Brakes applied to a car produces an acceleration of 6 m/^2.if the car takes 2 sec to stop after the application of brakes . Calculate the distance travelled by the car during this time?

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Solution

Acceleration a = -6 m/s^2
Time t = 2 s
Final velocity v = 0 m/s

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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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.

The brakes applied to a car produce an acceleration of 6 ms-² in opposite direction to the motion. If the car takes 2 s to stop after the application of rakes, calculate the distance it travels during this time.

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Brakes applied to a car produces an acceleration of 6 m/^2.if the car takes 2 sec to stop after the application of brakes . Calculate the distance travelled by the car during this time?

Acceleration a = -6 m/s^2 Time t = 2 s Final velocity v = 0 m/s 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

Acceleration a = -6 m/s^2
Time t = 2 s
Final velocity v = 0 m/s

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