A car weighs 1000 kg. It is moving with a uniform velocity of 72 km/h towards a straight road. The driver suddenly presses the brakes. The car stops in 0·2 s. The retarding force applied on the car to stop it is
100 N
1000 N
10 kN
100 kN
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-2 – 2024
The initial velocity is u = 72 km/h. To convert to m/s, multiply by 5/18: u = 72 * (5/18) m/s = 4 * 5 m/s = 20 m/s.
The final velocity is v = 0 m/s (since the car stops).
The time taken is t = 0.2 s.
Using the kinematic equation v = u + at, we find the acceleration (a):
0 = 20 + a * 0.2
a * 0.2 = -20
a = -20 / 0.2 = -100 m/s². The negative sign indicates retardation.
The retarding force is given by F = ma:
F = 1000 kg * (-100 m/s²) = -100,000 N.
The magnitude of the retarding force is 100,000 N.
Since 1 kN = 1000 N, 100,000 N = 100 kN.