Question

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the starting position? give the magnitude and direction relative to due east. (treat this as a positive x-axis)

Answer 1

Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

$D=\sqrt{9^2+12^2} \\\\D=15\ km$

For direction,

$\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= 53.13^{\circ}$

Hence, this is the required solution.