Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How far must the

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Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How far must the faster car travel before it has a 15-min lead on the slower car

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Nick 3 years 2021-07-28T11:51:45+00:00 1 Answers 202 views 0

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  1. vankhanh
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    2021-07-28T11:53:32+00:00
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    Answer:

    The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

    Explanation:

    Given;

    speed of the faster car, v₁ = 60 mi/h

    speed of the slower car, v₂ = 55 mi/h

    Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

    \frac{x}{55} - \frac{x}{60} = \frac{15}{60}

    Note: divide 15 mins by 60 to convert to hours for consistency in the units.

    \frac{x}{55} - \frac{x}{60} = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles

    Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

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