DISTANCE AND DISPLACEMENT · Distance : The distance covered by a moving object is the actual length of the path followed by the object. Distance is a scalar quantity. SI unit of

MOTION AND REST NOTES

A body is said to be in the state of rest when it remains in the same position with respect to time.

e.g.: The position of trees around the building is not changing with respect to the building. Then the trees are at rest.

A body is said to be in the state of motion, when it continuously changes its position with respect to time

e.g.: When we are cycling or running, we are changing position with respect to trees and buildings. we are said to be moving

DISTANCE AND DISPLACEMENT

Distance is the length of the path covered by a body. It has no sense of direction so it is a scalar quantity. Its SI unit of measurement is meter or m

The distance between the two places is not the same; it depends upon the path chosen.

Displacement: The shortest distance, or distance travelled along a straight line, is known as displacement.

If we draw a semicircle of radius 10cm. Then,

Displacement = r + r = 20cm

Distance = 2πr = 10 x 22/7 cm = 31.4cm

UNIFORM MOTION AND NON UNIFORM MOTION

If an object covers equal distances in equal intervals of time, it is said to be in

uniform motion.

If an object covers unequal distance in equal intervals of time, it is said to be in

non-uniform motion.

MEASURING THE RATE OF MOTION

Speed :It is the distance travelled in one second (or) rate of distance travelled.

A car starts from Samastipur and reaches Patna in 6 hour. A bus takes 8 hour to travel the same distance. Which has moved faster? Why?

The car travels faster than the bus, because it covers the distance in a short time.

When a body covers a distance in a short time, it is said to be fast. If it takes more time to cover the distance, it is said to be slow.

Hence, Speed is the quantity used to say whether the motion is slow or fast.

Speed = Total Distance travelledt Time taken

Speed has no sense of the direction of motion so it is a scalar quantity

SI Units: Speed is measured in mts (or) mts It can also be expressed in kmthour (or) kmh-1

Velocity : The speed with direction is known as velocity.

So to measure the velocity, you should consider displacement instead of distance.

Velocity is the displacement made in one second (or) rate of change of displacement.

Rate of change means, change per second.

Velocity = DisplacementtTime

It S I Unit is also mts

Uniform Velocity: Equal displacement covered by a body in equal intervals of time is known as uniform velocity.

Acceleration is the change in velocity of an object per second or rate of change of velocity.

Acceleration = Change in velocitytTime taken. The unit of acceleration is mts2 or ms-2

There are two cases of of change of velocity

If the velocity of the body increases with time, the acceleration is positive, and the kind of motion is called accelerated motion.

If the velocity of the body decreases with time, the acceleration is negative (retardation), and the motion is called decelerated motion.

Uniform Acceleration: If an object travels in a straight line and its velocity increases or decreases by equal amount in equal intervals of time, then the acceleration of the object

is uniform.

We can derive a formula for acceleration.

a= [v – u] tt

Where u - initial velocity, v - final velocity t – time

Distance - time graph

We can easily understand the relation between time and distance by using a graph.

Taking a suitable scale, a graph is drawn bytaking time along the x axis and distance along the y axis. The graph is known as distance – time graph.

In the graph for uniform speed is a straight line and curved for Non uniform speed.

9th Notes Motion and Rest

In the physical world, one of the most common phenomena is motion. The branch of Physics, which deals with the behavior of moving objects, is known as mechanics.

Mechanics is further divided into two sections namely Kinematics and Dynamics.

Kinematics deals with the study of motion without taking into account the cause of motion.

Dynamics is concerned with the cause of motion, namely force.

☼ An object is said to be in motion if it changes its position with respect to its surroundings in a given time

An object is said to be at rest if it does not change its position with respect to its surroundings.

☼ We have observed that the position of stars and planets change while you remain stationary. In reality the earth is moving too. Thus, an object which appears to be at rest, may actually be in motion. Therefore, motion and rest are relative terms .

There are three types of motion:☼ Translatory motion ☼ Rotatory motion ☼Vibratory motion

TranslatoryMotion : - In translatory motion the particle moves from one point in space to another. This motion may be along a straight line or along a curved path.

Rectilinear motion : Motion along a straight line is called rectilinear motion. Example: A car moving on a straight road

Curvilinear motion : Motion along a curved path is called curvilinear motion.

Example: A car negotiating a curve

Rotatory Motion : In rotatory motion, the particles of the body describe concentric circles about the axis of motion.

Vibratory Motion : In vibratory motion the particles move to and fro about a fixed point. Example Simple Pendulum

Distance : The distance covered by a moving object is the actual length of the path followed by the object. Distance is a scalar quantity. SI unit of distance is Metre .

Displacement is the shortest distance covered by a moving object from the point of reference (initial position of the body), in a specified direction.

Note: But the displacement when the bus moves from A B and then from B A is zero. SI unit of displacement is metre. Displacement is a vector, i.e., the displacement is given by a number with proper units and direction.

☼ When a body covers equal distances in equal intervals of time then the body is said to describe uniform motion.

☼ When a body moves unequal distances in equal intervals of time or vice-versa, then the body is said to describe non-uniform motion.

☼ Speed can be defined as the distance covered by a moving object in unit time

Speed = distance t time = stt where S is the distance covered and t is the time taken. SI unit of speed is mts or m s-1. Speed is a scalar quantity.

Uniform Speed : An object is said to be moving with uniform speed if it covers equal distances in equal intervals of time.

Non-uniform : An object is said to be moving with variable speed or non-uniform speed if it covers equal distances in unequal intervals of time or vice-versa.

Average speed : When we travel in a vehicle the speed of the vehicle changes from time to time depending upon the conditions existing on the road. In such a situation, the speed is calculated by taking the ratio of the total distance travelled by the vehicle to the total time taken for the journey. This is called the average speed.

Instantaneous speed : When we say that the car travels at an average speed of 60 kmth it does not mean that the car would be moving with the speed of 60 kmth throughout the journey. The actual speed of the car may be less than or greater than the average speed at a particular instant of time.

The speed of a moving body at any particular instant of time, is called instantaneous speed.

Velocity is defined as the distance travelled in a specified direction in unit time. The distance travelled in a specified direction is displacement.

Therefore, velocity can be defined as the rate of change of displacement .

Acceleration : When the train starts from rest its speed increases from zero and we say that the train is accelerating. After sometime the speed becomes uniform and we say that it is moving with uniform speed that means the train is not accelerating. But as the train is nearing Mysore it slows down, which means the train is accelerating in negative direction. Again the train stops accelerating when it comes to a halt at Mysore .

☼ Acceleration is a vector quantity

Positive Acceleration If the velocity of an object increases then the object is said to be moving with positive acceleration. Example: A ball rolling down on an inclined plane.

Negative Acceleration : If the velocity of an object decreases then the object is said to be moving with negative acceleration. Negative acceleration is also known as retardation or deceleration. Example: (1) A ball moving up an inclined plane. 2) A ball thrown vertically upwards is moving with a negative acceleration as the velocity decreases with time

Zero Acceleration : If the change in velocity is zero, i.e., either the object is at rest or moving with uniform velocity, then the object is said to have zero acceleration. Example: a parked car, a train moving with a constant speed of 90 kmthr

Uniform Acceleration : If the change in velocity in equal intervals of time is always the same, then the object is said to be moving with uniform acceleration. Example: a body falling from a height towards the surface of the earth.

Non-uniform or Variable Acceleration : If the change in velocity in equal intervals of time is not the same, then the object is said to be moving with variable acceleration.

Uniform velocity : A body is said to be moving with uniform velocity if it covers equal distances in equal intervals of time in a specified direction.

Variable velocity A body is said to be moving with variable velocity if it covers unequal distances in equal intervals of time and vice-versa in a specified direction or if it changes the direction of motion.

Circular motion : Motion along circular track is called circular motion .

An object moving along a circular track with uniform speed is an example for a non - uniform motion because the direction of motion of the object goes on changing at every instant of time. Example - A car negotiating a curve with uniform speed

A circle can be considered as a polygon with infinite sides and hence motion along a circular path is classified as non-uniform motion.

Describe the terms ‘Rest’ and ‘Motion’.

Rest - A body is said to be at rest if it does not change its position with respect to its surroundings. For example, a table lying in a room is at rest with respect to the walls of the room.

Motion -A body is said to be in motion if it changes its position with respect to its surroundings. For example, a car running on a road is in motion with respect to trees on roadside.

Describe the various types of motions observed in bodies.

Translatory motion - When a body moves as a whole along a straight or curved path, it is said to be in translatory motion. Translatory motion is of two types:

(a) Rectilinear motion: Here a body moves as a whole along a straight path. For example, a train moving on straight rails has translatory rectilinear motion.

(b) Curvilinear motion: Here a body moves as a whole along a curved path. For example, a bicycle taking a turn along a curved path.

Rotatory t Rotational motion - When a body rotates about a fixed point or axis, it has a rotatory motion. For example, motion of flywheel about a shaft.

10m

10 m

Vibratory or Oscillatory motion - When a body moves to and fro about a mean position again and again, it has vibratory or oscillatory motion. For example, the motion of the pendulum of a wall-clock.

Complex motion - Sometimes, the motion of a body may be a combination of more than one type of motion. For example, a ball rolling down an inclined plane has both translatory and rotatory motions.

What are scalar quantities? Give examples.

Ans: The physical quantities which require only magnitude and not the direction for their complete description are called scalars or scalar quantities. Distance, speed, time, area, mass, volume, density, work, energy etc are all scalar quantities.

What are vector quantities? Give examples.

Ans: The physical quantities which need both magnitude and direction for their complete description are called vectors or vector quantities. Displacement, velocity, force, acceleration, momentum, weight etc. are all vector quantities.

Ql. An object has moved through a distance. Can it have zero displacement ? If yes, support your answer with an example.

Ans. Yes, it is possible that an object has moved through a distance but has zero displacement. A student goes to school from his home and then return back to his home. Here, he has covered a finite distance but the displacement is zero.

A minute hand of a clock completes, one complete revolution in 1 hour. During this time, it has covered finite distance but displacement is 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 min 20 s ?

Ans. As the farmer moves once along the boundary of field in 40 s, hence he will complete n revolutions in time t = 2 min 20 s = 140 s, whereIt means that if farmer had started from point A, then aftercompleting 3 — revolutions he will reach at diagonally opposite point 2

C and magnitude of his displacement

= AD2 + DC2=(10)2 +(10)2 =10√2 m

Note : The answer given above is not unique. Magnitude of displacement also depends upon the initial position of the farmer.

Q.3. Which of the following is true for displacement ?

It cannot be zero.

Its magnitude is greater than the distance travelled by the object. Ans. (a) Wrong, the displacement can be zero.

Wrong, magnitude of displacement can never be greater than the distance travelled by the object. Magnitude of displacement can either be equal to or less than the distance travelled by the object.

Q.1. Distinguish between speed and velocity.

Ans. Difference between speed and velocity.

Q.3. What does the odometer of an automobile measure ?

Ans. Odometer measures the total distance covered by the automobile.

Q.4. What does the path of an object look like when it is in uniform motion ?

Ans. In a uniform motion, the path of an object is 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 x 108 m s-1.

page number 103 of the Textbook.

Q.l. When will you say a body is in (t) uniform acceleration ? (if) non-uniform acceleration ? Ans. (a) If velocity of a body changes by equals amounts in equal intervals of time, howsoever small or large these intervals may be, the acceleration is said to be uniform.

(b) If velocity changes by unequal amounts in equal intervals of time or if the direction of motion changes then the acceleration is said to be non-uniform.

Q.2. A bus decreases its speed from 80 km/ h to 60 km/ h in 5 s. Find the acceleration of the bus.

Speed Veloc i ty

1. It is the distance covered per unit time.

2. Speed has only magnitude i . e . , it is a scalar. 3. Speed is always positive.

1. It is the displacement covered per unit time.

2. Velocity has both magnitude as well as

direction i . e . , it is a vector.

3. Velocity can be positive or negative.

Q.2. Under what condition is the magnitude of average velocity of an object equal to its average

speed ?

Ans. Magnitude of average velocity of an object will be equal to its average speed when motion is

along a given line and direction of motion throughout remains same.

Q.3. 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.

|page number 107 of the Textbook.

Q.l. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object ?

Ans. For uniform motion of an object, the distance-time graph is a straight line inclined to the timeaxis.For non-uniform motion of an object the distance-time graph is a curve, which may have any shape other than a straight line.

Q.2. What can you say about the motion of an object whose distance time-graph is a straight line parallel to the time axis ?

Ans. The object is at rest.

Q.3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis Ans. The motion of the object is uniform linear motion.

Q.4. What is the quantity which is measured by the area occupied below the velocity-time graph ?

Ans. Area under the velocity-time graph gives the magnitude of displacement covered by the object in that time.

page number 109-110 of the Textbook.

Q.l. A bus starting from rest moves with a uniform acceleration of 0.1 m /s2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Ans. Here w = 0, a - 0.1 m s~2and time t = 2 minutes = 2 x 60 s = 120 s

The final speed acquired v = u + at = 0 + 0.1 x 120 = 12 m s_1

The total distance travelled in given time

5 = ut + -at2 = 0 x 120 + - x 0.1 x (120)2 2 2

= 0 + 720 = 720 m

Q.2. A train is travelling at a speed of 90 km IT1. 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.

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 ?

Fig.

8.02

Ans. Here initial velocity u = 0, acceleration a = 2 cm s-2 = 0.02 m s~2, and time t = 3 s .•. Final velocity of the trolley v=u + at = Q + (0.02) x 3 = 0.06 m s-' or 6 cm s-1

Q.4. A racing car has a uniform acceleration of 4 m s"2. What distance will it cover in 10 s afterstart ?

Q.5. A stone is thrown in a vertically upward direction with a velocity of 5 ni s If the acceleration of the stone during its motion is 10 in s 2 in the downward direction, what will be the height attained by the stone and how much time will it tsike to reach there ?

TEXTBOOK EXERCISES

Q.l. 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. Here diameter of circular track D = 200 m. As the athlete completes one round of track in 40 s, hence in time t = 1 minutes 20 s = 140 s, the number of rounds completed by athlete

(a) As in 1 round athlete covers distance equal to circumference of circle i.e., 2nR = , hence in 2 minutes 20 s, the distance covered will be

(b) if athlete starts running from point A then after completing 3 rounds, he will be at point B, diagonally opposite end of diameter AOB. Hence, his displacement

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 ?

Q.3. Abdul, while driving to school, computes the average speed for his trip to be 20 km IT1. On his return trip along the sameroute, there is less traffic and the average speed is 40 km h-1. What is the average speed for Abdul's trip ?

Q.4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of n s"2 for 8.0 s. How far does the boat travel during this time ?

Q.5. A driver of a car travelling at 72 km ir1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 36 km IT1 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 travelled farther after the brakes were applied ?

Thus, both cars stopped after covering equal distances when brakes are applied. The

speed-versus time graph for.

[There appears to be some printing error in Q. No. 5 in NCERT Textbook. We have, therefore, suitably corrected it.]

Q.6. Fig.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 ? (t)) Are all three ever at the same point on the road ? (c) How far has C travelled when B passes A ? ((I) How far has B travelled by the time it passes CI Ans. (a) The object B is travelling the fastest because slope of its distance-time graph is maximum. (6) All three are not at the same point on the road at any instant.

Q.7. 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 ?

Q.8. The speed-time graph for a car is shown in

Ans. (a) From graph we find that the car has travelled nearly 15 m in first 4 seconds. The shaded area representing the distance travelled by the car during the period has been shown in . (b) The graph after ti

4 6

time

((s)

Fig. 8.04

B is passing ,4 at a time t = 1.1 h and at that time, C was at a position 8 km.

As at start position of C was at 2 km, hence, distance travelled by C = 8 km -

2 km = 6 km

B is passing C at a time 0.6 s and at that time B had covered 0.5 km.

me t = 6 s represents uniform motion of car.

.

Q.9. State which of the following situations are possible and give an example for each of these :

an object with a constant acceleration but with zero velocity.

an object moving in a certain direction with an acceleration in the perpendicular direction.

Ans. (a) Yes possible. Consider a ball thrown upwards. The ball goes up to a maximum height h andthen falls back on ground. At the highest point, the velocity of ball is zero but a constant acceleration (a = g = 9.8 m s"2 downwards) is acting on it.

(b) -Yes possible. A particle in uniform circular motion has its speed along the tangent drawn at a point on circle and acceleration is along radius and the two are mutually perpendicular to each other.

.'. speed of satellite v =2πr/ t

Q.10. An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Ans. Here radius of orbit r = 42,250 km and revolution time t = 24 h = 24 x 60 x 60 s = 86400 s circumference of orbit 2ur time t

Very Short Answer Type Questions (One Mark Each)

Q.l.When is a body in a state of motion ?

AnsWhen its position changes with time.

Q.2.Give an example where motion is inferred through indirect means.

AnsMotion of air is inferred by observing movement of leaves and branches of trees.

Q.3.Are the walls of your classroom at rest or in motion ?

AnsThe walls of our classroom are at rest with respect to ground.

Q.4.Name different types of motion, which are possible in nature.

Ans(t) Translational motion, (tt) rotational motion, (Hi) vibrational motion.

Q.5.Define displacement.

AnsSee point number 7 under the heading "Chapter At A Glance".

Q.6. Does displacement of a particle depend on the actual path along which the particle moves between two points ?

Ans. No, it does not depend on the actual path followed.

Q.7. Under what condition will the distance covered and displacement of moving object will have the same magnitude ?

Ans.When the object moves along a straight line along a given direction.

Q.8. Does displacement of a particle in a given interval of time depend upon the choice of origin ?

Ans. No, it does not depend on the choice of origin (or the reference point).

Q.9. Smita went from her home to school 2.5 km away. On finding her school closed, she returned to her home immediately. What is her net displacement ? What is the distance covered by her ?

Ans.- Total displacement = 0Total distance covered = 2.5 km + 2.5 km = 5 km

Q.10. A student considered earth as a point object while considering its motion around the sun. Is he justified ?

Ans. Yes, because size of earth is very- very small as compared to its distance from the sun.

Q.ll. Is absolute position of rest or absolute motion ever possible ?

Ans. No, it is never possible because both rest and motion are relative terms.

Q.12. Is it possible that the train in which you are sitting appears to move while it is at rest ?

Ans. Yes. if a railway train on adjacent line is running slowly then we feel that our train is moving in the opposite direction of that train, although our train is at rest.

Q.13. Samir takes 15 minutes to walk from his house to his school. If his average walking speed is 4 km h-1, estimate the distance of the school from his house.

Ans. Distance of school from houses = average speed x time = (4 km h~') x (15 minutes)

= 4 km h = 1 km 60

Q.14. A train is running at a constant speed of 54 km h-1- What is its speed in m s"1 ?

Ans. Speed of train = 54 km

Q.15. Give the mathematical formula for average speed of an object.total distance travelled

Ans. Average speed = total distance /total time

Q.16. An object travels 16 m in first 4 seconds and then another 16 m in 3 seconds. Is his motion uniform or non-uniform ?Why ?

Ans. Non-uniform motion because he is covering equal distances in unequal intervals of time.

Q.17. Define velocity.

Ans.

Q.18. Can speed of a particle be negative ?

Ans. No, speed of a particle is always positive.

Q.19. Which instrument measures the total distance covered by a vehicle during a given time ?

Ans. Odometer.

Q.20. Which instrument, fitted in automobiles, measures its instantaneous speed ?

Ans.The speedometer.

Q.21. What is the common unit in which speeds of automobiles are measured ?

Ans.A kilometre per hour (km h ' 1.

Q.22. What is the other name of negative acceleration.

Ans. Retardation.

Q.23. What is the real meaning of negative acceleration ?

Ans. Direction of acceleration is opposite to that of velocity.

Q.24. Define acceleration of a moving body ?

Q.25. Give the SI unit of acceleration ?

Ans.A metret/second- (m s~2).

Q.26. The velocity of an object is decreasing with passage of time. What conclusion do you draw about the acceleration of the object ?Ans. The acceleration is negative.

Q.27. Give an example of uniform motion.

Ans. Motion of the hands of a wall clock.

Q.28. Give an example of non-uniform motion.

Ans .Motion of a railway train between any two stations.

Q.29. If velocity of a body changes regularly from u to v in time t, what is the average velocity of thatbody?Ans. Q.30. Ans.

avg. velocity = u + v/ t

In Q. No. 29, what is the value of acceleration of the body ?

Acceleration = v-u /t

Q.31. What will be the position time graph of a DTC bus standing at rest at a depot ?

Ans. A straight line parallel to the time axis.

Q.32. What is the nature of the distance-time graph for an object moving uniformly along a

straightlong road?

Ans. A straight line inclined to the time axis (as well as distance axis).

Q.33. What is the shape of velocity-time graph for uniform motion along a straight line ?

Ans. A straight line parallel to the time axis.

Q.34. What is the shape of velocity-time graph for uniformly accelerated motion along a

straightpath?

Ans. A straight line inclined to time as well as velocity axes.

Q.35. What is the value of acceleration for a body at rest ?

Ans. Zero.

Q.36. Does the graph shown in Fig. above represent uniformmotion ?

Ans. No, it does not represent uniform motion.

Q.37. Does the graph shown in Fig above.represent motion along a straight line path ?

Ans. No, it does not represent motion along a straight line path.

Q.38. How can yon find magnitude of velocity from adisplacement-time graph of an object ?

Ans. The slope of displacement-time graph gives the magnitude of velocity.

Q.39. In the Fig.1 two displacement-time graphs have been drawn. Which represents a higher velocity and why ?

Ans. B represents higher velocity because slope of B is more than that of A.

Q.40. What is the shape of displacement-time graph for a non-uniform linear motron ?

Ans.A curved line.

Q.41. Which graph A or B represents higher velocity in Fig2.and why ?

Ans. A represents higher velocity because its slope is more.

Q.42. Name the physical quantity whose SI unit is m s~2.

Ans. Acceleration. —

Q.43. How can you find the value of displacement covered by a particle during a given time from its velocity-time graph ?

Ans. The area under the velocity-time graph between two points gives the magnitude of displacement in the time interval corresponding to given points.

Q.44. How can you find the v.alue of acceleration from a velocity-time graph ?

Ans.By finding slope of velocity-time graph.

Q.45. In Fig. 3, two velocity-time graphs have been shown. Which represents higher acceleration ?Why ?

Ans. Acceleration of both is same because two graphs are parallel lines and their slopes are equal.

Q.46. What is the difference between graphs A and B shown in Fig.4

Ans. Initial velocity of A is zero but initial velocity of B is non-uniform

Q.47. Is speed-time graph shown in Fig.possible ?Why ?

Ans .no, speed cant be zero.

Q.49. Give an example of uniformly accelerated motion.

Ans. Motion of a freely falling body.

Q.50. What is the shape of velocity-time graph for a non-uniformly accelerated motion ?

Ans. Any shape other than a straight line.

Q.51. State the equation of motion correlating initial velocity w, Final velocity v and total distancecovered s.

Ans. v2 - u2 = 2as, where a = uniform acceleration.

Q.52. Give three examples of circular motion.

Ans. Motion of hands of a wall clock, motion of moon around the earth, motion oHhe blades of a running electric fan etc.

Q.53. What is the uniform circular motion ?

Ans. Motion along the circumference of a circle with constant speed.

Q.54. An athlete completes one round of a circular track of radius r in time t. What is his averagespeed ?

Ans. Average speed =

Q.55. Does velocity of an object remain constant in uniform circular motion ?Ans. No, because direction of motion is continuously changing.

Q.56. Is uniform circular motion an accelerated motion ?

Ans. Yes.it is an example of accelerated motion.

Q.57. What is the direction of acceleration acting on a particle having uniform circular motion ?

Ans. Acceleration is directed perpendicular to the direction of motion.

Q.58. If a particle being revolved uniformly along a circular path is let loose at an instant, in whatdirection does it move ?

Ans. The stone moves along a straight line tangential to the circular path.

Two Marks Each

Q.l. Define motion. Give examples.

Ans. Motion is the change in position of an object with respect to its surroundings with passage of time. A train running on a railway track, a car running on a road, a boy walking in a park, a flying bird, a satellite revolving round the earth etc. are examples of motion.

Q.2. Define displacement of a particle in linear motion. Does it depend upon the origin ?

Ans. Displacement of a particle in a given time is the shortest (straight) distance measured from the initial to the final position.

Value of displacement does not depend on the choice of origin (or the reference point).

QJ. Define distance covered by a particle in a given time. Is it equal to displacement ?

Ans. The actual path length covered in a given time by a particle in motion is known as the distance covered by it.

The distance covered in a given time is either equal to or greater than the value of displacement for that time.

Q.4. What do you mean by linear motion of an object ? Give three examples.

Ans. By the term linear motion of an object we mean its motion along a straight line.

Motion of a car or cycle along a straight level road, motion of a railway train along a straight track, motion of a body falling freely under gravity, motion of a lift in a multistorey building are examples of linear motion.

Q.5. What do you mean by uniform and non-uniform motion (or velocity) ?Ans. See point numbers 13 and 14 under the heading "Chapter At A Glance".

Q.6. Examine the data given below for motion of two different objects A and B carefully and state whether the motion of the objects is uniform or non-uniform.

Time Distance travelled by object A in m t

Distance travelled by object B in in

9.30 a.m. 10 12

9.45 a.m. 20 19

10.00 a.m. 30 23

10.15 a.m. 40 35

10.30 a.m. 50 37

10.45 a.m. 60 41

11.00 a.m. 70 44

' Ans. A careful analysis shows that motion of object A is uniform because it covers a distance of 10 km in every 15 minutes.

However, motion of object B is non-uniform because it is covering unequal distances in equal intervals of time.

Q.7. Define the term 'average speed1. Give a formula for finding average speed.

Ans. Average speed of a moving object is that uniform speed with which the object covers same distance in a given time interval as it does with actual variable speed during that time.

The average speed is obtained by dividing the total distance travelled by the total time taken.

Average speed =total distance travelled (s) t total time (t)

Q.8. An object moves with a uniform speed v1 for time t and then moves with a uniform speed v2 for further time t. Find his average speed.

Ans. Distance covered by the object while moving with a speed v, for time t

S1= v1t

and distance covered by the object while moving with a speed v2 for'time t

s2 = v2t

:. Total distance covered s = s1 + s2 = v1t + v2t => and total time = 2t

v, + v7

Average speed = total distance/ total time= v1t + v2t/ 2t = v1+ v2 / 2

Q.9. Sumit drives his car at a constant rate of 40 km h_1 for one hour and then at a constant rate

of 60 km h_1 for one more hour. Find his average speed.

Ans.

Q.10. What causes variation in velocity of a particle ?

Ans. The velocity of a particle changes due to either of the following three causes :

1)Change in magnitude of velocity,

2)Change in direction of motion, and

3) Change in magnitude as well as direction of the motion.

Q.ll. What do you mean by 'average velocity' ? How is it calculated ?

Ans.

Q.12. What is the physical significance of zero time, positive time and negative time while considering motion ?

Ans. For considering motion we consider that instant of time, from which we have started observing motion, as the zero time. Time interval after the zero time is considered as positive but time interval prior to zero time is considered as negative time.

As an example, a bus starts from Delhi at 8 a.m. for Agra. Then the time 8 a.m. is taken as zero (or reference) time. At 9 a.m. the time interval from start of motion of bus is +1 hour. If we consider a time 7.30 a.m., then its interval from the reference time is 1/2 hour.

Q.13. Ashok covered a distance of 40 m due east. Then he turned towards north and further covered a distance of 30 in. What is the total distance covered by him ? What is his total displacement ?

Ans. As shown hi figure total distance covered by Ashok = AB + BC = 40 m + 30 m = 70 m. But the magnitude of displacement = Distance between initial and final point = AC

√(AB)2 +{DC)2 =√ (40)2 + (30)2 = √1,600 + 900 = √2,500 = 50 m.

Q.14. The minute-hand of a wall clock is 10 cm long. Find its displacement and the distance covered from 10.00 a.m. to 10.30 a.m.

Ans. Length of minute hand t= 10 cm. From 10.00 a.m. to 10.30 a.m. the tip of minute hand moves to diametrically opposite point on the clock. Displacement =AOB = diameter = 2t = 20 cm and the total distance covered by the minute hand

Q.15. The velocity of sound waves at room temperature on a particular day is 340 m s_1. How much time will it take to cover a distance of 102 m ?

Q.16. Define acceleration. What do you mean by positive and negative acceleration ?

Q.17. What is the SI unit of acceleration ? How is it obtained ?

Q.18. Can a particle be accelerated (t') if its speed is constant, (tt') if its velocity is constant ? Give reason.

Q.l9. Draw position-time graph for a stationary object.

Q.20. Fig. 8.17 shows displacement-time curves of two particles I and II. What conclusions do you draw from these curves ?

. Q.21. A train 100 m long is to cross a river bridge of length 800 m. What time will it take to cross the bridge ? Given that the train moves with a constant velocity of 36 km hf1.

Q.22. Distinguish between uniform and non-uniform acceleration. Ans. Difference between uniform and non-uniform acceleration.

Uniform acceleration Non-uniform acceleration

1. The velocity of the moving object regularly increases or decreases with time.

2. Velocity-time graph is a straight line inclined to the time axis.

1. The velocity of the object increases or decreases irregularly with time.

2. Velocity-time graph is a curve.

Q.23. A particle is moving with a uniform acceleration. Will it surely move in a straight line ?Ans. No, we are not sure about straight line motion.

The motion of the particle will be along a straight line only if the uniform acceleration is in the same direction as that of its velocity. If velocity and acceleration are in different directions them motion will not be along a straight line.

Q.24. An object starting with a velocity '«' is under a uniform acceleration and acquires a final velocity V in time t. Find its acceleration.

Q.25. An object is moving along a straight line with a constant acceleration of 4 m s~2. Calculate the change in velocity of the object in 5 seconds.

Q.26. At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes sometime to reach an observer after he sees the lightning. Why does it happen ?

Ans. The speed of light is extraordinary large (3 x 108ms~' or 30,000 km s"') whereas speed of sound is comparatively less (about 340-350 m s~'). As a result, lightning produced at a height h (about 2-3 kilometres) is seen immediately but thundering sound is heard sometime afterwards. The time gap depends upon the height h.

Q.27. During rainy season, Mohit noted that thundering of cloud was heard 6.5 seconds after the lightning was seen by him. If speed of sound in air be 340 m s_1, find the distance of the point where lightning has taken place.

Ans. Here velocity of sound v = 340 in s~', and time noted between occurrence of lightning and thundering of cloud t = 6.5 s

.'. Distance of lightning place s = vt = 340 x 6.5 = 2210 m = 2.21 km Q.28. Draw velocity-time graph for uniform motion along a straight line. How can you find distance covered by a particle from this graph ?

Q.29. In Fig. 8.19, five distance-time graphs have been drawn. What type of motion do these graphs represent ?

Ans. (a) Particle is at rest and no motion is taking place.

Uniform linear motion.

And (d) Non-uniform motion. (e) Uniform linear motion.

Q.30. In Fig. 8.20 four speed time graphs have been drawn. What conclusion do you draw about acceleration of the particle in motion ?

(B-ll) Short Answer Type Questions (Three Marks Each)

Q.l. Distinguish between rest and motion. Give two examples of each. Ans. See point numbers land 2 under the heading "Chapter At A Glance".

Q.2. Show that rest and motion are both relative and absolute rest or absolute motion is not possible.

Ans. We call an object to be in a state of rest or motion with respect to its surroundings i.e., it is relative to the surroundings. As an example, a passenger sitting in a running train is in a state of rest as compared to his fellow passengers but in motion with respect to the earth. Again our home is at rest with respect to earth but in motion with respect to the sun and so on. Thus, it is clear that rest and motion both are relative terms and absolute motion or absolute rest positions arc not possible.

Q.3. Distinguish between displacement and the distance covered in a given time. Ans. Difference between displacement and distance covered.

Displacement Distance covered

1. It is the shortest path length along a straight line drawn from initial position to final position.

2. Displacement has both magnitude and direction i.e..it is a vector.

3. Displacement of a particle in motion may be positive, negative or even zero.

4. Displacement of a particle in a given time can be either equal to or less than the distance covered.

1. It is the total path length covered by the particle in motion in given interval of time.

2. Distance covered has only magnitude i.e., it is a scalar.

3. Distance covered is always positive.

4. Distance covered is equal to or more than the displacement of a moving particle in a given time.

Q.4. An object covers a distance s1 with a constant speed v1 and then covers a further distance s2 with a constant speed v2. Find an expression for his average speed. Ans. Here total distance covered by the moving object 5 = s, + s2 Now time taken by the object to cover a distance s} with a constant speed

Q.5. Satvik went on his bike from Delhi to Faridabad at 40 kmth and in the evening returned back at a speed of 60 kmth. What is his average speed for the entire journey ? What is his average velocity ?

Q.6. What do you mean by motion in one, two and three dimensions ? Give examples.

Ans. Translational motion is of following three types :

I. One-dimensional or linear motion, in which the object moves along a given straight line. Motion of a train along a straight railway track or motion of a car along a level straight road are one dimensional motion.

Two-dimensional motion, in which the object moves along a plane.Motion of a child crawling on a floor, motion of an insect on ground, circular motion are examples of this type of motion.

Three-dimensional motion, in which the object may move in all possible directions.Motion of a bird, motion of an aeroplane, motion of a car on a hillroad are examples of this type of motion.

Q.7. Rajeev went from Delhi to Chandigarh on his motor bike. The odometer of bike read 4,200 km at the start of trip and 4,460 km at the end of his trip. If Rajeev took 4 h 20 minutes to complete his trip, find the average speed in km h"1 as well as m s_1.

Q.8. In your everyday life you come across a range of motions in which

acceleration is in the direction of motion.

acceleration is against the direction of motion.

acceleration is uniform.

acceleration is non-uniform.

Identify one example each of the above type of motion ?

Ans. (a) Engine of a train standing at a platform applies force and train is accelerated in same direction

as the direction of motion.

(b) Engine of a moving train applies brakes so as to stop the train at the next railway station and

produces an acceleration in a direction opposite to that of motion.

'(c) Acceleration of a freely falling body is uniform.

(d) Acceleration of a moving bird is non-uniform.

Q.9. Starting from a stationary position, Anil paddles his bicycle to attain a velocity of 10 m s~l in

25 s. Then, he applies brakes such that he again comes to rest after next 50 s. Calculate the acceleration

of the bicycle in both cases. Also find the total distance covered by Anil.

Q.10. Define velocity and acceleration in terms of time rate of change. Give their units too. Ans. Velocity of an object is defined as the time rate of change of its displacement. SI unit of velocity is m s~'.

Acceleration of an object is defined as the time rate of change of its velocity. SI unit of acceleration is m s"2.

Q.ll. Draw distance-time graph of an object moving with uniform speed. Use the graph to determine the speed of the object.

Ans. The distance-time graph of an object, starting from origin, moving with uniform speed has been shown in the adjoining Fig. 8.21. The graph is a straight line inclined to the axes. «> To determine speed of the object we consider two points A and B § on the graph. Draw

normals AC and BD from thesepoints'respectively to on time axis. Similarly draw normals AE and BF on the distance axis. Y Extend the line EA so that it meets BD at point H. <

Q.12. Draw distance-time graphs for (t) uniform motion, (if) non-uniform motion. Ans. (t) The distance-time graph for uniform motion (i.e., for an object moving with uniform speed) has been shown in Fig. 8.21. The graph, is a straight line inclined to the axes.

Q.13. Prove that for uniform motion area under the velocity-time graph is equal to the magnitude of the displacement.

Ans. The velocity-time graph for a particle in uniform is as shown ^ in Fig. 8.23. E We know that for a particle moving with uniform velocity the > product of velocity and time gives the displacement of the particle g ($ = vt). Hence, to know the distance moved by the particle between time % f, and t2 using velocity-time graph, we draw perpendiculars from the | points A and B corresponding to the time t( and t2 on the graph. Let these perpendiculars be AC and BD respectively. Velocity v is represented by ^ AC or BD and time (t2 - *|) is represented by AB (or CD). Fjg 8 23

—►time

t(s) Fig.

( 25)

■ time

t(s) Fig.

8.24

.'. Magnitude of displacement s of the particle in time (t2 - t1) will be s = v(t2-t]) = ACx(AB) = area of the rectangle ABDC

= area enclosed by velocity-time graph and the time axis = area enclosed under the velocity-time graph. Q.14. Find the value of distance covered by an object in a given time from its velocity-time graph, if the motion is uniformly accelerated one.

Ans. Velocity-time graph for uniformly accelerated motion of an object is as shown in Fig. 8.24. Select two points A and B corresponding to times ^and t2 respectively on the graph. Draw normals AC and BD on the time axis. The distance travelled by the object in time (t2 - t,) will be represented by area under the graph i.e., area ABDC. To find the area draw a normal AE from point A on BD. Now,

Distance covered s - area ABEDC

= area of the rectangle AEDC + area of the triangle ABE

= AC*AE+ -AExEB. 2

Q.15. State three equations of motion. Which of them describes (;') velocity-time relation, (ii) position-time relation ?

Ans. Let an object starts motion with an initial velocity u and a uniform acceleration a. Let after time t its final velocity becomes v and during this time it covers a total distance s. Then, the three equations of motions are :

(i) v = u + at,

(tt) s = ut + — at. and (Hi) v2 - ii2 = 2 as

Out of these equations, the first equation (v = it + at) describes the velocity-time relation and the l i

second equation (s = ut + ~at~) describes the position-time relation.

Q.16. Shatabdi Express starting from rest attains a velocity of 108 km h-1 in 2 minutes. Assuming the acceleration to be uniform, find the value of (t) acceleration, (if) the distance travelled by the train for attaining this velocity.

Ans. Here initial velocity a = 0, time t = 2 minutes = 120 s,

108x5

= 0.25 m s~

and final velocity v = 108 km h =

m s = 30 m

s 30-0

( 26)

■u

(0 •'• Acceleration of the train a =

t 120

(if) If the total distance travelled by the train for attaining a velocity v be s, then

2 as = v- u

or vir (30)--(oy

2 x 0.25

900 - 0

1800 m

( 27)

,20 m

s-1

Q. 17. 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 3 s to come to rest after the application of brakes, calculate (t) the initial velocity of the car before applying the brakes, and (if) the distance covered by the car before stopping. Ans. Here acceleration a = - 6 m s~~, time ( = 3 s and final velocity v = 0 (t) From the relation v = u + at, the initial velocity is given by

(t't') Using the relation s ■

zt = v-at = 0-(-6)x3=0-(-18) = 18 ms"' 1

ut + — at2, we have ?

distance covered s = 18 x 3 + — x (- 6) x (3)2 = 54 - 27 = 27 m

Q.18. The velocity-time graph of an object under linear motion is shown in Fig. 8.25. What is its acceleration ? Find the total distance covered by the object in 8 s.

—time t(s) Fig. 8.25

Ans. Acceleration of the object = slope of v-t graph

-20 m s

8s

-l

= - 2.5 m s 2

As the graph is sloping downward, hence, the slope has been taken negative. Total distance covered by the object in 8 s = area of triangle formed by v-t curve

= — x base x height 2 5

= - x(8 s)x(20ms"')

80 m

( 28)

Q.19. A car is moving along a straight line, say OP in Fig. 8.26. It moves from O to P in 18 s and

returns from P to Q in 6 s. What are the average velocity and average speed of the car in going

(a) from O to P and (b) from O to P and back to Q ?

O P

—

H

Q

—

I—

-

+-

-

+

-

40 m 80 m 120 m 160 m 200 m 240 m 280 m 320 m 360 m 400 m

Fig. 8.26

Ans. (a) In motion from O to P :

. . total displacement 360 m „„ ,

Average velocity = - — - = - = 20 m s'

total time 18 s

0

( 29)

total path length 360 m _, and average speed = — = —-—= 20 m s

Average velocity average speed =

total time 18 s

(b) For motion from O to P and then back to Q :

total displacement 240 m

240 m

= 10 m s"

total time (18 + 6)s 24 s

total path length 360 m + (360 - 240) m

(18 + 6)s

20 m s"1

and

total time 360 + 120 480

-m s

24 24

Q.20. A 100 m sprinter uniformly increases his speed from rest at the rate of 1 m s~2. How much time does he take to complete the race ? What is his final speed ?

Ans. Here total distance s = 100 m, initial velocity u = 0 and uniform acceleration a = + 1 m s"2.

1 9

Using the relation s = ut +— at, we have

100 = 0xt + ixlxr2 = -r2 2 2

,2

100 x 2 = 200

( 30)

=> t = V200 = 14.14 s

Now from the relation v = u + at, we have final speed of sprinter v = 0 + 1 x 14.14 = 14.14.rn s"1 Q.21. The velocity-time graph of a particle moving along a straight line is as shown in

Fig. 8.27. Find the magnitude of displacement covered by it before coming to rest.

20-

-

( 31)

Q.22. Which of the two decides the direction of motion of an object: its velocity or the acceleration acting on it ? Explain by giving an example.

Ans. It is the direction of velocity of an object which determines the direction of motion of a particle and not the acceleration. As an example, let us consider a ball moving in vertically upward direction when projected upward with a finite velocity. In this case, an acceleration due to gravity is acting on the ball in vertically downward direction. However, the direction of motion is vertically upward which is same as the direction of velocity.

Q.23. The velocity acquired by a body moving with uniform acceleration is 12 m s_1 in 2 s and 18 m s_1 in 4 s. Find the initial velocity of the body.

Ans. Let initial velocity of the body beam s-1 and its constant acceleration be a. Then velocity after 2 s will bev=12 = zt + ax2

12 = u + 2a ' ...(;)

and velocity after 4 s will be 18 = u + a x 4

18 = u + 4a ....(ii) Subtracting (t) from (t;'), we have

6 = 2a 6 „

=> a = — = jms"

2

Substituting this value of 'a' in equation (t), we get

12 = u + 2a ~u + 2x3 = zt + 6 => it = 12 - 6 = 6 m s_1

Q.24. An object moving with uniform acceleration has the displacement of 9 m in 3 s but a displacement of 16 m in 4 s. Find the acceleration.

Ans. Let initial velocity and uniform acceleration of the object be u and a, respectively. It is given that when t, = 3 s, s,= 9 m and for t2 = 4 s, s2 = 16 m.

1 i

Using the equation s = ut + ~^ar~, we get

1 i 9

9 = M x 3 + — x a x (3)" = 3zt + —a

2 2

3

or 3 = it + — a ....(t)

2

( 32)

I i

and 16 = ux4 + — xax (4y = 4u + 8a

or 4 = u + 2a ....(ii)

Subtracting (t) from (tt), we get

1=^ 2

=> a = 2 in s 2

Q.25. A truck running along a straight road increases its speed uniformly from 30 m s~' to 60 m s_1 over a time interval of 1 minute. Find the acceleration. Also Find the distance travelled by the truck during this time interval.

Ans. Here u ~ 30 m sv = 60 m s 1 and t = 1 minute = 60 s

(33)

0.5 m s~2

acceleration a

v-u 60 - 30 30 60

t 60

Now using the relation v2 - it = 2as, we have (60)2-(30)2 = 2 x (0.5) x.? or 3600 - 900 = s

or a- = 2700 m

Q.26. A particle moving with constant acceleration along a straight line covers the distance between two points, 75 m apart, in 5 s. Its speed as it passes the second point is 20 m s_1.

What is its speed at the first point ?

What is its acceleration ?

Ans. Here s = 75 m, t = 5 s and v = 20 m s~'

(a) As distance covered s = average speed x time

U + v

> + 20\ c 75 = I —:— I x 5

:30

2

75x2

it + 20

or «=30-20 = 10 ms"

* , . v-u 20-10 10 „ ,

(b) :. Acceleration a = = = — = 2 m s

t 5 5

(C) Long Answer Type Questions (Five Mark Each)

Q.l. An object starts linear motion with a velocity u and under uniform acceleration a it acquires a velocity v in time t. Draw velocity-time graph. From this graph obtain the following equations :

(a) v = u + at,

(34)

(b) S: ut+ \atz. 2

Ans. The velocity-time graph has been shown in Fig. 8.28. ^ At t = 0, the initial velocity is it (at point A) and then increases to <t> v (at point B) in time t. ■£ (a) Draw perpendicular lines BC and BE from point B on time & and the velocity axes respectively, so that OA = u, OE = BC = v. o . Also draw a line AD parallel to time axis, so that OC = AD - t. ? Then change in velocity in time t = BC - OA = BC - CD = BD T But BC = v and CD = OA = it

Hence, BD - v - it c From the velocity-time graph, the acceleration of the object is given by

change in velocity _ BD _ BD _ v - it time taken AD OC t

(b) According to velocity-time graph the total distance covered by the object is obtained by the area under the graph.

Hence, distance s travelled by the object = area of the trapezium OABC

= area of rectangle OADC + area of triangle ADB

= OAxOC+ ~(^D x BD) But OA = w, OC = t, AD = OC = t and BD = change in velocity = v - u = at

1 -3 ■

s = ut + — (t x at) = ut + —ar. ....(ii)

2 W

Q.2. From the velocity-time graph of uniformly accelerated motion, establish the equation las = v2- h2 (equation for position-velocity relation).

Ans. As shown in Q. I., from velocity-time graph shown in Fig. 8.28, the distance 5 travelled by the object in time t is given by

s = area of the trapezium OABC

= ~(OA + BC) x OC

But OA '= w, BC = v and OC = t

s = —(v + u) t 2

From velocity-time relation v = v. + at, we have

...(f)

A

E

v

D u

C t

->■

-time (s)

Fig.

8.28

(35)

1 3 4

t = ...(ii)

Substituting value of time t from equation (if) into (A, we get

j(» + v)

a J 2a

2as = v2 - u2 ...(Hi) which is the requisite equation.

t

Q.3. Define uniform circular motion. Is it an accelerated motion ? If yes, what is the direction of the acceleration ?

Ans. An object is said to be in uniform circular motion if it describes a circular path with a constant speed. Motion of the moon around the earth, motion of a satellite in a circular orbit around the earth, motion of an electron around the nucleus in a circular orbit are some examples of uniform circular motion.

In uniform circular motion although speed of the moving particle remains constant, yet the direction of motion continuously goes on changing with time so as to compel the particle to move along a circular path instead of a straight line path. Hence, uniform circular motion is an accelerated motion.

Here the acceleration is changing only the direction of motion and the magnitude of the velocity (i.e., the speed) remains Fig. 8.29

unchanged. It is possible only when the direction of acceleration is perpendicular to the direction of velocity. As being shown in Fig. 8.29, the direction of velocity at any point is along the tangent drawn on the circle at that point and radius and tangent in a circle are mutually perpendicular to each other, therefore, it is clear that direction of acceleration in uniform circular motion is along the radius of the circle at that point and is pointed towards the centre of circle.

Q.4. A body starts from rest and moves along a straight line. It is uniformly accelerated at a rate of 2 m s-2 for 10 s. For next 20 seconds, the body moves with uniform velocity. Then a negative acceleration of,2 m s-2 acts on the body till the body comes to rest again. Draw velocity-time graph for the entire motion. Also find the total distance covered by the body during its motion. Ans. Here u = 0, a = + 2 m s~2 and t = 10 s .■. Velocity after 10 s

v = u + at = 0 + 2 x 10 = 20 m s"1

20

—► time(s) Fig. 8.30

Now fornext 20 s (i.e., from 10 s to 30

(36)

s) the body is travelling with uniform velocity of 20 m s"1. These two motions are being represented on velocity-time graph by lines OA and AB, respectively. Finally, a negative acceleration of - 2 m s~2 acts till the body comes to rest. Hence, for this motion : 20 m s_1, a' = - 2 m s"2 and v' = 0 From relation v = u + at, we get

0 = 20 + (-2) x t' = 20 - 2t'

2t' = 20 or t' =

20 = 10 s

This motion is being represented by line BC on the graph Total distance travelled by the body s = area OABC

(OC+AB^

x AD

40 + 20

x 20 = 30 x 20 = 600 m

Deepak sir 9811291604 Motion