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A bus starting from rest moves with a uniform acceleration of 0.1 m s−2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
(a) 12 m/s (b) 720 m
(a) Initial speed of the bus, u = 0 (since the bus is initially at rest)
Acceleration, a = 0.1 m/s2
Time taken, t = 2 minutes = 120 s
Let v be the final speed acquired by the bus.
∴v = 12 m/s
(b) According to the third equation of motion:
v2 − u2 = 2as
Where, s is the distance covered by the bus
(12)2 − (0)2 = 2(0.1) s
s = 720 m
Speed acquired by the bus is 12 m/s.
Distance travelled by the bus is 720 m.
A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.
Initial speed of the train, u = 90 km/h = 25 m/s
Final speed of the train, v = 0 (finally the train comes to rest)
Acceleration = −0.5 m s−2
According to third equation of motion:
v2 = u2 + 2 as
(0)2 = (25)2 + 2 (−0.5) s
Where, s is the distance covered by the train
The train will cover a distance of 625 m before it comes to rest.
A trolley, while going down an inclined plane, has an acceleration of 2 cm s−2. What will be its velocity 3 s after the start?
Initial velocity of the trolley, u = 0 (since the trolley was initially at rest)
Acceleration, a = 2 cm s−2 = 0.02 m/s2
Time, t = 3 s
According to the first equation of motion:
v = u + at
Where, v is the velocity of the trolley after 3 s from start
v = 0 + 0.02 × 3 = 0.06 m/s
Hence, the velocity of the trolley after 3 s from start is 0.06 m/s.
A racing car has a uniform acceleration of 4 m s−2. What distance will it cover in 10 s after start?
Initial velocity of the racing car, u = 0 (since the racing car is initially at rest)
Acceleration, a = 4 m/s2
Time taken, t = 10 s
According to the second equation of motion:
Where, s is the distance covered by the racing car
Hence, the distance covered by the racing car after 10 s from start is 200 m.
A stone is thrown in a vertically upward direction with a velocity of 5 m s−1. If the acceleration of the stone during its motion is 10 m s−2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Initially, velocity of the stone,u = 5 m/s
Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height)
Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2
(in downward direction)
There will be a change in the sign of acceleration because the stone is being thrown upwards.
Acceleration, a = −10 m/s2
Let s be the maximum height attained by the stone in time t.
According to the first equation of motion:
v = u + at
0 = 5 + (−10) t
According to the third equation of motion:
v2 = u2 + 2 as
(0)2 = (5)2 + 2(−10) s
Hence, the stone attains a height of 1.25 m in 0.5 s.
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