## Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 {\rm mi} away. He travels at a steady 47.0 mph. Beth leaves Los Angeles

Question

Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 {\rm mi} away. He travels at a steady 47.0 mph. Beth leaves Los Angeles at 9:00 a.m. and drives a steady 55.0 mph. Who gets to San Francisco first? How long does the first to arrive have to wait for the second?

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3 years 2021-08-16T22:27:14+00:00 1 Answers 23 views 0

1. Complete question is;

Alan leaves Los Angeles at 8:AM to drive to San Francisco, 400 mi away. He travels at a steady 47 mph. Beth leaves Los Angeles at 9:00 AM and drives a steady 55.0 mph.

a. Who gets to San Francisco first?

b. How long does the first to arrive have to wait for the second?

A) Beth will get there first

B) Beth will have to wait for 15 minutes for alan

Explanation:

We are given;

Distance between Los Angeles and San Francisco;d = 400 mi

Alan’s velocity;v_a = 47 mph

Beth’s velocity;v_b = 55 mph

we know that time is given by;

Time = distance/velocity

Time required for alan;

t_a = 400/47

t_a = 8.5106 hours = 8 hours 31 minutes

Time required for Beth;

t_b = 400/55

t_b = 7.2727 hours = 7 hours 16 minutes

So Alan will reach there at;

8:00 a.m + 8 hours 31 minutes = 4:31 p.m

Beth will reach there at;

9:00 a.m + 7 hours 16 minutes = 4:16 pm

Thus, beth will arrive 4:31 pm – 4:16pm = 15 minutes before alan

A) Beth will get there first

B) Beth will have to wait for 15 minutes for alan