numerical of motion for exam
Answers
Question 1
A train accelerates from 36 km/h to 54 km/h in 10 sec.
(i) Acceleration
(ii) The distance travelled by car.
Answer a) Acceleration is given by
a
=
Δ
v
Δ
t
a=ΔvΔt
So a=.2 m/s2 b) Distance is given by
S
=
u
t
+
1
2
a
t
2
S=ut+12at2 So s=125m
Question 2
A body whose speed is constant
(a) Must be accelerated
(b) Might be accelerated
(c) Has a constant velocity
(d) Cannot be accelerated.
Answer might be accelerated
Question 3
A truck traveling at 54 km/h is slow down to 36 km/h in 10 sec. Find the retardation
Answer Acceleration is given by
a
=
Δ
v
Δ
t
a=ΔvΔt
So a=-.5 m/s2
Negative sign implies retardation
Question 4
A particle is moving in a circle of diameter 20m. What is its distance and as per the table given below
Rounds Displacement Distance
1
1.5
2
2.5 Answer
Rounds Displacement Distance
1 0
20
π
20π
1.5 20 m
30
π
30π
2 0
40
π
40π
2.5 20m
50
π
50π
Question 5
A scooter travelling at 10 m/s speed up to 20 m/s in 4 sec. Find the acceleration of train.
Answer Acceleration is given by
a
=
Δ
v
Δ
t
a=ΔvΔt
So a=1.25 m/s2
Question 6
A train starts from rest and accelerate uniformly at the rate of 5 m/s2 for 5 sec. Calculate the velocity of train in 5 sec.
Answer 25m/s
Question 7
A object moves with uniform positive acceleration. Its velocity-time graph will be
(a) A straight line parallel to the time axis
(b) A straight line inclined at an obtuse angle to the time axis
(c) A straight line inclined at an acute angle to the time axis
(d) None of these.
Answer Answer is (c)
Question 8
The maximum speed of a train is 90 km/h. It takes 10 hours to cover a distance of 500 km. Find the ratio of its average speed to maximum speed?
Answer Average speed=500/10=50 km/hr
Ratio of average speed to maximum speed= 50:90=5:9
Question 9
9 A car start from rest and acquire a velocity of 54 km/h in 2 sec. Find
(i) the acceleration
(ii) distance travelled by car assume motion of car is uniform?
Answer a) Acceleration is given by
a
=
Δ
v
Δ
t
a=ΔvΔt
So a=7.5 m/s2
b) Distance is given by
S
=
u
t
+
1
2
a
t
2
S=ut+12at2 and can be calculated using the above formula.
Question 10
An object dropped from a cliff falls with a constant acceleration of 10 m/s2. Find its speed 5 s after it was dropped.
Answer V=u+at
u
≥
0
u≥0
v
=
10
×
5
=
50
m
/
s
v=10×5=50m/s
Question 11
11 A ball is thrown upwards and it goes to the height 100 m and comes down
1) What is the net displacement?
2) What is the net distance?
Answer As it comes down to the initial point
Net displacement is zero
Net distance=200 m
Practice questions on motion
Question 12
Two cars A and B race each other. The Car A ran for 2 min at a speed of 7.5 km/h, slept for 56 min and again ran for 2 min at a speed of 7.5 km/h. find the average speed of the car A in the race.
Answer We know that
Distance = speed�time
Distance travelled in first 2 min = 7.5�2/60 = 0.25 km
Distance travelled in last 2 min = 7.5�2/60= 0.25 km
Total distance = .25+.25 = 0.5 km
Total time = 2+2+56 = 60min = 1 hr
Average speed = 0.5/1
= 0.5 km/hr
Question 13
Anand leaves his house at 8.30 a.m. for his school. The school is 2 km away and classes start at 9.00 a.m. If he walks at a speed of 3 km/h for the first kilometer, at what speed should he walk the second kilometer to reach just in time?
Answer 6 km/hr
Question 14
An object moves along a straight line with an acceleration of 2 m/s2. If its initial speed is 10 m/s, what will be its speed 2 s later?
Answer 14 m/s
Question 15
A bullet hits a Sand box with a velocity of 20 m/s and penetrates it up to a distance of 6 cm. Find the deceleration of the bullet in the sand box. Question 16
A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance D1 in the first 10 seconds and distance D2 in the next 10 seconds then,
(a) D2 = D1
(b) D2 = 2D1
(c) D2 = 3D1
(d) D2 = 4D1