Advertisements
Advertisements
Question
A racing car, initially at rest, picks up a velocity of 180 kmh−1 in 4.5 s. Calculate
- acceleration
- distance covered by the car.
Advertisements
Solution
Initial velocity of car = u = 0
Final velocity of car = v = 180 kmh−1 = `180xx5/18` ms−1
v = 50 ms−1
Time = t = 4.5 s
v = u + at
50 = 0 + a (4.5)
4.5a = 50
a = `50/4.5` = 11.11 ms−2
(Distance) S = ut + `1/2` at2
S = `0(4.5)+1/2xx50/4.5xx(4.5)^2`
S = 0 + 25 × 4.5
S = 112.5 m
APPEARS IN
RELATED QUESTIONS
A train moving with a velocity of 20 ms−1 is brought to rest by applying brakes in 5 s. Calculate the retardation.
A space craft flying in a straight course with velocity of 75 km s-1 fires its rocket motors for 6.0 s. At the end of this time its speed is 120 km s-1 in the same direction.
Find
(i) The space craft's average acceleration while the motors were firing
(ii) The distance travelled by the space craft in the first 10 s after the rocket motors were started, the motors being in action for only 6 s.
Explain the following concept in your own words with everyday examples:
Acceleration
When is the acceleration due to gravity positive?
The change in velocity of a motorbike is 54 kmh−1 in one minute. Calculate its acceleration in (a) ms−2 (b) kmh−2.
From the velocity – time graph given below, calculate deceleration in region BC.
Distinguish between uniformly and non-uniformly accelerated motions.
Negative acceleration is called ______.
A boy throws a ball up and catches it when the ball falls back. In which part of the motion the ball is accelerating?
Acceleration is a scalar.
