Motion Question Bank
Calculate the average speed of a train, which covers 120 km in 2 hours.
Distance covered by the train = 120 km
Time taken = 2 h
How far does a car travel in 2.5 h if its average speed is 60 km/h ?
Time taken by the car = 2.5 h
Average speed = 60 km/h.
Here we have to calculate the total distance covered by the car in 2.5h.
Convert 1 km/h to m/s.
A carriage moves 30 km in one hour. Calculate its speed in m/s. Is it average or uniform speed?
The speed is average speed.
You are asked to run in a straight line. If you cover a distance of 5 m in each second, what is your (a) speed (b) velocity?
speed = 5 m/s
velocity = 5 m/s along the straight path.
Draw the velocity time graph of a body moving with uniform velocity of 5m/s.
Write the equation of motion connecting S, u, v and a where the symbols have their usual meanings.
What do you infer, if the distance-time graph is a straight line?
The straight line graph indicates that the speed is uniform.
The distance (S) in meters traveled by a particle is related to time(t) in seconds by the equation of motion . What is the initial velocity of the particle?
initial velocity = 0 [because t is not 0]
On a 60 km track, a train travels the first 30 km at a uniform speed of 30 km/h. How fast must the train travel the next 30 km so that the average speed is 40 km/h for the entire trip?
Given, average speed = 40 km/h
The train must travel the next 30 km with a speed of 60 km/h so that the average speed is 40 km/h.
A train travels a distance with a speed of 30 km/h and returns with a speed of 50 km/h. Calculate the average speed of the train.
Imagine that the train is traveling from A to B and back to A.
Let x be the distance between the two stations A and B.
Which one of the three is having maximum and the least average speed?
1) Bicycle moving with 12 km/h
2) A man running with 5m/s
3) A car moving with 1500 m/min.
The speed of bicycle, man and car is given in different units. In order to compare and find out who is faster we have to express the speed in the same system of units.
Let us express the speed in the SI.
Speed of the man = 5m/s.
Car is moving with maximum speed and the bicycle is moving with least speed.
A car moving with an initial velocity of 25 m/s is brought to rest in 20 seconds. Calculate the acceleration.
Final velocity v = 0 (the car is brought to rest)
Initial velocity = u = 25 m/s
Time = 20 s.
Acceleration = – 1.25 m/
Negative sign shows that, the retardation is produced in the car.
What conclusions do you draw about the nature of motion of the body from the following velocity-time graph?
a) The object is moving with uniform acceleration (since the graph is a straight line. Straight line graph means change in velocity in equal intervals of time)
b) The object is having zero acceleration (since the object is moving with uniform velocity)
c) The object is moving with negative acceleration or retardation (since the velocity decreases and finally becomes zero)
d) The object is moving with variable acceleration (since the change in velocity is not equal)
The figure below gives the v-t graph of a car. What is the distance covered by the car when it is moving with a uniform speed of 15 m/s.
The distance covered by the car when it is moving with a uniform speed of 15 m/s
= area of the rectangle ABDE
A car, which was at rest, attains a speed of 90 km/h in 10 seconds. Calculate the acceleration.
Initial velocity (u) = 0 [the car is initially at rest]
A body starts from rest and moves with a uniform acceleration of 2m/s2. What will be its velocity and displacement at the end of 10 seconds?
Initial velocity (u) = 0 (body is at rest)
acceleration (a) = 2m/s2.
time = 10 s
Final velocity (v) =?
According to the first equation of motion
Final velocity = 20 m/s
Velocity at the end of 10 seconds = 20 m/s.
Displacement is nothing but distance traveled in a particular direction.
According to the second equation of motion
A body moves with an initial velocity of 2 m/s and uniform acceleration of 3m/s2. Calculate the velocity when it has traveled a distance of 84 meters.
Initial velocity (u) = 2 m/s
Acceleration (a) = 3m/s2
Distance (s) = 84 m
Final velocity (v) =?
We make use of the third equation of motion
Calculate the distance traveled in 10 seconds by an object
Time (t) = 10 seconds
Initial velocity (u) = 0 (object was initially at rest)
Acceleration (a) = 10 m/s2
Distance traveled = 500m.
A ball starting from rest slides down an inclined plane of length 10 m with a speed of 10 m/s. What is the acceleration produced in the ball?
Initial velocity (u) = 0 (the ball is starting from rest)
Length of the inclined plane = distance covered by the ball = s = 10 m.
Final velocity (v) = 10 m/s.
Acceleration produced in the ball = a
Acceleration is calculated using III equation of motion
acceleration produced in the ball = 5 m/s2.
A ship is moving at a speed of 56 km/h. One second later it is moving at 58 km/h. What is its acceleration?
Initial velocity (u) = 56 km/h
Final velocity (v) = 58 km/h
Change in velocity v-u = (58-56)km/h
A car traveling at 20 km/h speeds up to 60 km/h in 6 seconds. What is its acceleration?
Initial velocity (u) = 20 km/h
Final velocity (v) = 60 km/h
What is the distance covered by a bus starting from rest and moving with a uniform acceleration of 1 m/s2 for 120 seconds?
Initial velocity (u) = 0 (bus is initially at rest)
Acceleration (a) = 1 m/s2
Time = 120 s.
To calculate the distance traveled we make use of II equation of motion
Find the initial velocity of a train, which is stopped in 25 seconds. The retardation produced in the train is 1.5 m/s2.
Final velocity (v) = 0 ( train is stopped)
Time (t) = 25 s
Initial velocity (u) =?
Plot a distance-time graph for the given data and calculate
(a) the corresponding physical quantity and,
(b) the distance covered by the car at the end of 2.5 s and 6.5 s.
The physical quantity, which we get from the distance-time graph, is speed.
To calculate the distance covered at the end of 2.5 s draw a line vertically up from 2.5 s till it meets the distance-time graph. From that point draw a line to the y-axis. Check where the line meets. The corresponding reading on the y-axis gives the distance traveled. Similarly we can find distance covered at the end of 6.5 s.
The distance covered by the car at the end of 2.5 s = 25 m.
The distance covered by the car at the end of 6.5 s = 65 m
A car travels 30 km at a uniform speed of 40 km/h and the next 30 km at a uniform speed of 20 km/h. Find its average speed.
Plot a graph for the following data and compute the corresponding physical quantity from it.
The physical quantity which we get from v- t graph is acceleration.
A car is moving with a uniform acceleration for the first ten seconds and then it moves with a uniform velocity of 30 m/s for the next ten seconds. The brakes are applied and the car comes to rest with a uniform acceleration in 5 seconds. Draw a v-t graph to show the nature of motion. How far does the car travel after the brakes are applied?
Here, you are supposed to calculate the distance traveled by the car after the brakes are applied. i.e., the distance traveled between the time interval 20 – 25 s.
The area of the shaded part gives the distance traveled.
Distance covered after the application of brakes = 75 m.
For the given v-t graph find,
a) the acceleration when the object is moving along AB, BC and CD.
b) displacement in each part and
c) total distance covered by the object.
Acceleration along BC = 0 (as there is no change in velocity)
(velocity reduces to zero from 40 m/s)
the object is decelerating along CD.
Displacement of the object in the first 6 seconds = area of ABF
Displacement of the object when the object is moving with uniform velocity = Area of rectangle BCEF
Displacement of the object in the last two seconds = area of CDE
Total distance covered by the object
Plot a displacement-time graph for the given data and calculate the speed (a) at the end of 5 seconds (b) in the next two seconds (c) in the last two seconds.
Scale: x axis – 1 unit = 1 s y axis 1 unit = 1 min.
Displacement – time graph
To calculate speed at the end of 5 seconds.
The slope OA will give the speed at the end of 5 seconds.
(0.6 m/s is the speed in a particular direction i.e., it is the velocity).
To calculate the speed in the next two seconds.
In the next two seconds there is no displacement and hence the speed is zero.
(C) To calculate the speed in the last two seconds
The slope BC gives the speed in the last two seconds.
A car is traveling at 54 km/h. If brakes are applied so as to produce uniform retardation of 5m/s2. Calculate the time taken by it to stop.
Initial velocity (u) = 54 km/hr
Final velocity (v) = 0 (because the car is brought to rest)
Retardation = 5m/s2
Negative acceleration = -5m/s2
To find the time taken
use the first equation of motion
A driver of a car A traveling at a uniform speed of 54 km/h for 10 secs applies the brakes and decelerates uniformly. The car stops in 5 seconds. Driver of car B going at 36 km/h for 10 sec applies the brakes slowly and stops after 10 seconds. Plot the speed-time graph of both cars and find out which of the two cars traveled further after the brakes were applied.
Since the time is given in seconds, the speed is converted to m/s.
Here we have to find out which of the two cars traveled more distance after the brakes were applied. For that we calculate the area of the shaded part and compare.
Distance covered by car A after the brakes were applied = area of ABC
Distance covered by car B after the brakes were applied = area of PQR
Hence car B covers more distance.
What is the average speed of a car which covers half the distance with a speed of 20m/s and other half with a speed of 30m/s?
Let x be the distance covered by the car.
Plot a velocity-time graph for the following data and calculate the corresponding physical quantity.