Advertisements
Advertisements
प्रश्न
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.
उत्तर
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
संबंधित प्रश्न
Define acceleration. State its unit.
State its value in C.G.S. as well as in S.I. system.
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.
Diagram is given below shows velocity – time graph of car P and Q, starting from the same place and in the same direction. Calculate the Acceleration of car P.
A motorbike, initially at rest, picks up a velocity of 72 kmh−1 over a distance of 40 m. Calculate
- acceleration
- time in which it picks up above velocity.
Write the SI unit of acceleration and retardation.
Out of energy and acceleration, which one is a vector?
The distance covered by a body is directly proportional to the square of the time elapsed. What can you say about its acceleration?
What is meant by negative acceleration?
m/s2 is the unit of ______.
