Request a call back

a car travelling along a straight road at 14ms-1 is approaching traffic lights.The driver applies the brakes and the car comes to rest with constant deceleration.The distance from the point where the brakes are applied to the point where the car comes to rest is 49m.Find the deceleration of the car
A car is travelling along a straight road at u =14 m/s
The distance from the point where the brakes are applied to the point where the car comes to rest is s = 49m
We know that
speed = distance / time
time = distance / speed = 49 / 14  = 3.5 seconds.

Thus, this car travells for t= 3.5 seconds.
During the course of applying the brakes and finally stopping the motion (v = 0 m/s)
The deceleration can be found using the below equation of motion:

v = u + at
Where, v is final velocity (zero, as the car stops)
t is the time of travel (3.5 seconds, as calculated)
u is the initial velocity ( 14 m/s, as given)

Substituting the values,
0 = 14 + a (3.5)
a = -14 / 3.5
a = - 4 m/s2

The negative symbol above symbolises that the car undergoes deceleation (gradual decrease in acceleration)
Thus, the deceleration of the car is - 4 m/s2 .
Answered by Abhijeet Mishra | 23 Aug, 2018, 03:37: PM

## Application Videos

CBSE 9 - Physics
Asked by janhavisoni2099 | 02 Oct, 2023, 05:20: PM
CBSE 9 - Physics
Asked by sagrawal0081 | 18 Aug, 2023, 05:04: PM
CBSE 9 - Physics
CBSE 9 - Physics
Asked by rsaranya658 | 30 Jun, 2022, 02:44: PM
CBSE 9 - Physics
Asked by sanskargunjkar01 | 26 Jun, 2022, 09:13: AM