle = 30°

Joe's total displacement 15.63 m [W 30° S].