Two cars leave from the same location but travel in opposite directions. The first car travels an average of 50 mi/hr and leaves 15 min before the second car, which travels an average of 55 mi/hr. How long each car travel before they are 200 mi apart?

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Two cars leave from the same location but travel in opposite directions. The first car travels an average of 50 mi/hr and leaves 15 min before the second car, which travels an average of 55 mi/hr. How long each car travel before they are 200 mi apart?

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Answer 1 · 2016-04-27T09:41:29

1.79 hours

Explanation:

If we define the time t=0 as when the second care leaves.

Before the second care leaves, the first car travels a distance of:
$0.25 \times 50 mi/hr = 12.5 mi$ $0.25 times 50"mi/hr"=12.5"mi"$
(since 15min = 0.25hr)

The total distance car 1 travels will be:
$= 12.5 + 50 \cdot t$ $=12.5 +50*t$

And car 2 travels a distance in the opposite direction:
$= 55 \cdot t$ $=55*t$

The total distance between them will be:
$12.5 + 50 \cdot t + 55 \cdot t = 12.5 + 105 \cdot t$ $12.5+50*t+55*t=12.5+105*t$

We want to know t when this equals 200 miles. So:

$12.5 + 105 t = 200$ $12.5+105t=200$

Solving for t:

$105 t = 200 - 12.5 = 187.5$ $105t=200-12.5=187.5$
$t = 1.79 h r s$ $t=1.79 hrs$

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Two cars leave from the same location but travel in opposite directions. The first car travels an average of 50 mi/hr and leaves 15 min before the second car, which travels an average of 55 mi/hr. How long each car travel before they are 200 mi apart?

1 Answer

Explanation:

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