Advertisements
Advertisements
Question
A cyclist driving at 36 kmh−1 stops his motion in 2 s, by the application of brakes. Calculate
- retardation
- distance covered during the application of brakes.
Advertisements
Solution
Initial velocity = u = 36 kmh−1
= u = `36xx5/18` ms−1 = 10 ms−1
Final velocity = v = 0
Time = t = 2s
(i) Retardation = a = ?
v = u + at
0 = 10 + a (2)
2a = −10
a = `(-10)/2` = −5 ms−2
(ii) Distance covered S =?
v2 − u2 = 2aS
(0)2 − (10)2 = 2(−5) S
−10S = −100
S = `100/10`
S = 10 m
APPEARS IN
RELATED QUESTIONS
Define acceleration. State its unit.
Figure given below shows a velocity-time graph for a car starting from rest. The graph has three parts AB, BC and CD.
Which part of graph shows motion with uniform
(a) velocity (b) acceleration (c) retardation ?
A racing car is moving with a velocity of 50 m/s. On
applying brakes, it is uniformly retarded and comes to rest in 20 seconds. Calculate its acceleration.
The speed of a car increases from 10 km/h to 64 km/h in 10 seconds. What will be its acceleration?
The graph shows how the velocity of a scooter varies with time in 50 s.
Work out: Deceleration
Correct your friend who says that acceleration gives the idea of how fast the position changes.
Negative acceleration is called ______.
The acceleration of the body that moves with a uniform velocity will be ______.
Assertion: A positive acceleration of a body can be associated with the slowing down of the body.
Reason: Acceleration is a vector.
Assertion: When a body is subjected to a uniform acceleration, it is always moving in a straight line.
Reason: Motion may be straight-line motion or circular motion.
