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
State the type of motion represented by the following sketches in Figures.
Give an example of each type of motion.
Figure shows the velocity-time graphs for two cars A and B moving in the same direction. Which car has greater acceleration? Give reasons to your answer.
Draw a graph for acceleration against time for a uniformly accelerated motion. How can it be used to find the change in speed in a certain interval of time?
A vehicle is accelerating on a straight road. Its velocity at any instant is 30 km/h. After 2 s, it is 33.6 km/h, and after further 2 s, it is 37.2 km/h. Find the acceleration of the vehicle in m s-2. Is the acceleration uniform?
Give an example of an accelerated body, moving with a uniform speed.
Draw velocity – time graph for the following situation:
When a body is moving with variable velocity and variable acceleration.
Convert the following acceleration:
1/36 m/s2 into km/h2
It is possible to have objects moving with uniform speed but variable acceleration.
What is the difference between uniform acceleration and non–uniform acceleration?
