Displacement: Magnitude and Direction

[kraaaaamos]

Homework Statement

Jim walks 100m due south, 60m due east, and then 20m due north. What is his net
displacement (both magnitude and direction)?

Homework Equations

I know how to find the net displacement by the magnitude of the distance...
using the d = sq. rt ( (x2-x1)^2 + (y2-y1)^2 )

but i don't know how to express the direction?

PLEASE HELP!

The Attempt at a Solution

 

[mgb_phys]

Draw the path on graph paper then draw a line back to the start - the direction is just the angle of this line ( usually given from the horizontal x-axis )
The magnitude is the length of this line.

 

[m2003]

Hello,

To find the net displacement in terms of both magnitude and direction, we can use the Pythagorean theorem as you have stated, but we can also use trigonometric functions to determine the direction. In this case, we can use the inverse tangent function to find the angle of the net displacement.

First, we can draw a diagram to represent Jim's movements. We can see that his net displacement is a diagonal line connecting the starting point to the end point. Using the Pythagorean theorem, we can calculate the magnitude of this diagonal line to be approximately 117.3m.

To find the direction, we can use the inverse tangent function. We can take the opposite side (60m) and the adjacent side (100m) and use the formula tanθ = opposite/adjacent to find the angle θ. In this case, tanθ = 60/100, so θ = tan^-1(0.6) ≈ 31.6°. This means that Jim's net displacement is 117.3m at an angle of approximately 31.6° east of south.

I hope this helps! Let me know if you have any further questions.