Question

Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 46.0 mph . Beth leaves Los Angeles at 9:00 a.m. and drives a steady 55.0 mph .How long does the first to arrive have to wait for the second?

Answer 1

Answer:

25 min, 48 sec

Explanation:

Alan:

t = d/v

  = 400 mi / 46 mph = 8.70 hr

So it took Alan 8.70 hrs; from 8 am, that's 4:42 pm that he arrives.

t = d/v

  = 400 mi / 55 mph = 7.27 hr

So it took Beth 7.27 hrs; from 9 am, that's 4:16:12 that she arrives.

4:42 - 4:16:12 = 25.8 minutes = 25 minutes, 48 sec