1.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?
2. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
3. 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?
4. A car accelerates uniformly from 36 km h–1 to 54 km h–1 in 10 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
5.The brakes applied to a car produce an acceleration of 10 m s-2 in the opposite direction to the motion. If the car takes 5 s to stop after the application of brakes, calculate the distance it travels during this time.
6. A train starting from rest attains a velocity of 54 km h–1 in 5 minutes. Assuming that the
acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
Answers
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1
Answer:
1)Initial velocity is: u=0
Time is: t=3s
Acceleration is: a=2 cm s
−2
From first equation of motion:
v=u+at
v=0+2×3=6 cm s
−1
2) Initial velocity is: u=0
Acceleration is: a=4 ms
2
Time is: t=10 s
Using second equation of motion:
s=ut+
2
1
at
2
s=0×10+
2
1
×4×10
2
=0+200=200 m
3)
Velocity=5m/s,a=10m/s
2
So, the height is:
v
2
−u
2
=2as
s=
2a
v
2
−u
2
=
2×(−10)
0−25
=1.25 since v=0
Since a is negative so its direction is opposite to the direction of motion.
So,
t=
a
v−u
=
−10
0−5
=
2
1
time=0.5s
4) final velocity= 54 km/h
= 15m/s
initial velocity = 36km/h
= 10m/s
time taken= 10second
acceleration= v- u/t
= 15-10/10
= 5/10
= 1/2
= 0.5
distance covered= v²- u² = 2as
= 15² - 10² = 2×0.5× s
= 225- 100= 10× s
= 125 = 10s
s = 12.5
4)
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