The brakes applied to a car produce an acceleration of 6m/s×s in the opposite direction to the motion. If the car takes 2s to stop after the application of brakes find the distance it travelled during this time.

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Solution

Acceleration a = -6 m/s^2
Time t = 2 s
Final velocity v = 0 m/s (since after 2 secnds the car is stopped)

Let initial velocity be u
Let distance be s

v = u + at
So, 0 = u + (-6)(2)
So, u = 12 m/s

Now, s = ut + (1/2) at^2
So, s = 12(2) + (1/2)(-6)(2^2)
So, s = 24 - 12
So, s = 12 m

Thus, distance travelled is 12 m

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Q. The brakes applied to a car produce an acceleration of

6 {m/s}^{2}

in the opposite direction to the motion. If the car takes

2 s

to stop after the application of brakes, calculate the distance it travels during this time.

The brakes applied by a car produces an acceleration of $6 m / s^{2}$ in opposite direction of motion. If the car takes 2 sec to stop after the application of brakes. Calculate distance during this time.

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