The driver of a motor car moving with a speed of 90 km/hr applies brakes such that it comes to rest in next 10 sec.Find the acceleration of the car.

Open in App

Solution

Initial velocity of the car is, u = 90 km/h
1 km/h = 5/18 m/s
Therefore, u = 90 Ã— 5/18 = 25 m/s
The car comes to rest in, t = 10 s
Using, v = u + at
=> 0 = 25 + a Ã— 10
=> a = -2.5 m/s²
The negative sign indicates that the direction of acceleration is opposite to the motion of the car.

Suggest Corrections

Similar questions

Q. A car is moving on a straight road with some speed. The driver applies brakes so that car comes to rest with a retardation of

5 m / s^{2}

. The distance traversed by the car in last second of its motion is :

A motor car moving with a uniform speed of $20 m s^{- 1}$ comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

Q. A car moving at a constant speed of

54 k m / h

comes to rest in

2.5 s e c o n d s

. The car covers

5 m

after driver applies the brakes. The total force applied to stop the car is

1000 N

. Calculate the power dissipated in stopping the car.

A motor car moving with a uniform speed of $20 m s^{- 1}$ comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

Q. A man travelling in a car at a constant speed of 10 m/sec, applies brakes and the car come to rest, after moving a distance of 10 m. The time taken to stop the car assuming uniform restardation is

Join BYJU'S Learning Program

The driver of a motor car moving with a speed of 90 km/hr applies brakes such that it comes to rest in next 10 sec.Find the acceleration of the car.

Initial velocity of the car is, u = 90 km/h1 km/h = 5/18 m/sTherefore, u = 90 Ã— 5/18 = 25 m/sThe car comes to rest in, t = 10 sUsing, v = u + at=> 0 = 25 + a Ã— 10=> a = -2.5 m/s2The negative sign indicates that the direction of acceleration is opposite to the motion of the car.

Initial velocity of the car is, u = 90 km/h
1 km/h = 5/18 m/s
Therefore, u = 90 Ã— 5/18 = 25 m/s
The car comes to rest in, t = 10 s
Using, v = u + at
=> 0 = 25 + a Ã— 10
=> a = -2.5 m/s²
The negative sign indicates that the direction of acceleration is opposite to the motion of the car.