Your Cart

01_MOTION_ class 9

 

Introduction

  • Look around yourself at this fascinating and vast cosmos. We see some objects at rest and others in motion. We often perceive an object to be in motion when its position changes with time. At times, we infer motion through indirect evidences. e.g. the motion of the earth is indicated by the phenomenon of sunrise, sunset and changing of the seasons. In order to understand this beauty of Physics, let us take a step forward by understanding Motion. In this chapter, we shall study what is motion and how do the physical quantities like velocity and acceleration help to describe it. We shall study uniform and non-uniform motion / acceleration, retardation, graphical representation of motion, equations of motion and uniform circular motion.

Describing Motion

  • Rest and motion are relative terms. An object may appear to be moving for one observer and stationary for another observer. e.g. for the passengers in a moving train, the trees outside it appear to move backward. A person outside the trains find the trains as well as the passengers moving. However, a passenger inside the train finds this follow passengers to be at rest (Assuming that they are sitting on their berths). So, there is a need for the term reference point.
  • Reference Point :The point from which location is observed and measured is called the reference point. It is called origin. We describe the location of an object by specifying a reference point. e.g. a mall is 2 km east of the cinema hall. We are specifying the position of the mall with respect to the cinema hall. Here, cinema hall is the reference point.
  • Rest : An object is said to be at rest if it doesn’t change its position with respect to surroundings (reference point) as the time passes e.g. a book placed on the table does not change its position (if it is not disturbed by any body) with respect to the table (reference point) and with the passage of time, so it is considered to be at rest.
  • Motion :An object is said to be in motion if it changes its position with respect to surrouondings (reference point) and/or time. e.g. a school bus is said to be in motion if it changes its position with respect to your home or school (reference point) and with the passage of time.
  • Scalar Quantity :Is described completely by its magnitude alone. It has no direction e.g. distance, mass, speed, energy etc.
  • Vector Quantity :Has both magnitude and direction and obeys the laws of vector addition. e.g. displacement, velocity, momentum, force etc.
  • Motion along a Straight Line (or Rectilinear motion) :If the motion of a body is described by only one position coordinate, it is known as rectilinear motion. e.g. a bus moving along a straight line path.
  • Distance :It is the total path length covered by body during its complete journey without considering its direction. So, it is a scalar quantity and it is always positive.

e.g. consider a body starting its journey from 0, along a straight line OX, let it go from O to C and come back to B along the same straight line.

Distance travelled = OC + CB = 100 + (100 – 40) = 160 km.

Value of distance can never be zero or negative.

  • Displacement :Is the shortest distance between the initial position and final position of the body in a particular direction. It is a vector quantity.

In the example given above, displacement

Displacement may be positive, negative or zero

          In many cases, body covers some distance but displacement = 0, e.g. (i) if we move along a circular path and come back to initial position, displacement = 0, (ii) A body thrown up comes back to the initial position then displacement = 0

Difference between distance and displacement

Distance Displacement
(i) It is the actual path length covered by a body without considering its direction (i) It is the shortest distance between the initial and final position is a particular direction
(ii) It is a scalar quantity (ii) It is a vector quantity
(iii) It depends upon the path (iii) It is independent of the path
(iv) It is always positive (iv) It may be positive
(v) Its magnitude is always greater than or equal to the displacement (v) Its magnitude is always less than or equal to the distance

Example 1:

A body starts from O, travels 5 km towards west, represented by OA 5 km towards north represented by AB and 10 km towards east, represented by BC. Find the distance and displacement.

Solution :

Distance = OA + AB + BC = 5 + 5 + 10 = 20 km

Displacement =

Magnitude of displacement

Example 2:

A body travels 4 km east, then 3 km south. Find the distance and displacement.

Solution :

          Distance travelled = OA + AB = 4 + 3 = 7 km.

Magnitude of displacement

Example 3:

An athlete completes one round of a circular track of diameter 200 m in 40 sec. What will be the distance covered and the displacement at the end of 2 minute 20 sec?

Solution :

Time taken = 2 × 60 + 20 = 140 sec

No. of rotations

If A is the initial position, after 3.5 rounds, the body is at B.

Displacement covered = AB = 2r = 200 m

Distance covered in one complete rotation

 

Distance covered in 3.5 rotation

 

 

Uniform Motion and Non-Uniform Motion

Uniform Motion :A body is said to be in uniform motion when it covers equal distance is equal internal of time, howsoever small the internal may be.

e.g. a car covers 120 km in first hour, another 120 km in second hour and so on. The car is in uniform motion as it covers equal distance i.e. 120 km in equal interval of time i.e. 1 hour. The graph between distance and time is a straight line. In the strict sense, it should cover 60 km in each half hour, ….and 2 km in every one minute. Only then, the motion is said to be uniform. Other examples can be movement of hands of a clock, movement of Earth about its axis.

Non-Uniform Motion :

          A body is said to be in non-uniform motion if it covers unequal distances in equal intervals of time, howsoever small the interval may be e.g. the motion of a train starting from a railway station or approaching a station, a freely falling body under the action of gravity, a car moving on a crowded road.

Measuring the Rate of Motion

  • Different bodies take different time to cover a given distance as some of them move fast and some move slowly. One of the ways of measuring the rate of motion is to find out the distance covered by the body in unit time.

Speed

  • Speed is defined as the distance travelled per unit time i.e.

Its S.I. unit is metre/second. It is written as m/s or ms–1. The other units include cm s–1 and km h–1. Speed is a scalar quantity. It has only magnitude. It can be zero or positive. It can never be negative.

Uniform speed (or Constant speed)

  • The speed is said to be uniform speed if the body covers equal distances is equal intervals of time, no matter howsoever small these interval may be.

Non-Uniform speed (or Variable speed)

  • The speed of body is said to be non-uniform if it covers unequal distances is equal intervals of time howsoever small these intervals may be. e.g. motion of a freely falling body, moving of a car on a crowded road. In most of the cases, bodies move with non-uniform speed. So, their rate of motion is described in terms of their average speed.

Average Speed

  • Average speed is defined as the total distance travelled by the body per unit time i.e.

 

If the body covers a distance s in time t, then average speed,

Instantaneous Speed

  • Speed of a body at any instant is called its instantaneous speed. It is measured by finding the distance covered by the body in a very short time interval and then calculating the ratio of distance and time interval. The speedometer of a vehicle measures its instantaneous speed.

Example 4:

Sohan needs 10 minutes to cover a distance of 1.5 km on his bicycle. Calculate his average speed in

(a)  km/h             (b)  m/s          (c)    km/min.

 

Solution :

(a)

(b)

(c)

 

Example 5:

A body covers 20 m in 4s and next 20 m in 8 s. Find its average speed.

Solution :

 

 

 

Example 6:

Convert (a) 72 km/h into m/s (b) 15 m/s into km/h

Solution :

(a)

          short-cut

(b)

short-cut

 

Speed with Direction

  • If we know the speed of a body we can find the distance covered in a given time interval. But, to find its position at the end of the time interval, we need to know the direction in which the body has moved, when speed and direction both are specified, we get its velocity. Velocity of a body is the quantity that gives us its speed as well as its direction of motion or velocity of a body is the distance travelled by the body in a specified direction in a unit time interval.
  • We can also say that velocity is the rate of change of displacement i.e., the displacement of the body divided by the time interval. It is a vector quantity, represented by .

 

Unit

  • Its S.I. unit is same as that of speed. i.e., ms–1.

Uniform Velocity

  • If velocity of the body doesn’t change with time it is said to move with uniform velocity. Or if a body covers equal distances is equal intervals of time in a particular direction, it is said to move with uniform velocity.

Example a car travels over a straight horizontal road covering 40 km in every hour,  we can say that car has uniform velocity of 40 km/h.

Non-Uniform Velocity (or Variable Velocity)

  • If the body covers unequal distances in a particular direction in equal intervals of time or the direction of motion of the body changes even with the body covering equal distances is equal intervals of time, it is said to move with non-uniform velocity.

e.g. – a car moving on a crowded road; a car rounding a curve at constant speed has variable velocity as its direction of motion changes continuously, motion of a freely falling body has variable velocity.

Average Velocity

  • When a body is moving along a straight line at a variable speed, the magnitude of its rate of motion can be expressed in terms of average velocity.

When velocity of the body is changing at a uniform rate, the average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time. i.e.,

 

 

 

where vav is the average velocity, u is the initial velocity & v is the final velocity.

In general, average velocity is defined as the total displacement of a body divided by the total time taken.

i.e.

 

when vav is average velocity, s is total displacement and t is total time taken.

 

Instantaneous Velocity

  • For a body moving with variable velocity, the velocity of the body at any instant is called the instantaneous velocity. It is measured by finding the ratio of the displacement and the time interval but the time interval should be so small that there is no change in direction of motion of the body during that time interval.
Difference between Speed and Velocity
Speed Velocity
(i) It is defined as the distance travelled by the body per unit time. (i) It is defined as the distance travelled by the body per unit time in a particular direction.
(ii) It is a scalar quantity. (ii) It is a vector quantity.
(iii) It is always  positive. (iii) It can be positive or negative.
(iv) Average speed is not zero in one rotation. (iv) Average velocity is zero in one rotation.

 

Example 7:

A body travels along a straight path 20 m at a speed of 40 m/s and returns but to same location with a speed of 60 m/s. Find the average speed and average velocity for the entire journey.

Solution :

 

 

 

Analysis :

Let S be the distance, t1 be the time taken when body is travelling with speed v1 and v2 be the time taken when body is travelling with speed v2.

 

 

 

(as displacement is zero).

Note :  If distances are covered is equal time intervals with speeds v1 and v2,  then

         

Example 8:

A body moving in a straight line covers half of the total distance with speed of
3m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s., respectively. What is the average speed of the body.

Solution :

 

Let total distance be 2 s.

AB = BD = s

Distance BC and CD are covered is equal time intervals

So, average speed for BD is

 

Distance AB is covered with speed 3 m/s.

Distance BD is covered with speed 6 m/s.

Equal distances are covered with speed v1 and v2

so, average speed

Example 9:

Ram swims in a 90 m long swimming pool. He covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Calculate his average speed and velocity.

Solution :

 

 

 

 

Rate of Change of Velocity (Acceleration)

  • Consider a train starting from a station. Its velocity is zero when it just starts. Its velocity is just few km/h after few seconds. Its velocity becomes very high within a few minutes. As the train approaches next station, where it is scheduled to stop, its velocity gradually decreases, finally becoming zero. So, its velocity changes (increases or decreases) during its motion. To express the change in velocity, we need to introduce another physical quantity called acceleration which is a measure of the change in the velocity of a body per unit time.

i.e.

If the velocity of a body changes from initial value u to final value v in time t, then

Its S.I. unit is m/s2 or ms–2.

This type of motion is known as accelerated motion.

Note :

  1. The acceleration is considered positive if it is in the direction of velocity and negative if it is opposite the direction of velocity.
  2. It is a vector quantity.
  3. It may be positive, negative or zero.
  4. If velocity of a body increases, the velocity and acceleration both have same sign. If velocity of a body decreases the velocity and acceleration they have opposite sign.

Retardation

  • If magnitude of velocity of a body decreases, it is said to be retarded. Here, (v < u) final velocity is less than the initial velocity. e.g. on applying brakes on a moving train, its velocity decreases and finally becomes zero.

Graphical Representation of Motion

Graphs provide a convenient method to represent basic information about a variety of events. For example, in the telecast of 1 Day Cricket Match, vertical bar graphs show the run rate of a team in  each over. As you have studied in mathematics, a straight line graph helps in solving a linear equation having two variables.

To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one physical quantity, such as distance or velocity, on another quantity, such as time.

Distance Time Graph

          The nature of the motion of an object can be studied by plotting a graph between the distance covered and time. Sucha graph is called a distance-time graph.

Consider a car on a long journey. It moves with a constant speed, covering
50 km every 20 time. Table shows the distance of the car from the starting position at a different instants of time.

In figure, we have taken the distance covered on the y -axis and time on the
x-axis, and plotted each set of values. The resulting graph is the distance-time graph for the car. The points on the graph fall on a straight line. This happens when an object covers equal distances in equal time intervals, i.e., when it moves with a uniform speed.

So, the distance-time graph of an object moving with a uniform speed is a straight line. Conversely, if the distance-time graph of an object is a straight line, the object is moving with a uniform speed.

Note that if the distance-time graph is a straight line, it does not mean that the object is moving along a straight line. The 500-km drive described in table need not be along a straight road, although the distance-time graph is a straight line.

Speed from distance-time graph

Consider an object moving with a uniform speed. The distance-time graph is represented by a straight line as shown in figure. We can calculate the speed of the object from the distance-time graph. From the graph shown in figure, we find that the distance covered till time t1 is d1 and till time t2 it is d2. Thus, the object covers a distance of d2 – d1 in the time intervals t2 – t1. The speed of the object is, therefore,

 

The ratio is called the slope of the line. Thus, if the distance-time graph of an object is a straight line, the speed of the object is equal to the slope of the straight line.

The slope of a line tells us how steeply it is inclined to the horizontal axis (drawn from left to right). If the line is parallel to the horizontal axis, the slope is zero. As the line gets more and more inclined to this axis, its slope increases. Thus, a more steeply inclined distance-time graph indicates greater speed.

Distance-time Graph for Non-uniform Speed

          If an object moves with a non-uniform speed, its distance-time graph is not a straight line. Figure shows an example. The inclination of the graph is different at different places, and hence, it does not have a unique slope.

 

Velocity-Time Graph

          Consider an object moving along a straight line. We choose and fix a particular direction along the line as the positive direction. If the object moves in the positive direction, its velocity is positive. In fact, it is the same as its speed. If the object moves in the negative direction, its velocity is negative. In this case, the velocity is the negative of its speed.

If a graph is plotted taking the velocity of an object moving along a straight line on the vertical axis and time on the horizontal axis, we get a velocity-time graph.

If the particle moves with a constant velocity v, the velocity-time graph will be a straight line parallel to the time-axis, as shown in Figure 2.24. The displacement s in time t is given by

            s= vt = OA.OC

                     = area of the rectangle OABC, which is the area under the graph.

So, the area under the velocity-time graph of an object gives its displacement. We have already seen that the area under the speed-time graph gives the distance covered.

Let us take an example. A ball is dropped from a height. We take the downward direction as positive. As the ball falls, its velocity increases. Table shows the velocity of the ball at different instants. The velocity versus time graph is shown in Figure.

We see that the plotted points fall on a straight line. What is the displacement of the ball in the time interval 0 to 0.6 s? It is equal to the area under the velocity-time graph from t = 0 to t = 0.6 s. This area is in the shape of a triangle. The area is

 

 

 

 

The ball has fallen through 1.8 m in 0.6 s.

Next consider the situation shown in Figure. A ball placed on a smooth floor moves towards a wall, hits the wall at time t = 7s and returns (Table). We take the direction towards the wall as positive and the direction away from the wall as negative. From time 0 to 7 seconds, the velocity is +10 cm/s, and from 7 to 14 seconds it is -10 cm/s. What will the velocity-time graph look like for the whole period 0-14 s?

Figure shows the graph. The ‘height’ of the rectangle OABC is positive and that of the rectangle CDEF is negative.

The area under the curve for 0-7 s is

 

which is the displacement in the period 0 – 7 s.

The area under the curve for 7-14 is

                  

which is the displacement in the period 7-14 s.

The total area under the curve for 0-14 s is

(70 cm)+ (-70 cm) = 0.

The displacement in the period 0-14 s is indeed zero, as the ball returns to its original position.

 

Acceleration from Velocity-Time Graph

Suppose a particle moves with a uniform acceleration of 2 m/s2 along a straight line. This means that the velocity increases by 2 m/s in one second. Also, suppose its velocity at t = 0 is 10 m/s.

Let us plot the velocity-time graph for this situation. We first find the values of the velocity at certain instants. At t = 0, the velocity is 10 m/s. At t = 1 s it will become 10 m/s + 2 m/s = 12 m/s. At t =2s, it will become 14 m/s, and so on. These values are given in Table 2.7 and the velocity-time graph is shown in Figure.

We see that the graph is a straight line. When the acceleration is uniform, the velocity-time graph is a straight line. We will now show that the slope of the velocity-time graph gives the acceleration.

Suppose the velocity-time graph of a particle moving along a straight line is as shown in Figure 2.29. The graph is a straight line. At time tx, the velocity is i>j, and at time 12, it is v2. These values are represented by the points A and B on the graph.

The acceleration of the object is

                  

            As defined earlier, the ratio is called the slope of the line. Thus, we have the following.

The slope of the velocity-time graph gives the acceleration for an object moving along a straight line.

Velocity-time graph for non-uniform acceleration

If the acceleration of an object moving along a straight line is not constant, the velocity-time graph is not a straight line. The graph in Figure represents a motion in which acceleration increases with time.

Equation of Motion by Graphical Method

  • For a body moving along a straight line with uniform acceleration, relation between velocity, acceleration and covered by a body in a particular time can be established by a set of equation known as the equation of motion. There are three such equations :-
  1. v = u + at

Where ‘u’ is initial velocity of the body moving with uniform acceleration ‘a’ for time ‘t’, v is the final velocity and ‘s’ is the displacement covered in time ‘t’.

Equation (1) represents velocity-time relation

Equation (2) represents position-time relation

Equation (3) represents position-velocity relation

 

Derivation of Equation of Motion by Graphical Method :

Equation for velocity-time Relation

  • Consider a body moving along a straight line with uniform acceleration a. Let its initial velocity be u (at t = 0). Its velocity-time graph is represented by AB.

At t = 0, initial velocity  = OA = u.

Its velocity increases at a constant rate, a = (acceleration) in going from A to B.

At t = OC, final velocity = CB = v

Draw AD ^ BC and BE ^ OY. LetÐBAD be q.

We know that, acceleration of body = slope of velocity-time graph

i.e.,

 

 

 

or    v – u = at

or  v = u + at ; which is the required velocity-time relation.

Equation of Position time relation

  • Consider a body moving along a straight line with uniform acceleration a. Let +ve initial velocity be u at t = 0. The velocity-time graph of the body is represented by AB as shown in figure. Let S be the displacement (it will also be equal to the distance as there is no change in direction of motion of the body) travelled by the body in time is going from A to B.

Displacement, S = Area under velocity-time graph

= Area OABC

 

 

 

 

 

(as v – u = at)

 

or                   Which is one required position-time relation.

Equation for Position-Velocity relation

  • Consider a body moving along a straight line with uniform acceleration a. Let its initial velocity be u at t = 0. The velocity-time graph of the body is represented by AB as shown in the figure. Let s be the displacement (it will also be equal to the distance as there is no change in direction of motion of the body) travelled by the body is going from A to B.

Displacement, s = Area under Velocity-time graph

= (Area of trapezium OABC)

(Sum of parallel sides) × distance between the parallel sides.

 

 

 

 

or         v2 – u2 2as     which is the required position-velocity relation.

 

Example 10:

A train starting from rest from a station attains a velocity of 90 km h–1 in 5 minutes. Find   (i) The acceleration (ii) The distance travelled by the train in 5 minutes (assume that the acceleration is uniform).

Solution :

u = 0,

t = 5 min = 5 × 60= 300 s

(i)

(ii) v2 – u2 = 2as

as   a = 0

 

= 3750 m = 3.75 km

 

Example 11:

A car accelerates uniformly from 72 km h–1 to 90 km h–1 in 5 sec. Find (i) the acceleration (ii) the distance travelled by the car in 5 s.

Solution :

 

 

(i)

(ii)

 

= 112.5 m

 

Example 12:

Brakes are applied to a train travelling at a speed of 90 km h–1so as to produce a uniform acceleration of 0.5 ms–2. Find how far the train will travel before coming to rest.

Solution :

 

v = 0, a = -0.5 ms-2

usingv2 – u2

s = 625 m

 

 

Example 14:

A train starts from a station and moves with a constant acceleration for 2 minutes. It cross a distance of 400 m in this interval. Find its acceleration.

Solution :

t = 120 s, u = 0, s = 400 m

 

 

 

 

Uniform Circular Motion

  • Consider a body moving along a circular path in such a way that it covers equal distances in equal intervals of time. Its speed is constant. Is its velocity also constant? The change in velocity can be due to change in its magnitude or the direction of motion or both. Here, we have to find whether or not the body has changed its direction of motion?

                                    (fig. a)                               (fig. b)

  • Let us first consider a hexagonal path, ABCDEFA (fig.a) with all sides equal. Consider the body moving with a uniform speed along this path considering a complete round, and to stay on the path, it has to change its direction 6 times. Now, if it moves along an octagonal path (fig. (b)) it has to change its direction 8 times during a complete round. If we make the number of sides very large and the length of each side very small, the polygon becomes almost a circle. If the body moves along such a path, i.e., along a circle, it has to change its direction continuously at every instant.
  • The direction of motion of a body moving along a circular path can be found in a simple way. Draw a tangent to the circle from the body’s position on the circle. The direction of motion at the instant is along this tangent. The arrow in the figure shows the directions of motion of the body at different positions. A body moving along a circular path changes its direction continuously. Its velocity is not constant, even if its speed is constant circular motion is therefore an example of accelerated motion.
  • If v is the speed of a body moving along a circular path of radius r.

(circumference is 2pr and t is the time taken to complete one revolution)

  • When a body moves in a circular path with uniform speed, its motion is calleduniform circular motion. e.g. A piece of stone tied to a thread and rotated in a circle with uniform speed, moon moving around the earth uniformly in a circular orbit, the tip of seconds hand, minutes hand and hour hand of watch.

Example 15:

A particle moves along a circular path of radius 20 cm with constant speed. It takes 1 minute to move from a point on the path to the diametrically opposite point. Find

(i)  the distance covered

(ii) the displacement

(iii) the speed

(iv) the average velocity

Solution :

Suppose A is the initial position and the particle moves along ACB to reach B in 1 minute.

(i)  Distance covered in 1 minute =  = 3.14 × 20 = 62.8 cm

(ii) Displacement = AB = 2r = 40 cm in the direction A to B

(iii)

(iv)

in the direction A to B.

 

qqqqq

 

 

KEY POINTS

 

  1. State of Rest :An object is said to be at rest if it doesn’t change its position w.r.t. surroundings (reference point) and/or time.
  2. Motion :An object is said to be in motion if it changes its position w.r.t. surroundings (reference point) and/or time.
  3. Scalar Quantity :A physical quantity which is described completely by its magnitude alone.
  4. Vector Quantity :A physical quantity which has magnitude as well as direction and obeys the laws of vector addition.
  5. Distance :It is the total path length covered by a body during its complete journey without considering its direction.
  6. Displacement :It is the shortest distance between the initial and final position of a body in a particular direction.
  7. Uniform Motion :A body is said to be in uniform motion when it covers equal distances is equal intervals of time howsoever small the interval may be.
  8. Non-Uniform Motion :A body is said to be in non-uniform motion if it covers unequal distances is equal intervals of time howsoever small the interval may be.
  9. Speedis defined as the distance travelled per unit time.
  10. Average speedis defined as total distance travelled per unit time.
  11. Instantaneous speedis defined as the speed of a body at any instant.
  12. Velocity is defined as the rate of change of displacement with time.
  13. Uniform Velocity :If a body covers equal distance is equal intervals of time in a particular direction, it is said to move with uniform velocity.
  14. Non-Uniform Velocity : If a body covers unequal distances in a particular direction in equal intervals of time or the direction of motion of the body changes even with the body covering equal distances is equal intervals of time, it is said to move with non-uniform velocity.
  15. Average Velocity :Is defined as the total displacement per unit time.
  16. Instantaneous Velocity :Velocity of the body at any instant is called Instantaneous Velocity.
  17. Acceleration :The rate of change of velocity of a body with respect to time is called its acceleration.
  18. Retardation :When acceleration of a body is opposite to its velociy, it is called retardation.
  19. Uniform Acceleration :If a body travels in a straight line and its velocity increases or decreases by equal amounts is equal intervals of time, it is said to have uniform acceleration.
  20. Non-Uniform Acceleration :If velocity of the body change by unequal amount in equal intervals of time, the acceleration is said to be non-uniform acceleration.
  21. Uniform Circular Motion :When a body moves in a circular path with a constant speed, the motion is said to be uniform circular motion. It is an accelerated motion.

 

qqqqq

 

 

 


EXERCISE

 

  1. Very Short Answer Type Questions :
  2. Can distance be negative?
  3. Can displacement be negative?
  4. Can displacement and distance ever be equal?
  5. What does the slope of velocity time graph indicate?
  6. What does the slope of velocity time graph indicate?
  7. What does the slope of distance time graph indicate?
  8. What does the slope of displacement time graph indicate?
  9. What is the S.I. unit of speed? Is it same for velocity?
  10. What is the S.I. unit of acceleration?
  11. What is the S.I. unit of acceleration of a body moving with uniform velocity.
  12. Is the motion of moon around the earth uniform or accelerated?
  13. What is the relation between the distance and time for the motion with uniform velocity
  14. If the displacement-time graph of a body is a straight line parallel to the time axis, what does it tell us about its motion?
  15. Can displacement-time graph of a body be parallel to displacement axis? Give reason.
  16. Define displacement give its S.I. unit.
  17. Define speed give its S.I. unit.
  18. Define velocity give its S.I. unit.
  19. Define acceleration give its S.I. unit.
  20. Can the distance travelled by a body be smaller than the magnitude of its displacement?
  21. When is the distance covered by a body equal to the magnitude of its displacement?
  22. Can the equations of motion be used for a body moving with non-uniform acceleration?
  23. A body is moving with a uniform velocity. Is it necessary that it is moving along a straight line.
  24. Can a body have an acceleration when it is momentarily at rest? Give one example, if any.
  25. Can the motion of a body be accelerated even when it is moving uniformly? Give one example, if any.
  26. Unit of time is stated twice in unit of acceleration why? How is the distance related with time for the motion under uniform acceleration as the motion of a freely falling body?
  27. Short Answer Type Questions
  28. Convert a speed of 90 km/h into m/s.
  29. Convert a speed of 30 m/s into km/h.
  30. What can you say about the motion of a body if

(a) its displacement time graph is a straight line with some slope?

(b) its velocity time graph is a straight line with some slope?

  1. Name the two quantities, the slope of whose graph gives

(a) speed      (b) acceleration

  1. Distinguish between acceleration and retardation.
  2. Sketch the shape of the velocity-time graph for a body moving with

(a) uniform velocity

(b) uniform acceleration.

    

Short Answer Type Question

  1. A person moves along the boundary of a square of side 10 cm is 40 sec. Find the magnitude of displacement at the end of 3 minutes from his initial position.
  2. Distinguish between speed and velocity.
  3. Distinguish between distance and displacement.
  4. A motor cycle covers first 30 km with a uniform speed of 60 km/h and next
    30 km with a uniform speed of 40 km/h. Find

(i) the total time taken

(ii) the average speed

  1. A bus travels first 10 km in 20 minutes and another 10 km in 30 minutes. Find the average speed of the bus in m/s.
  2. A boy while going to the market from his home, observes the average speed of his car to be 20 km/h. On his return along the same way the average speed is 30 km/h. Find the average speed of the car in the entire journey.
  3. Derive graphically the equation v = u + at.
  4. Derive graphically the velocity-time relation.

 

  1. A girl leaves her home at 6.00 pm for her dance class which is at a place 2 km away and class starts at 6:30 pm. If she walks at a speed of 3 km/h for the first kilometre, at what speed should she wall  the second kilometer to reach just in time?

The motion of a body is given in velocity-time graph as shown in figure. Find the distance travelled between

(a) 0 – 20 s

(b) 20 – 50 s

  1. A body moving with uniform acceleration travels 84 m in 6 s and 264 m in 11 s. Find

(i) the initial velocity

(ii) the acceleration.

Long Answer Type Questions

  1. Using velocity time graph, establish the relation where the symbols have their usual meanings.

OR

Drive graphically equation for position-time relation.

  1. Drive graphically equation for position-velocity relation.

OR

Use graphical method to derive the relation v2 – u2 = 2as, where the symbols have their usual meanings.

  1. A bus travels with a uniform velocity of 20 ms–1for 5s. The brakes are applied and the bus is uniformly retarded. It comes to rest in further 8s. Draw velocity time graph use it to find

(i)  distance travelled in first 5 s

(ii) distance travelled after application of  brakes

(iii)     total distance travelled

(iv)     Acceleration during first 5 s and last 8sec.

 

 

  1. From the given graph

(a) Find the velocity of the body as it moves from (i) 0 – 5 s    (ii) 5 – 7s, (iii) 7 – 9 s

(b) Find the average velocity during the interval 5 – 9s.

  1. Based on the velocity-time graph of a body show in the figure (moving in a straight line)

(i)  Is the motion uniform.

(ii) Is the motion uniformly accelerated?

(iii)     Does the body charge its direction of motion

(iv) Is the distance travelled by the body from 0 to 4 s same as from4 to 6s?
If no compare then

(v) Find the acceleration from 0 – 4 s and retardation from 4 – 6 s.

  1. A ball is thrown vertically upwards. It goes to a height of 19.6 m and then comes back to the ground. Find

(i)  its initial velocity

(ii) total time of journey

(iii)its final velocity when it strikes the ground (g = 9.8 m/s2).

 

 

WORKSHEET – 1

 

  1. In what condition is the distance equal to the magnitude of the displacement of the body?
  2. A body completes are round of a circular path of diameter 14 m in 10 s. Calculate the distance and displacement at the end of

(a)  12.5 s       (b) 25 s               (c)      30 s

  1. A body moves 200 m west and then 50 m due east. Calculate the distance and displacement.
  2. A body thrown up vertically reaches maximum height h. Then, it returns to the ground. Find the distance and displacement
  3. is

(a)  ³ 1            (b) £ 1              (c)         always > 1   (d)      always < 1

 

 

WORKSHEET – 2

 

  1. On a 120 km track, a train travels the first 30 km at a uniform speed of
    30 km/h. How fast must it travel the next 90 km so as to average 60 km/h for the entire trip?
  2. A train 100 m long moving on a straight track passes a pole in 5 s. Find

(a) its speed (b) time taken by it to cross a bridge 500 m long.

  1. A bus travels at 30 km/h for 0.24 h., at 60 km/h for next 0.52 h and at 70 km/h for the next 0.7h. Calculate its average speed and average velocity.
  2. A bus travels along a straight path for first half time with speed 80 km/h and the second half time with 60 km/h. Find its average speed.
  3. An insect moves along a circular path of radius 20 cm with a constant speed. It takes 1 minute to move from a point on the path to the diametrically opposite point. Calculate

(a) The distance covered            (b)      The speed

(c)The displacement                  (d)      Average velocity

 

WORKSHEET – 3

 

  1. A motor cycle acquires a velocity of 36 km/h in 10 seconds just after the start. It takes 20 seconds to stop. Find the acceleration in the two cases.
  2. A motorcycle travelling at 36 km/h speeds upto 72 km/h in 5 seconds. Find its acceleration. If it stops in 20 seconds, find the acceleration in that case.
  3. Motion of a body falling freely under gravity is _____.
  4. A boy throws a ball up and catches it when the ball falls back. In which part of the motion is the ball decelerating?
  5. Which of the following could be a unit of acceleration?

(a) km/s2                         (b) cm/s2

(c) km/s                          (d) m/s2

 

WORKSHEET – 4

 

  1. A train accelerates from 20 kmh–1to 80 kmh–1 in 4 minutes. Find the distance covered in this interval. Assume that it travels in a straight line.
  2. A ball is dropped freely in the river from a bridge. If its velocity increases uniformly at the rate of 9.8 ms–2and it takes 5 s to reach the water surface. Find (a) the height of the bridge from the water level. (b) the distance covers by the ball in 2 seconds.
  3. A body with an initial velocity of 18 km h–1 accelerates uniformly at the rate of 9 cm s–2over a distance of 200 m. Find its final velocity in ms–1.
  4. A bicycle moves with a constant velocity of 5 km h–1for 10 minutes and then accelerates at the rate of 1 km/h2till it stops. Find the total distance covered.
  5. The brakes applied to a car produce an acceleration of 6 m/s2in the  opposite direction to the motion. If it takes 2s to stop. Find the distance travelled by the car during this interval.

 

WORKSHEET – 5

 

  1. Calculate the speed of a minute hand of a clock of length 5 cm.
  2. Calculate the speed of a seconds hand of a clock of length 1.5 cm.
  3. A cyclist complete a round of circular track of diameter 210 m in 10 minutes. Find his speed.
  4. In a circular motion the

(a)  direction of motion is fixed

(b)  direction of motion changes continuously

(c)  acceleration is zero

(d)  velocity is constant

  1. A cyclist goes around a circular track once every two minutes. If the radius of the circular track is 105 m. Calculate his speed ( Given 🙂

 

notes

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

 

notes

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

 

 

notes

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

Leave a Reply

Your email address will not be published.

X