
01.Matter in Our Surroundings
11
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11


02.Is Matter Around Us Pure
11
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11


03.Atoms and Molecules
7
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7


04.Structure of The Atom
7
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7


05.Cell  Fundamental Unit of Life
11
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11


06.Tissues
8
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8


07.Diversity in Living Organisms
8
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8


08.Motion
11
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11


09.Force and Newtons Laws of Motion
12
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11

Lecture9.12


10.Gravitation
9
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9


11.Work and Energy
7
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7


12.Sound
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Why do We Fall Ill
7
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7


14.Natural Resources
11
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11


15.Improvements in Food Resources
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Motion
Intext Questions
Q.1 An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Sol. Yes, zero displacement is possible if an object has moved through a distance.
Suppose a ball starts moving from point A and it returns back at same point A, then the distance will be equal to 20 meters while displacement will be zero.
Q.2 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?
Sol.
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 mTherefore, in 1s distance covered by farmer = 4040m=1m
Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 m
Now, number of rotation to cover 140 along the boundry = TotaldistancePerimeter
=140m40m=3.5round
Thus after 3.5 round farmer will at point C of the field.
Therefore, Displacment AC=(10m)2+(10m)2−−−−−−−−−−−−−−√
=100m2+100m2−−−−−−−−−−−−−√
=200m2−−−−−√
=2×100m2−−−−−−−−−√
=102–√m Thus, after 2 minute 20 second the displacement of farmer will be equal to 102–√m north east from initial position.
Q.3 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.
Sol. None
Page 102
Q.1 Distinguish between speed and velocity.
Sol. Speed has only magnitude while velocity has both magnitude and direction.
Q.2 Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Sol. The magnitude of average velocity of an object will be equal to its average speed in the condition of uniform velocity.
Q.3 What does the odometer of an automobile measure?
Sol. In automobiles, odometer is used to measure the distance.
Q.4 What does the path of an object look like when it is in uniform motion?
Sol. In the case of uniform motion the path of an object will look like a straight line.
Q.5 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, 3×108ms−1.
Sol. Here we have, speed = 3×108ms−1
Time = 5 minute = 5×60s=300 second
We know that, Distance = Speed × Time
⇒ Distance = 3×108ms−1×300s=1800×108m=1.8×1011m
Page 103
Q.1 When will you say a body is in
(i) uniform acceleration?
(ii) nonuniform acceleration?
Sol. (i) A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.
(ii) A body is said in nonuniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.
Q.2 A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
Sol. Here we have, u = 80 km/h, v = 60km/h, t = 5s
Therefore, acceleration, a = ?
We know that, v = u + at
⇒ 60 km/ h = 80km/h + a × 5s
⇒ 60km/h – 80km/h = a × 5s
⇒ – 20km/h = a × 5s
⇒ a = −20km/h5s
⇒ a = – 4km/h/s
Therefore, Acceleration = –4 km/h/s or, –1.1 m/s2
Q. 3 A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.
Sol. Here we have,
Initial velocity, u = 0,
Final velocity, v = 40km/h = 11.11m/s
Time (t) = 10 minute = 60 x 10=600s
Acceleration (a) =?
We know that, v = u + at
⇒ 40km/h = 0km/h + a × 10m
⇒ 11.11 m/s = a × 600s
⇒ a = 11.11m/s600s=0.0185m/s2
Page 107
Q.1 What is the nature of the distancetime graphs for uniform and nonuniform motion of an object?
Sol. (a) The slope of the distancetime graph for an object in uniform motion is straight line.
(b) The slope of the distancetime graph for an object in nonuniform motion is not a straight line.
Q.2 What can you say about the motion of an object whose distancetime graph is a straight line parallel to the time axis?
Sol. When the slope of distancetime graph is a straight line parallel to time axis, the object is moving with uniform motion.
Q.3 What can you say about the motion of an object if its speedtime graph is a straight line parallel to the time axis?
Sol. When the slope of a speed time graph is a straight line parallel to the time axis, the object is moving with uniform speed.
Q.4 What is the quantity which is measured by the area occupied below the velocitytime graph?
Sol. The quantity of distance is measured by the area occupied below the velocity time graph.
Page 109
Q.1 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 travelled.
Sol. Here we have,
Initial velocity (u) = 0
Acceleration (a) = 0.1ms2
Time (t) = 2 minute = 120 second
(a) The speed acquired:
We know that, v = u + at
⇒ v = 0 + 0.1m/s2 x 120 s
⇒ v = 120 m/s
Thus, the bus will acquire a speed of 120 m/s after 2 minute with the given acceleration.
(b) The distance travelled:
We know that, s=ut+12at2
=0×120s+12×0.1m/s2×(120s)2
=12×14400m=7200mor7.2km
Thus, bus will tavel a distance of 7200 m or 7.2 km in the given time of 2 minute.
Page 110
Q.2 A train is travelling at a speed of 90 km/h. 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.
Sol. Here,we have,
Initial velocity, u = 90km/h
=90×1000m60×60s=25m/s
Final velocity, v = 0
Acceleration,a = – 0.5 m/s2
Therefore, distance travelled = ?
We know tha, v2=u2+2as
⇒0=(25m/s)2+2×−0.5m/s2×s
⇒0=625m2s−2−1ms−2s
⇒1ms−2s=625m2s−2 s=625m2s−21ms−2=625m
Therefore, train will go 625 m be fore it brought to rest.
Q.3 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?
Sol. Here we have,
Initial velocity, u = 0
Acceleration (a) = 2cm/s^{2} = 0.02m/s^{2
} Time (t) = 3s
Therefore, Final velocity, v = ?
We know that, v = u + at
Therefore, v = 0 + 0.02m/s2× 3s
⇒ v = 0.06 m/s
Therefore the final velocity of trolley will be 0.06 m/s after start.
Q.4 A racing car has a uniform acceleration of 4 m s^{2}. What distance will it cover in 10 s after start?
Sol. Here we have,
Acceleration, a = 4m/s2
Initial velocity, u = 0
Time, t = 10s
Therefore, Distance (s) covered =?
We know that, s=ut+12at2
⇒s=0×10s+12×4m/s2×(10s)2
⇒s=12×4m/s2×100s2
⇒s=2×100m=200m
Thus, racing car will cover a distance of 200m after start in 10 s with given acceleration.
Q. 5 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?
Sol. Here we have,
Initial velocity (u) = 5m/s
Final velocity (v) = 0 (Since from where stone starts falling its velocity will become zero)
Acceleration (a) = – 10m/s^{2
} (Since given acceleration is in downward direction, i.e. the velocity of the stone is decreasing, thus acceleration is taken as negative)
Height, i.e. Distance, s =?
Time (t) taken to reach the height =?
We know that, v2=u2+2as
⇒0=(5m/s)2+2×−10m/s2×s
⇒0=25m2s2−20m/s2×s
⇒20m/s2×s=25m2s2 s=25m2s220m/s2
⇒s=1.25m
Now, we know that , v = u + at
⇒0=5ms−1−(10ms−2)×t
⇒0=5ms−1−10ms−2×t
⇒10ms−2×t=5ms−1
⇒t=5ms−110ms−2=12s=0.5s Thus, stone will attain a height of 1.25m And time taken to attain the height is 0.5s
Exercise
Q.1 An athlete completes one round of a circular track of diameter 200 m in 40s. What will be the distance and displacement at the end of 2 minutes 20 s ?
(i) Distance covered in 40 s
=2×π×100
(r=d2=200m2=100m)
Distance covered in 1 s
=200π40
Distance covered in 2 min 20s or 140 s
=200π40×140
=5π×140
=700π
=700×3.14
=2198m
(ii) Displacement in 2 min 20 s = 0 m (initial and final position is the same).
Q.2 Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 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 ?
(a) From A to B
Distance = 300 m
Time = 2 min 50 s = 170 s
verage speed = Total distance / Total time
=300m170s=1.76ms−1
Average velocity = Displacement / Total time
=300m170s=1.76ms−1
(b) From A to C
Distance = (300 + 100)m = 400 m
Time = (170 + 60)s = 230 s
Average speed = Total distance / Total time
=400m230s=1.73ms−1
Average velocity = Displacement / Total time
=200m230s
=0.86ms−1
Q.3 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 40 kmh ^{–1}. What is the average speed for Abdul’s trip ?
Let the distance between starting point and school be x km.
Average speed from the starting point to school
= 20 km h^{–1}
Time for onward journey =x20h
Average speed from the school to the starting point
= 40 km h^{–1}
Time for return journey =x40h
Average speed for total trip
= Total distance / Total time
=2xx20+x40
2x21x40
=2x×4021x=8021=3.8km/h
Q.4 A motor boat starting form 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 ?
u = 0
A = 3ms^{–2}
T = 8s
According to second equation of motion,
s=ut+12at2
s=0×8+12×3×(8)2
s=96m
Q.5 A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 s^{–2}, with what velocity will it strike the ground ? After what time will it strike the ground
s = 20 m
u = 0
a = 10 ms2 (since velocity increases at a uniform rate)
By second equation of motion
s=ut+12at2
s=0+12×10×t2
⇒20=5t2
⇒205=t2
ort=4–√=2s
By first equation of motion,
v = u + at
⇒ v = 0 + 10 × 2
⇒ v = 20 ms^{–1}
Q.6 How do movements in the living and nonliving objects differ ?
The movements in living organisms appear to take place on their own. Nonliving bodies need some outside agency to set them into motion. For example, to run a car we need the chemical energy of petrol / diesel.
Q.7 What is motion ? Give some examples of motion in daily life.
(i) A body is said to be in motion if it changes its position in relation to a reference (fixed) point.
(ii) Some examples of motion in daily life are :
• Cars moving on the roads.
• Aeroplanes flying in the air.
• Blades of a fan going round.
• Different parts of machines moving in different ways.
• Fishes swimming in water.
Q.8 What is meant by the term distance?
The actual of the path covered by the body irrespective of the direction is called the distance. Distance is a scalar quantity
Q.9 What is displacement ?
Change in the position of an object in a particular direction is called displacement. If the position of an object changes from x_{1} to x_{2}, then the displacement is given by
d = (x_{2} – x_{1})
Displacement is the shortest distance (straight distance) between the initial position and the final position of a moving body.
Displacement is a vector quantity, and it is the distance travelled by a body in a particular direction.
Q.10 When do we say that the position of a body has changed?
If the distance, or direction, or both, of a body relative to a reference point changes, then we say that the position of the body has changed.
Exemplar
Multiple Choice Questions
Q.1 A particle is moving in a circular path of radius r. The displacement after half a circle would be:
(a) Zero
(b) π r
(c) 2 r
(d) 2π r
Sol. (c)
Q.2 A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) u/g
(b) u^{2}/2g
(c) u^{2}/g
(d) u/2g
Sol. (b)
Q.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
Sol. (d)
Q.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
Sol. (b)
Q.5 From the given v – t graph figure, it can be inferred that the object is
(a) In uniform motion
(b) At rest
(c) In nonuniform motion
(d) Moving with uniform acceleration
Sol. (a)
Q.6 Suppose a boy is enjoying a ride on a merrygoround which is moving with a constant speed of 10 m s^{–1}. It implies that the boy is
(a) At rest
(b) Moving with no acceleration
(c) In accelerated motion
(d) Moving with uniform velocity
Sol. (c)
Q.7 Area under a v – t graph represents a physical quantity which has the unit
(a) m^{2}
(b) m
(c) m^{3}
(d) m s^{–1}
Sol. (b)
Q.8 Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in Figure.
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.
Sol. (b)
Q.9 Which of the following figures represents uniform motion of a moving object correctly?
Sol. (c)
Q.10 Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed
Sol. (a)