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2 years ago

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Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of

traffic. Beth leaves Los Angeles at 9:00 A.M. and surprisingly manages to also drive at a constant speed, in this case 60 mph. (Knight 2.1) a. Who gets to San Francisco first? (Beth) b. How long does the first to arrive have to wait for the second? (20 minutes)

1 answer:

Fudgin [204]2 years ago

3 0

Answer:

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

Explanation:

let 'd' be distance b/w Los Angeles and San Francisco i.e 400 mi

considering ,

Alan's speed $v_A$ =50mph

Beth's speed $v_B$ =60mph

->For Alan:

The time required $t_A$ = d/ $v_A$ = 400/50 => 8h

-> For beth:

The time required $t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h$ => 6h 40m

Alan will reach at 8:00 a.m +8h = 4:00p.m.

Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

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