First slide

Rectilinear Motion

Question

A car is moving at a certain speed. The minimum distance over which it can be stopped is x. If the speed of the car is doubled, what will be the minimum distance over which the car can be stopped for the same retardation?

Moderate

A

4x
B

2x
C

x/2
D

x/4

Solution

Let a be the maximum retardation the brakes of the car can produce.
Then $2 ax = v^{2}$ .
If v is doubled to 2v, then the minimum distance x is given by $2 {ax}^{} = (2 v)^{2} = 4 v^{2}$ .
Thus, $x^{} = 4 x$ .

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