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A motorist traveling at 10m/s, was able to bring his car to rest in a distance of 10m. If he had been traveling at 30m/s, in what distance could he bring his cart to rest using the same breaking force?

30 90 45 60

Toshali Mohapatra by toshali mohapatra , 22, India

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3 solutions

Syed Imad Azeem
May 2, 2014

simple just use 3rd equation of motion twice
v^2 = u^2 + 2as
let u= 0
then v^2 = 2as
first find a then s

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Jayant Sharma
Feb 27, 2014

kinetic energy in first case k = 1/2 X m X v^{2} = 1/2 X m X 100 =50m kinetic energy in second case k' = 1/2 X m X 900 = 450m mass of cart will be same in both cases. now k'/k = 450m/50m = 9 so it will take 9 times more time to stop. hence ans is 90 sec.

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You might think that the answer is 30m. This is wrong.

To comprehend this question, you need to sketch two speed-time graphs of the two cases, one for the first scenario and another for the next.

For the first scenario, in order to find the time taken to bring the car to rest in a distance of 10m, we let the time taken be t seconds. So, 1/2 x 10 x t = 10 → t = 2.

Then, we have to find the braking force needed.Letting the force be F, F = (0-10)/2 → F = -5N.

Subsequently, to find the time taken to bring the car to rest, which was traveling at 30m/s, we will form this equation: -5N = (0-30)/t → t = 6.

Finally, the distance traveled by the car will be: D = 1/2 x 30 x 6 = 90m. There you go.

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