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

4.34 mi at
`36.3^(\circ)` north of east

Step-by-step 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=√(x^2+y^2)=√(3.50^2+2.57^2)=4.34 mi`

We can also find the direction using the equation:

`tan \theta = (y)/(x)`

And therefore,

`\theta=tan^(-1)((y)/(x))=tan^(-1)((2.57)/(3.50))=36.3^(\circ)`