After applying a brake on a car, it generates acceleration of 6ms² in the opposite direction of the speed, if the car takes time of 2sec to stop after the brake is applied, then calculate the distance traveled in that time.

Open in App

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

Suggest Corrections

Similar questions

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.

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?

Join BYJU'S Learning Program

After applying a brake on a car, it generates acceleration of 6ms2 in the opposite direction of the speed, if the car takes time of 2sec to stop after the brake is applied, then calculate the distance traveled in that 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

After applying a brake on a car, it generates acceleration of 6ms² in the opposite direction of the speed, if the car takes time of 2sec to stop after the brake is applied, then calculate the distance traveled in that 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