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3 years ago

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Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 45.0 mph . Beth leaves Los A

ngeles at 9:00 a.m. 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?

1 answer:

WARRIOR [948]3 years ago

6 0

Answer:

a.Beth

b.2232 s

Explanation:

We are given that

Distance,d=400 mi

Speed of Alan,v=45 mph

Speed of Beth,v'=55 mph

a.Time = $\frac{distance}{speed}$

Using the formula

Time taken by Alan= $\frac{400}{45}=8.89 hr$

Time taken by Beth= $\frac{400}{55}=7.27hr$

Alan will reach San Francisco at 4:53 PM

Beth will reach San Francisco at 4:16 PM

Beth will reach before Alan.

b.Difference between time=8.89-7.27=1.62 hr

t=1.62 hr

1.62-1=0.62 hr

0.62 hr= $0.62\times 60\times 60=2232 s$

Hence, Beth has to wait 2232 s for Alan to arrive .

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