Question

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:

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}v=

10.8−t

d

d = v(10.8-t)d=v(10.8−t)

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

549 = 65(10.8 - t)549=65(10.8−t)

10.8-t = 8.4510.8−t=8.45

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