Question

Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Office A is walking at 5 mph directly south while Office B is walking at 6 mph directly west. How long would they need to walk before they are 2 miles away from each other?

Answer 1

Answer:

0.256 hours

Explanation:

Vectors in the plane

We know Office A is walking at 5 mph directly south. Let $X_A$ be its distance. In t hours he has walked

$X_A=5t\ \text{miles}$

Office B is walking at 6 mph directly west. In t hours his distance is

$X_B=6t\ \text{miles}$

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

$D=\sqrt{X_A^2+X_B^2}$

$D=\sqrt{(5t)^2+(6t)^2}$

$D=\sqrt{61}t$

This distance is known to be 2 miles, so

$\sqrt{61}t=2$

$t =\frac{2}{\sqrt{61}}=0.256\ hours$

t is approximately 15 minutes