Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
What is meant by the term retardation? Name its S.I. unit.
Draw velocity – time graph for the following situation:
When a body is moving with variable velocity, but uniform retardation.
From the velocity – time graph given below, calculate deceleration in region BC.
A motor bike running at 5 ms−1, picks up a velocity of 30 ms−1 in 5s. Calculate
- acceleration
- distance covered during acceleration.
A car traveling at 60 km/h, stops on applying brakes in 10 seconds. What is its acceleration?
State if the following situation is possible:
A body moving horizontally with an acceleration in vertical direction.
A car is moving with a speed of 50 km/h. One second later, its speed is 55 km/h. What is its 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.
If a body starts from rest, what can be said about the acceleration of the body?
When a body is said to be in
- uniform acceleration
- non–uniform acceleration?
