Question

A jogger runs 20 mi West and then 6.0 mi North. Find the magnitude and direction of the resultant displacement.

Answer 1

Answer:

The magnitude of the resultant displacement is 21 mi and its direction is 16.7° north of west

Explanation:

Hi there!

Please see the figure for a better understanding of the problem. The total displacement vector will be the sum of both displacements:

The vector for the first displacement is:

First displacement = (20 mi, 0)

The second displacement:

Second displacement = (0, 6.0 mi)

The resultant displacement will be:

R = (20 mi, 0) + (0, 6.0 mi) = (20 mi + 0, 0 + 6.0 mi) = (20 mi, 6.0 mi)

The magnitude of this vector will be:

$|R| = \sqrt{(20 mi)^{2} + (6.0 mi)^{2}} = 21 mi$

The magnitude of the vector displacement is 21 mi.

To find the direction of the vector R, we have to apply trigonometry:

In a right triangle the following trigonometric rule applies:

cos θ = adjacent side to the angle/ hypotenuse

In this case:

cos θ = 20 mi / magnitude of R

θ = 16.7°

The direction of the vector is 16.7° north of west.