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A private airplane leaves an airport and flies due east at 192 km/hr. Two hours later, a jet leaves the same airport and flies due east at 960 km/hr. How long after the jet leaves will it overtake the plane?

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The answer is 0.5 hours.

A velocity (v) of an object is a distance (d) divided by time (t):
v = d ÷ t ⇒ t = d ÷ v

It is given:
v1 = 192 km/h
v2 = 960 km/h
t1 - t2 = 2h
We need to calculate t2.

Since they will travel the same distance before the jet overtakes the plane, we can say that d1 = d2 = d

Now, let's express t1 and t2:
t1 = d/192
t2 = d/960

Therefore: d/192 - d/960 = 2
The least common denominator is 960, so:
5d/960 - d/960 = 2
4d/960 = 2
4d = 2 · 960
4d = 1920
d = 1920 ÷ 4 = 480 km

Now t2 = d/960 = 480/960 = 0.5 hours.
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