Advertisements
Advertisements
प्रश्न
A motor bike running at 5 ms−1, picks up a velocity of 30 ms−1 in 5s. Calculate
- acceleration
- distance covered during acceleration.
Advertisements
उत्तर
Initial velocity of motor bike = u = 5 ms−1
Final velocity of motor bike = v = 30 ms−1
Time = t = 5 s
(i) Acceleration = a = ?
v = u + at
30 = 5 + a (5)
5a = 30 − 5 = 25
a = `25/5` = 5 ms−2
(ii) Distance covered S =?
S = ut + `1/2` at2
S = `5(5)+1/2(5)(5)^2`
S = `25+125/2` = 25 + 62.5
S = 87.5 m
APPEARS IN
संबंधित प्रश्न
Distinguish between acceleration and retardation.
When is the acceleration due to gravity positive?
Give an example of an accelerated body, moving with a uniform speed.
From the diagram given below, calculate deceleration.

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.
A cyclist driving at 5 ms−1, picks a velocity of 10 ms−1, over a distance of 50 m. Calculate
- acceleration
- time in which the cyclist picks up above velocity.
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.
Distinguish between uniformly and non-uniformly accelerated motions.
Can you use the relation [B1]?
The slope of a velocity-time graph gives ______.
