Question

A man walks 3.50 mi due east, then turns and walks 2.57 mi due north. How far
and in what direction is he from the starting point

Answer 1

Answer:

4.34 mi at $36.3^{\circ}$ north of east

Explanation:

The displacement of an object in motion is a vector connecting its initial position to the final position of motion.

In this problem, the man has 2 different motions:

- 3.50 mi due east

- 2.57 mi due north

We can take the east direction as positive x-direction and north as positive y-direction, so these two motions can be written as:

$x=+3.50 mi$

$y=+2.57 mi$

Since the two motions are perpendicular to each other, the resultant displacement can be found by using Pythagorean's theorem; therefore:

$d=\sqrt{x^2+y^2}=\sqrt{3.50^2+2.57^2}=4.34 mi$

We can also find the direction using the equation:

$tan \theta = \frac{y}{x}$

And therefore,

$\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{2.57}{3.50})=36.3^{\circ}$