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.
State the type of motion represented by the following sketches in Figures.

Give an example of each type of motion.
A motor bike running at 90 kmh−1 is slowed down to 18 kmh−1 in 2.5 s. Calculate
- acceleration
- distance covered during slow down.
From the given v-t graph it can be inferred that an object is

If a body starts from rest, what can be said about the acceleration of the body?
A car is being driven at a speed of 20ms-1 when brakes are applied to bring it to rest in 5 s. The deceleration produced in this case will be ______.
Acceleration is a scalar.
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.
Assertion: Position-time graph of a stationary object is a straight line parallel to the time axis.
Reason: For a stationary object, the position does not change with time.
What is the difference between uniform acceleration and non–uniform acceleration?
