Question

A vehicles was moving at the speed of 72 km/hr .when a baby suddenly appeared on the road at a distance of 50m , the drive jammed on the brakes and it came to rest at a distance 40m. what was the retardation of the car? how long did it take to come to rest?
[ans:: 5m/s², 4 s]​

Answer 1

Answer:

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Explanation:

Now to find the deceleration of the car we use the eq. v^2 = u^2 - 2as, where v is equal to 0 and u is equal to 20 and s is equal to 50 and a is the deceleration. We find a is equal to 4 m/s^2. So, acceleration of the car is( - 4 m/s^2) as acceleration and deceleration are in opposite direction.

Now further assume that the car is moving with speed 20m/s in a circular track of some radius and the diagram is as follows,

this case we also use the same eq., stated above, but here s is not equal to 50, s is equal to the distance between them along circular track.

Many other cases will be possible for the first question.

For the second question, the car is moving with uniform velocity of 20m/s in a straight line and in this case, the process will be same as the first process in first question and finally we get the acceleration of the car is ( - 4 m/s^2).

Answer 2

Answer:

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