Question

The paths of two runners cross at a stop sign. One runner is heading south at a constant rate of 6.5 miles per hour while the other runner is heading west at a constant rate of 7 miles per hour. How fast is the distance between them changing after 10 minutes?

Answer 1

Answer:

The distance between them changing after 10 minutes will be 9.553 mph.

Step-by-step explanation:

The paths of two runners cross at a stop sign (O). One runner is heading south at a constant rate of 6.5 miles per hour towards A while the other runner is heading west at a constant rate of 7 miles per hour towards B.

So, after 10 minutes the first runner covers a distance of $6.5 \times \frac{10}{60} = 1.083$ miles and the second runner covers a distance of $7 \times \frac{10}{60} = 1.167$ miles.

Therefore, after 10 minutes their distance will be $\sqrt{(1.083)^{2} + (1.167)^{2}} = 1.592$ miles.

Now, the distance between them is given by

AB² = OA² + OB²

Now, differentiating this equation with respect to time t (in hours) we get

$2(AB) \frac{d(AB)}{dt} = 2(OA) \frac{d(OA)}{dt} + 2(OB) \frac{d(OB)}{dt}$

⇒ $(AB) \frac{d(AB)}{dt} = (OA) \frac{d(OA)}{dt} + (OB) \frac{d(OB)}{dt}$

⇒ $1.592 \times \frac{d(AB)}{dt} = 1.083 \times 6.5 + 1.167 \times 7 = 15.208$

⇒ $\frac{d(AB)}{dt} = 9.553$ mph.

Therefore, the distance between them changing after 10 minutes will be 9.553 mph. (Answer)