SMART EDUCATIONS
SMART EDUCATIONS 1
MOTION
Q. What is motion?
Ans. Change of position of an object with respect to its surroundings in a given time, is called
motion of the object.
Q. How many types of motion are there?
Ans. Types of motion depends on some bases, like:-
● Path of motion
● Change of position per unit time
On the base of the path of motion, there are mainly four types of motion
1. Rectilinear Motion or simply Linear Motion
2. Circular Motion
3. Oscillatory Motion
On the base of the change of position per unit time, there are two types of motion
1. Uniform Motion
2. Nonuniform Motion
Q. What is Linear Motion?
Ans. When the motion of an object is in a straight path, then the motion is called a linear
motion.
Q. What is Circular Motion?
Ans. When the motion of an object is in a circular path, then the motion is called a circular
motion.
Q. What is Oscillatory Motion?
Ans. When the motion of an object is back and forth, then the motion is called an oscillatory
motion.
Q Define Distance.
Ans. Distance is the magnitude of the change of the position of an object along its path. It is
mainly measured in meter and kilometer but its SI unit is meter. It is a scalar quantity.
Q. Define Displacement.
Ans. Displacement is a distance along a straight line.
OR
Displacement is the shortest distance. Its SI unit is meter. It is a vector quantity.
Q. An object has moved through a distance. Can it have zero displacement? If yes, support your
answer with an example.
Ans. Yes, A moving object can have zero displacement. When an object moves to some
distance and returns to its origin point, then the shortest distance of the object from its origin to
the destination point is zero so the displacement of the object is zero.
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SMART EDUCATIONS 2
Q. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the
magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial
position?
Ans.
Let the square field is ABCD and the origin point is C
Side of the square = 10 m
Time taken to complete 1 revolution = 40 s
Displacement in 2 min 20 s(120 + 20) = ?
Sol:-
(To find displacement we need origin which ic C and
the end point, which we will have to find.)
Nu of revolution =
𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑖𝑛 1 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=
140
40
= 3.5 revolution
So, after completing 3.5 revolution the farmer will be at A(If starts at C)
Then the displacement = CA
= (by Pythagoras formula in )
𝐴𝐵
2
+ 𝐵𝐶
2
∆ 𝐴𝐵𝐶
=
10
2
+ 10
2
=
100 + 100
=
200
= m
10 2
So, displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is
m Ans.
10 2
Q. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the
distance covered and the displacement at the end of 2 minutes 20 s?
Ans.
Let the athlete moves around the given circle and starts at A diameter of the circle
= 200 m
r = 100 m
Time taken to complete 1 round = 40 s
Distance in 2 min 20 s(120+20) = ?
Displacement in 2 min 20 s(120+20) = ?
Sol
Nu of revolution =
𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑖𝑛 1 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=
140
40
= 3.5 revolution
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So, after completing 3.5 revolution the farmer will be at B(If starts at A)
Then the displacement = AB
= 200 m
And the distance =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑜𝑢𝑛𝑑𝑠 × 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒
=
3 . 5 × 2 π 𝑟
=
3 . 5 × 2 ×
22
7
× 100
= 2200 m
So, the distance covered is 2200 m and the displacement is 200m at the end of 2 minutes 20 s.
Q. What is the difference between Distance and Displacement?
Ans.
Distance
Displacement
It is a scalar quantity
It is a vector quantity
It cannot be 0 for a moving object
It can be 0 for a moving object
It cannot be less than Displacement
It cannot be more than Distance
Q. Define the speed of an object.
Ans. The rate of the distance traveled by an object is called the speed of the object.
𝑆𝑝𝑒𝑒𝑑 =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
It is mainly measured in Km/h and m/s but its SI unit is m/s. It is a scalar quantity.
Q. What does the odometer of an automobile measure?
Ans.
Odometer is a device which measures the total distance
traveled by the vehicle.
Q. What does the speedometer of an automobile measure?
Ans.
Speedometer is a device which measures the current
speed of the vehicle.
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SMART EDUCATIONS 4
Q. During an experiment, a signal from a spaceship reached the ground station in five minutes.
What was the distance of the spaceship from the ground station? The signal travels at the
speed of light, that is, 3 × 10
8
m/s
Ans.
Time = 5 min
= 5 60 s
×
= 300 s
Distance = ?
Speed = 3 10
8
m/s
×
Sol
Distance = Speed Time
×
= 3 10
8
300
× ×
= 9 10
10
m
×
Q. Define the uniform motion?
Ans. When an object travels the same distance in every second, then the object is said to be in
uniform motion.
OR
When an object moves with a constant speed, then the object is said to be in uniform motion.
Examples:- Motion of light in a given medium, Motion of every planet, Motion of the hands of a
watch etc.
Q. Define the nonuniform motion?
Ans. When an object travels the different distances in every second, then the object is said to
be in nonuniform motion.
OR
When an object keeps changing its speed in a journey, then the object is said to be in
nonuniform motion.
Examples:- motion of wind, motion of vehicles, motion of living beings etc.
When an object travels in a nonuniform motion then finding its actual speed in the whole journey
is a difficult task because it keeps changing its speed. Hence the term ‘Average speed’ is used
for this type of motion.
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
Q. An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the
object?
Ans.
Total Distance = 16 m + 16 m
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= 32 m
Total time = 4 + 2 s
= 6 s
Average speed = ?
Sol
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
=
32
6
= 5 . 33 𝑚 / 𝑠
Velocity is another important term for the chapter.
Q. Define Velocity.
Ans. Velocity is the rate of displacement of an object. Its SI unit is m/s.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑇𝑖𝑚𝑒
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 + 𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
2
Q. What is the difference between Speed and Velocity?
Ans.
Distance
Displacement
It is the rate of distance
It is the rate of displacement
It is a scalar quantity
It is a vector quantity
It cannot be zero for a moving object
It can be zero for a moving object
It cannot be less than the velocity of an object
It cannot be more than the speed of an object
Q. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30
seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are
Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Ans.
(a) from A to B
Distance = 300 m
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Time = 2 min 30 s
= 120 +30
= 150 s
Average Speed = ?
Velocity = ?
Sol:-
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
300
150
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = 2 𝑚 / 𝑠
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑇𝑖𝑚𝑒
(Shortest distance of AB = 300 m)
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
300
150
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 2 𝑚 / 𝑠
(b) from A to C
Distance = 300 m + 100 m
= 400 m
Time = 2 min 30 s + 1 min
= 120 +30 + 60
= 210 s
Average Speed = ?
Velocity = ?
Sol:-
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
400
210
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = 1 . 904 𝑚 / 𝑠
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑇𝑖𝑚𝑒
(Shortest distance of AC = 200 m)
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
200
210
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 0 . 952 𝑚 / 𝑠
Q. How to convert km/h to m/s and m/s to km/h?
Ans.
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SMART EDUCATIONS 7
(a) km/h to m/s
1 𝑘𝑚
1 ℎ𝑟
=
1000 𝑚
3600 𝑠
=
5
18
𝑚 / 𝑠
So, to convert km/h to m/s, we must multiply the magnitude of km/h to
5
18
(b) m/s to km/h
1 𝑚
1 𝑠
=
1
1000
𝑘𝑚
1
3600
𝑠
=
1
1000
×
3600
1
𝑘𝑚 / ℎ
=
3600
1000
=
18
5
𝑘𝑚 / ℎ
So, to convert m/s to km/h, we must multiply the magnitude of m/s to
18
5
Q. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On
his return trip along the same route, there is less traffic and the average speed is 30 km h–1.
What is the average speed for Abdul’s trip?
Ans.
Average speed while going = 20 km/h
Average speed while returning = 30 km/h
Average speed for the journey = ?
Sol
Let the distance of his school from his home =
𝑥 𝑘𝑚
a. While going
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
20 =
𝑥
𝑡
𝑡 =
𝑥
20
b. While returning
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
30 =
𝑥
𝑇
𝑇 =
𝑥
30
c. In the journey
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
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=
𝑥 + 𝑥
𝑡 + 𝑇
=
2 𝑥
𝑥
20
+
𝑥
30
=
2 𝑥
3 𝑥 + 2 𝑥
60
=
2 𝑥
5 𝑥
60
= 2 𝑥 ÷
5 𝑥
60
= 2 𝑥 ×
60
5 𝑥
Ans
= 24 𝑘𝑚 / ℎ
Q. The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip.
If the trip took 8 h, calculate the average speed of the car in km h
–1
and m s
–1
.
Ans.
Reading of the odometer before the journey = 2000 km
Reading of the odometer after the journey = 2400 km
So, the travelled distance = 2400 - 2000
= 400 km
Time = 8 h
Average speed = ?(km/h and m/s)
Sol
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
=
400
8
= 50 km/h Ans
m/s
= 50 ×
5
18
= 13.89 m/s Ans
Q. Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one
end to the other and back along the same straight path. Find the average speed and average
velocity of Usha.
Ans.
Distance = 180 m
DIsplacement = 0 (She returns at the same position)
Time = 1 min
= 1 60 s
×
= 60 s
Average speed = ?
Average velocity = ?
Sol
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SMART EDUCATIONS 9
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
=
180
60
= 3 m/s Ans
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
=
0
60
= 0 m/s Ans
Q. Define acceleration of an object.
Ans. Acceleration of an object is the rate of change in its velocity.
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
Acceleration = a
Final velocity = v
Initial velocity = u
Time = t
So,
𝑎 =
𝑣 − 𝑢
𝑡
The SI unit of Acceleration is m/s
2
. It is a vector quantity.
Let’s make the term ‘Acceleration’ clear by a simple definition. “When the velocity of an
object increases or decreases then the object is said to be accelerating.” Decreases in velocity
is called retardation or .In short change in velocity or change in direction is called acceleration,
So in a uniform motion, acceleration is 0 because there is no change in velocity or direction
except in a Uniform Circular Motion. First of all we will study linear motion then Uniform Circular
Motion.
Q A bus is traveling with a speed of 15 m/s and the driver increases its speed to 20 m/s which
takes 2 seconds. Find the acceleration of the bus.
Ans.
Initial velocity = u = 15 m/s
Final velocity = v = 20 m/s
Time = t = 2 s
Acceleration = a = ?
Sol
𝑎 =
𝑣 − 𝑢
𝑡
𝑎 =
20 − 15
2
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𝑎 =
5
2
a = 2.5 m/s
2
Ans.
Q. Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in
30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the
next 5 s. Calculate the acceleration of the bicycle in both the cases.
Ans
Case - I
Starting from stationary position means initial velocity = 0 m/s
v = 6 m/s
t = 30 s
a = ?
Case - I
u = 6 m/s
v = 4 m/s
t = 5 s
a = ?
Sol
Case - I
𝑎 =
𝑣 − 𝑢
𝑡
𝑎 =
6 − 0
30
𝑎 =
6
30
a = 0.2 m/s
2
Ans.
Case - II
𝑎 =
𝑣 − 𝑢
𝑡
𝑎 =
4 − 6
2
𝑎 =
− 2
2
a = -1 m/s
2
Ans.
Like motion, there are two types of acceleration also:-
1. Uniform acceleration
2. Nonuniform acceleration
Q. Define uniform acceleration.
Ans. When the rate of change in velocity is constant, then the acceleration is called a uniform
acceleration.
OR
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When an object changes its velocity uniformly, then the acceleration is called a uniform
acceleration.
Example:- Free fall is an example of uniform acceleration.
Q. What is free fall?
Ans. When an object falls without any external force except gravitational force, then the fall is
called free fall.
You might have noticed that, while free fall, velocity of an object keeps increasing.
Q. Define nonuniform acceleration.
Ans. When the rate of change in velocity is not constant, then the acceleration is called a
nonuniform acceleration.
OR
When an object changes its velocity nonuniformly, then the acceleration is called a nonuniform
acceleration.
Example:- Vehicles on the road.
Q. A bus decreases its speed from 80 km h
–1
to 60 km h
–1
in 5 s. Find the acceleration of the
bus.
Ans.
u = 80 km/h
= 80 m/s
×
5
18
= m/s
200
9
v = 60 km/h
= m/s
60 ×
5
18
= m/s
50
3
t = 5 s
a = ?
Sol
𝑎 =
𝑣 − 𝑢
𝑡
𝑎 =
50
3
−
200
9
5
𝑎 =
150 − 200
9
5
𝑎 =
− 50
9
÷ 5
𝑎 =
− 50
9
×
1
5
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𝑎 =
− 10
9
𝑎 =− 1 . 11 𝑚 / 𝑠
2
Q. A train starting from a railway station and moving with uniform acceleration attains a speed
40 km h
–1
in 10 minutes. Find its acceleration.
Sol
u = 0 m/s
v = 40 km/h
= 40
×
5
18
𝑚 / 𝑠
=
100
9
𝑚 / 𝑠
t = 10 min
= 10
× 60 𝑠
= 600 s
a = ?
Sol
𝑎 =
𝑣 − 𝑢
𝑡
𝑎 =
100
9
− 0
2
𝑎 =
100
9
÷ 2
𝑎 =
100
9
×
1
2
𝑎 =
50
9
a = 5.55 m/s
2
Ans.
Q. How many types of graphical representation are there for motion?
Ans. There are four types of graphical representation for motion:-
1. Distance - Time Graph
2. Displacement - Time Graph
3. Speed - Time Graph
4. Velocity - Time Graph
Q. Explain Distance - Time Graph.
Ans Distance - Time Graph is a type of graph of motion in which distance is taken in along the
y-axis and time is taken along the x-axis. The curve or slope of Distance - Time Graph
represents the speed of the object.
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SMART EDUCATIONS 13
Distance - Time Graph can be seen in 3 types:-
Rest Uniform motion Nonuniform motion
Q. Explain Displacement - Time Graph.
Ans Displacement - Time Graph is a type of graph of motion in which displacement is taken in
along the y-axis and time is taken along the x-axis. The curve or slope of Distance - Time Graph
represents the velocity of the object.
Displacement - Time Graph can be seen in 3 types:-
Rest Uniform motion Nonuniform motion
Q. Explain Speed - Time Graph.
Ans. Speed - Time Graph is a type of graph of motion in which speed is taken in along the
y-axis and time is taken along the x-axis. The curve or slope of speed - Time Graph represents
the acceleration of the object and the area represent the distance of the object.
SMART EDUCATIONS
SMART EDUCATIONS 14
Q. Explain Velocity - Time Graph.
Ans. Velocity - Time Graph is a type of graph of motion in which velocity is taken in along the
y-axis and time is taken along the x-axis. The curve or slope of speed - Time Graph represents
the acceleration of the object and the area represent the displacement of the object.
Q. How to find Speed from the distance - Time graph?
Ans.
If it is a graph of a uniform motion, then to find the speed of the object, take any point of
the slope and find its coordinate. Abscissa of the coordinate is time and the ordinate of the
graph is distance. To find the speed, find the ratio of distance to time. For example:-
Coordinate of A in the graph is (5,10), So the distance
traveled by the object is 10 m and the time taken is 5 s.
Hence
𝑆𝑝𝑒𝑒𝑑 =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
𝑆𝑝𝑒𝑒𝑑 =
10
2
𝑆𝑝𝑒𝑒𝑑 = 5 𝑚 / 𝑠
Q. Derive the equation for velocity - time relation
Ans. Let an object be in uniform acceleration and moving
with OC as initial velocity at the time of OA and at the time
of OB its velocity changes to OD.
So,
Initial velocity = OC = BF = u
FInal Velocity = OD = BE = v
Time taken = OB - OA = AB = t
Change in velocity = Final velocity - Initial velocity
EF = v - u (i)
Now the acceleration = The rate of change in its velocity.
SMART EDUCATIONS
SMART EDUCATIONS 15
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
a =
𝐵𝐸 − 𝐵𝐹
𝐴𝐵
𝑎 =
𝐸𝐹
𝑡
𝐸𝐹 = 𝑎𝑡
From (i)
𝑣 − 𝑢 = 𝑎𝑡
𝑣 = 𝑢 + 𝑎𝑡
This is velocity - time relation.
Q. Derive the equation for Position - time relation
Ans. Let an object be in uniform acceleration and moving
with OC as initial velocity at the time of OA and at the time
of OB its velocity changes to OD.
So,
Initial velocity = OC = BF = u
FInal Velocity = OD = BE = v
Time taken = OB - OA = AB = t
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
a =
𝐵𝐸 − 𝐵𝐹
𝐴𝐵
𝑎 =
𝐸𝐹
𝑡
(i)
𝐸𝐹 = 𝑎𝑡
Now the displacement = area of AMEFB
= area of EMF + area of rectangle AMFB
∆
=
1
2
× 𝑏𝑎𝑠𝑒 × ℎ + 𝑙 × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ
=
1
2
× 𝑀𝐹 × 𝐸𝐹 + 𝐴𝐵 × 𝐵𝐹
=
1
2
× 𝐴𝐵 × 𝐸𝐹 + 𝐴𝐵 × 𝐵𝐹
s = {EF = at from eq (i)}
1
2
× 𝑡 × 𝑎𝑡 + 𝑡 × 𝑢
s =
𝑢𝑡 +
1
2
𝑎 𝑡
2
This is position - time relation.
Q. Derive the equation for Position - velocity relation
Ans. Let an object be in uniform acceleration and moving
with OC as initial velocity at the time of OA and at the time
of OB its velocity changes to OD.
So,
SMART EDUCATIONS
SMART EDUCATIONS 16
Initial velocity = OC = BF = u
FInal Velocity = OD = BE = v
Time taken = OB - OA = AB = t
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
a =
𝐵𝐸 − 𝐵𝐹
𝐴𝐵
𝑎 =
𝐸𝐹
𝑡
(i)
𝐸𝐹 = 𝑎𝑡
Now the displacement = area of trapezium AMEFB
S =
1
2
( 𝑠𝑢𝑚 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑠𝑖𝑑𝑒𝑠 ) × ℎ𝑒𝑖𝑔ℎ𝑡
S =
1
2
( 𝐴𝑀 + 𝐵𝐸 ) × 𝐴𝐵
S =
1
2
( 𝑢 + 𝑣 ) × 𝑡
S = (From v = u +at, t = )
1
2
( 𝑢 + 𝑣 ) ×
( 𝑣 − 𝑢 )
𝑎
( 𝑣 − 𝑢 )
𝑎
S =
𝑣
2
− 𝑢
2
2 𝑎
𝑣
2
− 𝑢
2
= 2 𝑎𝑠
This is position - velocity relation.
Q. A driver of a car traveling at 52 km h–1 applies the brakes and accelerates uniformly in the
opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies
his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time
graphs for the two cars. Which of the two cars traveled farther after the brakes were applied?
Ans
Driver - I Driver - II
u = 52 km/h u = 3 km/h
v = 0 m/s v = 0 km/h
t = 5 s t = 10 s
Sol
Green graph is for driver- I and red graph is
for driver - II
Distance for driver-I = Area of green triangular
=
1
2
× 𝑏 × ℎ
=
1
2
× 5 ×( 52 ×
5
18
)
( 52 km/h = m/s )
52 ×
5
18
=
5
2
×
260
18
SMART EDUCATIONS
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=
1300
36
= 36.11 m
2
Distance for driver-II = Area of red triangular
=
1
2
× 𝑏 × ℎ
=
1
2
× 10 ×( 3 ×
5
18
)
( 3km/h = m/s )
3 ×
5
18
=
5 ×
15
18
=
75
18
= 4.166 m
2
So, the first driver travels farther than the second driver. Ans
Q. A train starting from rest attains a velocity of 72 km h
–1
in 5 minutes. Assuming that the
acceleration is uniform, find (i) the acceleration and (ii) the distance traveled by the train for
attaining this velocity.
Ans.
Starting from rest, so
u = 0 m/s
v = 72 km/h
= 72 m/s
×
5
18
= 20 m/s
t = 5 min
t = 5 60 s
×
t = 300 s
a = ?
s = ?
Sol
𝑎 =
𝑣 − 𝑢
𝑡
𝑎 =
20 − 0
300
𝑎 =
20
300
Ans(i)
𝑎 =
1
15
𝑚 / 𝑠
2
𝑠 = 𝑢𝑡 +
1
2
𝑎 𝑡
2
= 0 × 300 +
1
2
×
1
15
× 300
2
SMART EDUCATIONS
SMART EDUCATIONS 18
= 0 +
1
30
× 90000
= 0 + 3000
= 3000 𝑚
= 3 km Ans(ii)
Q. A car accelerates uniformly from 18 km h
–1
to 36 km h
–1
in 5 s. Calculate (i) the acceleration
and (ii) the distance covered by the car in that time.
Ans
u = 18 km/h
= 18 m/s
×
5
18
= 5 m/s
v = 36 km/h
= m/s
36 ×
5
18
= 10 m/s
t = 5 s
a = ?
s = ?
Sol
𝑎 =
𝑣 − 𝑢
𝑡
𝑎 =
10 − 5
5
𝑎 =
5
5
Ans(i)
𝑎 = 1 𝑚 / 𝑠
2
𝑠 = 𝑢𝑡 +
1
2
𝑎 𝑡
2
= 5 × 5 +
1
2
× 1 × 5
2
= 25 +
25
2
= 25 + 12 . 5
Ans(ii)
= 37 . 5 𝑚
Q. The brakes applied to a car produce an acceleration of 6 m s
-2
in the opposite direction to the
motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it
travels during this time.
Ans.
a = - 6 m/s
2
(its velocity is decreasing so it retardation)
t = 2 s
v = 0 m/s
s = ?
Sol
SMART EDUCATIONS
SMART EDUCATIONS 19
𝑎 =
𝑣 − 𝑢
𝑡
− 6 =
0 − 𝑢
2
− 6 × 2 = 0 − 𝑢
− 12 = − 𝑢
𝑢 = 12 𝑚 / 𝑠
𝑠 = 𝑢𝑡 +
1
2
𝑎 𝑡
2
= 12 × 2 +
1
2
×− 6 × 2
2
= 24 − 12
Ans
= 12 𝑚
Q. 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 traveled.
Ans
u = 0 m/s
a = 0.1 m/s
2
t = 2 min
= 2 60 s
×
= 120 s
v = ?
s = ?
Sol
𝑎 =
𝑣 − 𝑢
𝑡
0 . 1 =
𝑣 − 0
120
12 = 𝑣
Ans
𝑣 = 12 𝑚 / 𝑠
𝑠 = 𝑢𝑡 +
1
2
𝑎 𝑡
2
= 0 × 120 +
1
2
× 0 . 1 × 120
2
= 0 + 720
Ans
= 720 𝑚
Q. A train is traveling 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.
Ans.
u = 90 km/h
SMART EDUCATIONS
SMART EDUCATIONS 20
=
90 ×
5
18
𝑚 / 𝑠
= 25 m/s
a = - 0.5 m/s
2
s = ?
v = 0 m/s
Sol
0
2
− 25
2
= 2 ×− 0 . 5 × 𝑠
0 − 625 = − 1 × 𝑠
− 625 = − 𝑠
s = 625 m Ans
Q. 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?
Ans
a = 2 cm/s
2
=
2
100
𝑚 / 𝑠
2
=
1
50
𝑚 / 𝑠
2
v = ?
t = 3 s
u = 0 m/s
Sol
𝑎 =
𝑣 − 𝑢
𝑡
1
50
=
𝑣 − 0
3
3
50
= 𝑣
Ans
𝑣 = 0 . 06 𝑚 / 𝑠
Q. A racing car has a uniform acceleration of 4 m s
-2
. What distance will it cover in 10 s after
start?
Ans
a = 4 m/s
2
s = ?
t = 10 s
u = 0 m/s
Sol
𝑠 = 𝑢𝑡 +
1
2
𝑎 𝑡
2
= 0 × 10 +
1
2
× 4 × 10
2
SMART EDUCATIONS
SMART EDUCATIONS 21
= 0 + 2 × 100
Ans.
= 200 𝑚
Q. 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?
Ans
u = 5 m/s
a = - 10 m/s
2
(velocity of a stone is decreasing)
s/h = ?
t = ?
v = 0 m/s ( as the stone will stop at the height before falling)
Sol
𝑎 =
𝑣 − 𝑢
𝑡
− 10 =
0 − 5
𝑡
− 10 𝑡 = − 5
𝑡 =
− 5
− 10
Ans
𝑡 =
1
2
𝑠
𝑠 = 𝑢𝑡 +
1
2
𝑎 𝑡
2
= 5 ×
1
2
+
1
2
×− 10 ×
1
2
2
=
5
2
−
10
2
×
1
4
=
5
2
−
5
4
=
10 − 5
4
=
5
4
= 1.25 m Ans
Q. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0
m s
–2
for 8.0 s. How far does the boat travel during this time?
Ans.
u = 0 m/s
a = 3 m/s
2
t = 8 s
s = ?
Sol
𝑠 = 𝑢𝑡 +
1
2
𝑎 𝑡
2
= 0 × 8 +
1
2
× 3 × 8
2
SMART EDUCATIONS
SMART EDUCATIONS 22
= 0 +
3
2
× 64
= 0 + 96
= 96 m Ans
Q. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of
10 m s
-2
, with what velocity will it strike the ground? After what time will it strike the ground?
Ans.
u = 0 m/s
s = 20 m
a = 10 m/s
2
v = ?
t = ?
Sol
𝑠 = 𝑢𝑡 +
1
2
𝑎 𝑡
2
20 = 0 × 𝑡 +
1
2
× 10 × 𝑡
2
20 = 0 + 5 𝑡
2
20 = 5 𝑡
2
5 𝑡
2
= 20
𝑡
2
=
20
5
𝑡
2
= 4
t = 2 s Ans
𝑎 =
𝑣 − 𝑢
𝑡
10 =
𝑣 − 0
2
20 = 𝑣
v = 20 m/s Ans
Q. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it
takes 24 hours to revolve around the earth.
Ans
r = 42250 km
speed = ?
time = 24 h
Sol
𝑣 =
2π 𝑟
𝑡
× 𝑛𝑢 𝑜𝑓 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑣 =
2 ×
22
7
× 42250
24
× 1
SMART EDUCATIONS
SMART EDUCATIONS 23
𝑣 =
2 × 22 × 42250
7 × 24
= 2 ×
22
7
× 42250
= 265571.428 km. Ans
Q. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the
distance covered and the displacement at the end of 2 minutes 20 s?
Ans
d = 200 m
r = 100 m
Time to complete 1 revolution = 40 s
distance = ?
Total time = 2 min 20 s
= 2 60 + 20
×
= 120 + 20
= 140 s
Sol
No of revolution =
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 1 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=
140
40
= 3.5
Distance = Circumference of the circle number of revolution
×
= 2πr 3.5
×
=
2 ×
22
7
× 100 × 3 . 5
= 2200 m Ans
Q. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity (b) an object moving in a certain
direction with an acceleration in the perpendicular direction.
Ans(a)
An object with a constant acceleration but with zero velocity is only possible when an
object is hanged at some height. The object is at the state of rest, hence zero velocity but there
is constant acceleration due to gravity.
Ans(b)
An object moving in a certain direction with an acceleration in the perpendicular direction
is only possible in a circular motion. We know that in a circular motion, there is acceleration due
to change in direction. At every point an object changes its direction perpendicularly. Hence
acceleration is also in the perpendicular direction also.
