Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Initial velocity = u = 0
Final velocity = v = 72 km/h = `72xx5/18` m/s
v = 20 m/s
Distance = S = 40 m
(i) v2 − u2 = 2aS
(20)2 − (0)2 = 2a (40)
80a = 400
a = `400/80`
a = 5 ms−2
(ii) v = u + at
20 = 0 + 5t
5t = 20
t = `20/5`
t = 4 s
APPEARS IN
संबंधित प्रश्न
Distinguish between acceleration and retardation.
Give one example of the following motion:
Uniform retardation
The change in velocity of a motorbike is 54 kmh−1 in one minute. Calculate its acceleration in (a) ms−2 (b) kmh−2.
What does a positive acceleration mean?
State if the following situation is possible:
A body moving with constant acceleration but with Zero velocity.
A car accelerates to a velocity of 30 m/s in 10 s and then decelerates for 20 s so that it stops. Draw a velocity-time graph to represent the motion and find:
The acceleration.
A car accelerates to a velocity of 30 m/s in 10 s and then decelerates for 20 s so that it stops. Draw a velocity-time graph to represent the motion and find:
The Deceleration.
Negative acceleration is called ______.
Exercise Problem.
A racing car has a uniform acceleration of 4 ms–2. What distance it covers in 10 s after the start?
Assertion: A negative acceleration of a body can be associated with speeding up of the body.
Reason: Increase in the speed of a moving body is independent of its direction of motion.
