CBSE - Motion

Yes,an object can have zero displacement even when it has moved through a distance.This happens when final position of the object coincides with its initial position. For example,if a person moves around park and stands on place from where he started then here displacement will be zero.

Given, Side of the square field= 10m
Therefore, perimeter = 10 m x 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m

Therefore, in 1s distance covered by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 X 140 m = 140 m
Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter
= 140 m / 40 m = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.



Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.

Which of the following is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer

None of the statement is true for displacement First statement is false because displacement can be zero. Second statement is also false because displacement is less than or equal to the distance travelled by the object.

Speed	Velocity
Speed is the distance travelled by an object in a given interval of time.	Velocity is the displacement of an object in a given interval of time.
Speed = distance / time	Velocity = displacement / time
Speed is scalar quantity i.e. it has only magnitude.	Velocity is vector quantity i.e. it has both magnitude as well as direction.

The magnitude of average velocity of an object is equal to its average speed, only when an object is moving in a straight line.

The odometer of an automobile measures the distance covered by an automobile.

An object having uniform motion has a straight line path.

Speed= 3 × 10⁸ m s⁻¹
Time= 5 min = 5 x 60 = 300 secs.

Distance= Speed x Time
Distance= 3 × 108 m s⁻¹ x 300 secs. = 9 x 10¹⁰ m

(i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant.

Initial speed of the bus, u= 80 km/h = 80 x 5/18 m/s = 22.22 m/s

Final speed of the bus, v= 60 km/h = 60 x 5/18 m/s=16.66 m/s

Time take to decrease the speed, t= 5 s


Here, the negative sign of acceleration indicates that the velocity of the car is decreasing.

Initial velocity of the train, u= 0 (since the train is initially at rest)

Final velocity of the train, v= 40 km/h = 40 x 5/ 18 m/s = 11.11 m/s

Time taken, t = 10 min = 10 x 60 = 600 s



Hence, the acceleration of the train is 0.0185 m/s².

When the motion is uniform, the distance time graph is a straight line with a slope.

Graph of uniform motion

When the motion is non uniform, the distance time graph is not a straight line.It can be any curve.

Graph of non uniform motion

When an object is at rest, its distance - 'time graph is a straight line parallel to the time axis.

A straight line parallel to the x-axis in a distance - '-time graph indicates that with a change in time, there is no change in the position of the object. Thus, the object is at rest.

The area below velocity-time graph gives the distance covered by the object.

Initial speed of the bus, u= 0

Acceleration, a = 0.1 m/s²
Time taken, t = 2 minutes = 120 s

v= u + at
v= 0 + 0.1 × 120
v= 12 ms^–1

(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.

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.

Initial Velocity of trolley, u= 0 cms^-1
Acceleration, a= 2 cm s-2
Time, t= 3 s
We know that final velocity, v= u + at = 0 + 2 x 3 cms^-1
Therefore, The velocity of train after 3 seconds = 6 cms^-1

Initial Velocity of the car, u=0 ms^-1
Acceleration, a= 4 m s^-2
Time, t= 10 s
We know Distance, s= ut + (1/2)at²
Therefore, Distance covered by car in 10 second= 0 × 10 + (1/2) × 4 × 10²
= 0 + (1/2) × 4× 10 × 10 m
= (1/2)× 400 m
= 200 m

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
t= -5 / -10 = 0.5 s

According to the third equation of motion:

v² = u²+ 2as

(0)²= (5)²+ 2( - 10) s

s= 5² / 20 =1.25 m

Hence, the stone attains a height of 1.25 m in 0.5 s.

Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 � ( 22 / 7 ) � 100
Speed of the athlete (v) = Distance / Time
= (2 � 2200) / (7 � 40)
= 4400 / 7 � 40
Therefore, Distance covered in 140 s = Speed (s) � Time(t)
= 4400 / (7 � 40) � (2 � 60 + 20)
= 4400 / (7 � 40) � 140
= 4400 � 140 /7 � 40
= 2200 m

Number of round in 40 s =1 round
Number of round in 140 s =140/40
=3 1/2
After taking start from position X,the athlete will be at postion Y after 3 1/2 rounds as shown in figure

Hence, Displacement of the athlete with respect to initial position at x= xy
= Diameter of circular track
= 200 m

Total Distance covered from AB = 300 m
Total time taken = 2 × 60 + 30 s
=150 s

Therefore, Average Speed from AB = Total Distance / Total Time
=300 / 150 m s^-1
=2 m s^-1
Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s^-1
=2 m s^-1
Total Distance covered from AC =AB + BC
=300 + 200 m

Total time taken from A to C = Time taken for AB + Time taken for BC
= (2 × 60+30)+60 s
= 210 s
Therefore, Average Speed from AC = Total Distance /Total Time
= 400 /210 m s^-1
= 1.904 m s^-1

Displacement (S) from A to C = AB - BC
= 300-100 m
= 200 m

Time (t) taken for displacement from AC = 210 s

Therefore, Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m s^-1
= 0.952 m s^-1

The distance Abdul commutes while driving from Home to School = S
Let us assume time taken by Abdul to commutes this distance = t₁
Distance Abdul commutes while driving from School to Home = S
Let us assume time taken by Abdul to commutes this distance = t₂
Average speed from home to school v_1av = 20 km h^-1
Average speed from school to home v_2av = 30 km h^-1
Also we know Time taken form Home to School t₁ =S / v_1av
Similarly Time taken form School to Home t₂ =S/v_2av
Total distance from home to school and backward = 2 S
Total time taken from home to school and backward (T) = S/20+ S/30
Therefore, Average speed (Vav) for covering total distance (2S) = Total Distance/Total Time
= 2S / (S/20 +S/30)
= 2S / [(30S+20S)/600]
= 1200S / 50S
= 24 kmh^-1

Given Initial velocity of motorboat,  u = 0
Acceleration of motorboat,  a = 3.0 m s^-2
Time under consideration,  t = 8.0 s
We know that Distance, s = ut + (1/2)at²
Therefore, The distance travel by motorboat = 0 ×8 + (1/2)3.0 × 8²
= (1/2) × 3 × 8 × 8 m
= 96 m

As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh−1 and 3 kmh−1 respectively.

Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) � OR � OP
= (1/2) � 5s � 52 kmh⁻¹
= (1/2) � 5 � (52 � 1000) / 3600) m
= (1/2) � 5 � (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) � OQ � OS
= (1/2) � 10 s � 3 kmh⁻¹
= (1/2) � 10 � (3 � 1000) / 3600) m
= (1/2) � 10 x (5/6) m
= 5 � (5/6) m
= 25/6 m
= 4.16 m

Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d)How far has B travelled by the time it passes C?

Answer

(a) Object B
(b) No
(c) 5.714 km
(d) 5.143 km

Therefore, Speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.

(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.

(c)

7 square box = 4 km
∴ 1 square box = 4/7 km
C is 4 blocks away from origin therefore initial distance of C from origin = 16/7 km
Distance of C from origin when B passes A = 8 km
Thus, Distance travelled by C when B passes A = 8 - 16/7 = (56 - 16)/7 = 40/7 = 5.714 km

(d)

Distance travelled by B by the time it passes C = 9 square boxes
9�4/7 = 36/7 = 5.143 km

Let us assume, the final velocity with which ball will strike the ground be 'v' and time it takes to strike the ground be 't'
Initial Velocity of ball, u =0
Distance or height of fall, s =20 m
Downward acceleration, a =10 m s^-2
As we know, 2as =v²-u²
v² = 2as+ u²
= 2 x 10 x 20 + 0
= 400
∴ Final velocity of ball, v = 20 ms^-1
t = (v-u)/a
∴Time taken by the ball to strike = (20-0)/10
= 20/10
= 2 seconds

The speed-time graph for a car is shown is Fig. 8.12.


(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?

Answer

(a)

The shaded area which is equal to 1/2 � 4 � 6 = 12 m represents the distance travelled by the car in the first 4 s.

(b)

The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.

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.

Answer

(a) Possible
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.

(b) Possible
When a car is moving in a circular track, its acceleration is perpendicular to its direction.

Radius of the circular orbit, r = 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v = (2π r) / t
= [2× (22/7)×42250 × 1000] / (24 × 60 × 60)
= (2×22×42250×1000) / (7 ×24 × 60 × 60) m s^-1
= 3073.74 m s^-1

1. A particle is moving in a circular path of radius r. The displacement after half a
circle would be:
(a) Zero
(b) πr
(c) 2r
(d) 2πr
Ans. (c) 2r
Explanation: After half a circle, the particle will be diametrically opposite to its origin.
Hence, displacement is equal to diameter.

2. A body is thrown vertically upward with velocity u, the greatest height h to which it
will rise is,
(a) u/g
(b) u2/2g
(c) u2/g
(d) u/2g
Ans. (b) u2/2g
3. The numerical ratio of displacement to distance for a moving object is
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1
Ans. (d) equal or less than 1
Explanation: Displacement can be equal or less than distance but it will never be more than distance.

4. If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration
Ans. (b) uniform acceleration
Explanation: Velocity is measured in terms of distance per second, while acceleration is measured in terms of distance per square second.
Hence, uniform acceleration is the correct answer.

5. From the given v � t graph (Fig. 8.1), it can be inferred that the object is


(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration
Ans. (a) in uniform motion
Explanation: Straight horizontal line which is above zero on x-axis; shows uniform motion.

6. Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms^�1. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

Ans. (c) in accelerated motion
Explanation: A boy on a merry-go-around is in circular motion. Circular motion is an example of acceleration motion.

7. Area under a  v � t graph represents a physical quantity which has the unit
(a) m2
(b) m
(c) m3
(d) ms�1
Ans. (b) m
Explanation: The given area represents displacement. Unit of displacement is meter.

8. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in Fig. 8.2. Choose the correct statement
(a) Car A is faster than car D.
(b) Car B is the slowest.
(c) Car D is faster than car C.
(d) Car C is the slowest.
Ans. (b) Car B is the slowest.
Explanation: Graph for car B attains least height (on x-axis) in the given time. Hence, it is the slowest car

9. Which of the following figures (Fig. 8.3) represents uniform motion of a moving object correctly?






Ans. (a)
Explanation: Distance is increasing at a uniform rate; as shown by upward slanting
straight line. Hence, this graph shows uniform motion.
10. Slope of a velocity � time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed
Ans. (c) the acceleration

11. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun
Ans. (a) If the car is moving on straight road
Explanation: In all other class, displacement can be less than distance.

When the displacement is zero, it does not mean that distance is also zero. Displacement can be zero when the moving object comes back to its original position. Displacement is either equal to or less than distance but distance travelled is always more than zero.

When the object is moving with a uniform velocity, then ν  = � and a = 0. In this situation, equation for distance would be as follows:

s = ut and ν ² - � ² = 0

Ans. For the initial 50 second, velocity is 2 m/s. After that, velocity drops of zero; as shown by vertical line in graph.


For the next 50 second, velocity is taken in negative because displacement is becoming zero.

The distance travelled in first 8 s, x₁ = 0 + 1/2 (5)(8)² =160 m
At this point the velocity v = u + at= 0 + (5�8) = 40 m s^�1
Therefore, the distance covered in last four seconds, x₂ = (40�4) m = 160 m
Thus, the total distance  x = x₁  + x₂   = + = (160 + 160) m = 320 m

Let AB = x, So  t₁ = x /30 and t₂ = x/20

Total Time = t₁ + t₂ = 5x / 60 hr 

Average speed for entire journey = Total distance / Total Time

2x /5x /60 = 24 km hr^-1

Ans. (i) Since velocity is not changing, acceleration is equal to zero.
(ii) Reading the graph, velocity = 20 ms-1
(iii) Distance covered in 15 seconds, s= u � t= 20 � 15 = 300 m

When a stone is thrown upward, its velocity is at maximum. The velocity begins to drops as the stone attains height. Once the stone attains the maximum height, velocity becomes zero. After that, stone begins to fall down. At this points, velocity beings to rise. Velocity is at its maximum when the stone hits the ground.

Initial difference in height = (150�100) m = 50 m

After 2s, height at which the first body will be = h_1�=150 � 2g

After 2s, height at which the second body will be = h_2� = 100 � 2g

Thus, after 2s, difference in height = 150 � 2 g � (100 � 2g) = 50 m = initial difference in height

Thus, difference in height does not vary with time.

Since acceleration is the same, we have 1 v ms^-1

Using following data, draw time-displacement graph for a moving object:

Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s. 

Average velocity for first 4s.

(or as x remains the same from 4 to 8 seconds, velocity is zero)

Given initial velocity, u = 5 � 10⁴ m/s

and acceleration, a = 10⁴ m s^�2

(i) final velocity = v = 2 u = 2 � 5 �10 ⁴  m s^�1

=10 � 10⁴ m s^�1

Using the equation of motion 



Distance travelled in the interval between 4th and 5th second