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Question

A purple car is moving three times as fast as a yellow car. Each car slows down to a stop with the same constant acceleration. How much more distance is required for the purple car to stop ?

A
twice as much distance
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B
three times as much distance
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C
five times as much distance
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D
seven times as much distance
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E
nine times as much distance
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Solution

The correct option is E nine times as much distance
Let the initial velocity of purple car be 3u , so the initial velocity of yellow car will be u. Final velocities ( vp,vy )of both cars are zero , acceleration a is same for both cars .
Using, the third equation of motion v2=u22as
For purple car 0=(3u)22asp
where sp= distance required for purple car to stop ,
a= acceleration (deceleration)
Hence, sp=9u2/2a ,

For yellow car 0=u22asy
Hence, sy=u2/2a
where sy= distance required for yellow car to stop ,

now by dividing sp from sy
we get sp/sy=9
or sp=9sy

Hence, the purple car requires nine times as much distance as the yellow car to stop.

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