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

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

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  1. thienan
    0
    2021-08-16T22:28:49+00:00
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    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?

    Answer:

    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

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