Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
Distinguish between acceleration and retardation.
A body falls freely from a certain height. Show graphically the relation between the distance fallen and square of time. How will you determine g from this graph?
A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C?

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.
The acceleration of a moving body is constant in magnitude and direction. Must the path of the body be a straight line?
If not, given an example.
Distinguish between uniformly and non-uniformly accelerated motions.
The speed of a car increases from 10 km/h to 64 km/h in 10 seconds. What will be its acceleration?
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 defined as the rate of change of ______.
An object can be moving with uniform speed but with variable acceleration.
