Question

Velocity: You leave on a 549-mi trip in order to attend a meeting that will start 10.8 h after you begin your trip. Along the way you plan to stop for dinner. If the fastest you can safely drive is 65 mi/h, what is the longest time you can spend over dinner and still arrive just in time for the meeting?

Answer 1

Answer:

1.55 h

Explanation:

Let the time to spend over dinner be t.

Since I need to spend 10.8 h for the whole trip, then I have 10.8 - t hours left to drive.

Speed = distance/time

$v = \dfrac{d}{10.8-t}$

$d = v(10.8-t)$

The distance is 549 and maximum speed is 65 mi/h.

$549 = 65(10.8 - t)$

$10.8-t = 8.45$

$t = 10.8 - 8.45 = 1.55 \text{ h}$