CBSE Class 9 Science Motion Exam Notes

CBSE Class 9 Science Motion Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

MOTION

INTRODUCTION

• When a body does not change its position with time, it is said to be at rest; if it changes its position with time, it is said to be in motion.
• An object is said to be a point object if it changes its position by distances which are much greater than its size.
• A point or stationary object with respect to which a body continuously changes its position in the state of motion is known as the origin or reference point.

TYPES OF MOTION

According to direction

• One-dimensional motion: motion of a particle along a straight line.
• Two-dimensional motion: a particle moving along a curved path in a plane.
• Three-dimensional motion: a particle moving randomly in space.

According to state of motion

• Uniform motion: a body travels equal distances in equal intervals of time. The distance–time graph is a straight line.
• Non-uniform motion: a body travels unequal distances in equal intervals of time, e.g. a freely falling body. The distance–time graph is a curved line.

TERMS USED TO DEFINE MOTION

The terms are (i) distance and displacement, (ii) speed and velocity, (iii) acceleration.

(i) DISTANCE AND DISPLACEMENT

• The path length between the initial and final positions of a particle is the distance covered.
• The shortest distance between the initial and final positions of a body in a time interval is its displacement.
• Both are measured in metre (m) in the SI (M.K.S.) system.
• Distance is a scalar quantity; displacement is a vector quantity.
• When a body moves along the same straight line in the same direction, displacement equals the distance travelled. If the body changes direction, the displacement is smaller than the distance travelled.
• Displacement in an interval of time may be zero, positive or negative, whereas distance cannot be negative.

Displacement ≤ Distance

Example 1

A person travels 5 m towards east, then 4 m towards north and then 2 m towards west. (i) Calculate the total distance travelled. (ii) Calculate the resultant displacement.

Answer: (i) 11 m. (ii) 5 m.

(i) Total distance = 5 m + 4 m + 2 m = 11 m.

(ii) Taking east and north as the directions, the person is finally 5 − 2 = 3 m east and 4 m north of the starting point A. The displacement is the straight line AD from the starting point A to the final point D:

\[R=\sqrt{3^2+4^2}=5\ \text{m}\]

The displacement is 5 m, along AD.

Example 2

A body is moving in a straight line. Its distances from the origin are shown with time in the graph. A, B, C, D and E represent different parts of its motion. Find: (i) the displacement of the body in the first 2 seconds, (ii) the total distance travelled in 7 seconds, (iii) the displacement in 7 seconds.

Answer: (i) 40 m. (ii) 160 m. (iii) zero.

(i) The displacement in the first 2 s is 40 m.

(ii) From t = 0 to t = 7 s, the body has moved a distance of 80 m from the origin and it has again come back to the origin. Total distance covered = 80 × 2 = 160 m.

(iii) Since the body has come back to its initial position, the displacement is zero.

(ii) SPEED AND VELOCITY

• The distance travelled by a body in unit time is its speed. When the position changes in a particular direction, it is called velocity, i.e. the rate of change of displacement is velocity.
• Speed is a scalar quantity; velocity is a vector quantity.

\[\text{Speed}=\dfrac{\text{distance}}{\text{time}},\qquad \text{Velocity}=\dfrac{\text{displacement}}{\text{time}}\]

• Unit: m s⁻¹ (SI or M.K.S.); cm s⁻¹ (C.G.S.).
• If a distance–time graph is given, the speed at a given time is the slope of the line.
• The area under a velocity–time graph gives the displacement.

\(v=\dfrac{\Delta s}{\Delta t}=\text{slope}\)

Types of speed: (a) average and instantaneous speed

Average speed is the total distance travelled divided by the total time interval:

\[\text{Average speed}=\dfrac{\text{total distance}}{\text{total time}},\qquad \text{Average velocity}=\dfrac{\text{displacement}}{\text{total time}}\]

• Average speed is a scalar; average velocity is a vector.
• For a moving body the average speed is never negative or zero (unless the time is infinite), while the average velocity can be positive, negative or zero.
• In general, average speed is not equal to the magnitude of average velocity. It can be equal if the motion is along a straight line without change in direction.

If a particle travels distances \(L_1, L_2, L_3,\dots\) at speeds \(v_1, v_2, v_3,\dots\) respectively, then

\[v_{av}=\dfrac{L_1+L_2+\dots+L_n}{\dfrac{L_1}{v_1}+\dfrac{L_2}{v_2}+\dots+\dfrac{L_n}{v_n}}=\dfrac{\sum L_i}{\sum \dfrac{L_i}{v_i}}\]

If a particle travels at speeds \(v_1, v_2,\dots\) for intervals \(t_1, t_2,\dots\) respectively, then

\[v_{av}=\dfrac{v_1t_1+v_2t_2+\dots}{t_1+t_2+\dots}=\dfrac{\sum v_it_i}{\sum t_i}\]

Instantaneous speed is the speed of a body at a particular instant of time:

\[v=\lim_{\Delta t\to0}\dfrac{\Delta s}{\Delta t}=\dfrac{ds}{dt}\]

(b) Uniform and non-uniform speed. If an object covers equal distances in equal intervals of time, the speed–time graph is a straight line parallel to the time axis and the body moves with uniform speed. If the speed of a body changes with time, it moves with non-uniform speed.

Example 3

The distance between two points A and B is 100 m. A person moves from A to B with a speed of 20 m/s and from B to A with a speed of 25 m/s. Calculate the average speed and average velocity.

Answer: Average speed = 22.2 m/s; average velocity = 0.

Total distance = 100 + 100 = 200 m. Time from A to B: \(t_1=\dfrac{100}{20}=5\) s. Time from B to A: \(t_2=\dfrac{100}{25}=4\) s. Total time = 5 + 4 = 9 s.

\[\text{Average speed}=\dfrac{200}{9}=22.2\ \text{m/s}\]

The person returns to the starting point A, so the displacement is zero and the average velocity is zero.

Example 4

A car moves with a speed of 40 km/h for the first hour, then with a speed of 60 km/h for the next half hour and finally with a speed of 30 km/h for the next \(1\tfrac12\) hours. Calculate the average speed of the car.

Answer: 38.33 km/h.

\(s_1=40\times1=40\) km; \(s_2=60\times\tfrac12=30\) km; \(s_3=30\times\tfrac32=45\) km. Total distance = 40 + 30 + 45 = 115 km. Total time = \(1+\tfrac12+1\tfrac12=3\) h.

\[\text{Average speed}=\dfrac{115}{3}=38.33\ \text{km/h}\]

Example 5

The figure shows the distance–time graph of an object. Calculate: (i) which part of the graph shows that the body is at rest, (ii) the average speed in the first 10 s, (iii) the speeds in the different parts of the motion.

Answer: (i) Part BC. (ii) 10 m/s. (iii) AB = 50/3 m/s, BC = 0, CD = 30 m/s, DE = 20 m/s.

(i) The part BC is horizontal: the distance does not change, so the body is at rest.

(ii) In the first 10 s the distance travelled = 100 m, so average speed = 100/10 = 10 m/s.

(iii) Speed in AB = slope = 100/6 = 50/3 m/s ≈ 16.7 m/s. Speed in BC = 0. Speed in CD = (100 − 40)/(12 − 10) = 60/2 = 30 m/s. Speed in DE = (40 − 0)/(14 − 12) = 40/2 = 20 m/s.

Example 6

The velocity–time graph of a particle is shown in the figure. Calculate the distance travelled in the first 8 seconds.

Answer: 560 m.

The distance in the first 8 s is the area under the graph up to t = 8 s. This is the area of a rectangle (8 s × 60 m/s) plus the area of the triangle above it (base 4 s, height 40 m/s):

\[s=8\times60+\tfrac12\times4\times40=480+80=560\ \text{m}\]

(iii) ACCELERATION

The rate of change of velocity is called acceleration. It is a vector quantity.

\[a=\dfrac{v-u}{t}\]

Unit: m/s² or m s⁻².

Types of acceleration. Uniform acceleration: a body travelling in a straight line has uniform acceleration if its velocity changes by equal amounts in equal intervals of time, e.g. a freely falling body. Non-uniform acceleration: the velocity changes by unequal amounts in equal intervals of time. Instantaneous acceleration is the acceleration of a body at any instant.

• If the velocity of a body decreases, it has a negative acceleration, called deceleration or retardation.
• Acceleration is the slope of the velocity–time graph.
• If the velocity–time graph is a straight line, the acceleration is constant.
• If the slope of the straight line is positive, the acceleration is positive; if it is negative, there is retardation.

Example 7

The velocity–time graph of a body is a straight line from v = 20 m/s at t = 0 to v = 80 m/s at t = 4 s. Find its acceleration.

Answer: 15 m/s².

\[a=\dfrac{v_2-v_1}{t_2-t_1}=\dfrac{80-20}{4-0}=15\ \text{m/s}^2\]

Example 8

For the velocity–time graph of Example 6, find the instantaneous acceleration at (i) t = 3 s, (ii) t = 6 s, (iii) t = 9 s.

Answer: (i) 0; (ii) +10 m/s²; (iii) −50 m/s².

(i) At t = 3 s the point lies on AB, which is horizontal, so the slope and the acceleration are zero.

(ii) At t = 6 s the point lies on BC: \(a=\dfrac{100-60}{8-4}=+10\) m/s².

(iii) At t = 9 s the point lies on CD: \(a=\dfrac{0-100}{10-8}=-50\) m/s² (retardation).

Example 9

Starting from rest, Deepak paddles his bicycle to attain a velocity of 6 m/s in 30 seconds, then applies brakes so that the velocity comes down to 4 m/s in the next 5 seconds. Calculate the acceleration of the bicycle in both cases.

Answer: (i) 0.2 m/s²; (ii) −0.4 m/s².

(i) u = 0, v = 6 m/s, t = 30 s: \(a=\dfrac{v-u}{t}=\dfrac{6-0}{30}=0.2\) m/s², positive acceleration.

(ii) u = 6 m/s, v = 4 m/s, t = 5 s: \(a=\dfrac{4-6}{5}=-0.4\) m/s², which is retardation.

EQUATIONS OF MOTION

For motion under uniform acceleration, with initial velocity u, final velocity v, acceleration a, time t and distance s:

First equation. Since acceleration = change in velocity / time, \(a=\dfrac{v-u}{t}\), so

\[v=u+at\qquad\cdots(i)\]

Second equation. Distance = average velocity × time, and average velocity = (u + v)/2. Putting v = u + at:

\[s=\dfrac{u+v}{2}\,t=\dfrac{2u+at}{2}\,t\ \Rightarrow\ s=ut+\tfrac12at^2\qquad\cdots(ii)\]

Third equation. From (i), \(t=\dfrac{v-u}{a}\). Substituting in \(s=\dfrac{u+v}{2}\,t\):

\[s=\dfrac{(v+u)(v-u)}{2a}=\dfrac{v^2-u^2}{2a}\ \Rightarrow\ v^2=u^2+2as\qquad\cdots(iii)\]

Motion under gravity. Replace a by g. When the body moves towards the centre of the earth: \(v=u+gt\); \(h=ut+\tfrac12gt^2\); \(v^2=u^2+2gh\). When a body is thrown upwards, the earth's attraction produces a retardation, so a is replaced by −g: \(v=u-gt\); \(h=ut-\tfrac12gt^2\); \(v^2=u^2-2gh\).

BODY FALLING FREELY UNDER GRAVITY

For a freely falling body u = 0, so \(v=gt\), \(h=\tfrac12gt^2\), \(v^2=2gh\); if h is given, \(t=\sqrt{2h/g}\) and \(v=\sqrt{2gh}\); if v is given, \(t=v/g\) and \(h=v^2/2g\).

Graphs for a freely falling body: s–t is a curve rising upwards, v–t is a straight line with slope tanθ = g, a–t is a horizontal line at a = g.

Body projected vertically up (initial position as origin, upward direction positive): (a) at the highest point v = 0; (b) a = −g. If u is given: time to the highest point \(t=u/g\) and maximum height \(h=u^2/2g\). If t is given: \(u=gt\) and \(h=\tfrac12gt^2\). If h is given: \(u=\sqrt{2gh}\) and \(t=\sqrt{2h/g}\). The total time of flight is 2u/g.

Graphs for a body projected vertically up (left to right): s–t, u–t and a–t.

In motion under gravity: (a) the time taken to go up equals the time taken to fall down through the same distance; (b) the speed with which a body is projected up equals the speed with which it returns to the point of projection.

Example 10

A body starts moving with an initial velocity of 50 m/s and an acceleration of 20 m/s². How much distance will it cover in 4 s? Also calculate its average speed during this time interval.

Answer: 360 m; 90 m/s.

\[s=ut+\tfrac12at^2=50\times4+\tfrac12\times20\times4^2=200+160=360\ \text{m}\]

\[\text{Average speed}=\dfrac{360}{4}=90\ \text{m/s}\]

Example 11

A body is moving with a speed of 20 m/s. When a certain force is applied, an acceleration of 4 m/s² is produced. After how much time will its velocity be 80 m/s?

Answer: 15 s.

u = 20 m/s, a = 4 m/s², v = 80 m/s. From v = u + at: 80 = 20 + 4t, so 4t = 60 and t = 15 s.

Example 12

A body starts from rest and moves with a constant acceleration. It travels a distance s₁ in the first 10 s and a distance s₂ in the next 10 s. Find the relation between s₂ and s₁.

Answer: s₂ = 3 s₁.

In the first 10 s: \(s_1=0+\tfrac12a(10)^2=50a\). In 20 s: \(s=\tfrac12a(20)^2=200a\). Distance in the next 10 s: \(s_2=200a-50a=150a\).

\[\dfrac{s_2}{s_1}=\dfrac{150a}{50a}=3\ \Rightarrow\ s_2=3s_1\]

Example 13

A train is moving with a velocity of 400 m/s. With the application of brakes a retardation of 10 m/s² is produced. Calculate (i) after how much time it will stop, (ii) how much distance it will travel before it stops.

Answer: (i) 40 s; (ii) 8000 m = 8 km.

u = 400 m/s, a = −10 m/s², v = 0.

(i) 0 = 400 + (−10)t, so t = 40 s.

(ii) \(v^2=u^2+2as\): \(0=400^2+2(-10)s\), so 20s = 160000 and s = 8000 m = 8 km.

Example 14

A body is thrown vertically upwards with an initial velocity of 19.6 m/s. Taking g = −9.8 m/s², calculate (i) the maximum height attained, (ii) the time after which it will come back to the ground.

Answer: (i) 19.6 m; (ii) 4 s.

(i) u = 19.6 m/s, v = 0: \(0=19.6^2+2(-9.8)h\), so \(h=\dfrac{19.6\times19.6}{2\times9.8}=19.6\) m.

(ii) Time to reach the top: 0 = 19.6 + (−9.8)t gives t = 2 s. The same time is taken to come down, so the total time = 2 × 2 = 4 s.

Example 15

From the top of a tower of height 490 m a shell is fired horizontally with a velocity of 100 m/s. At what distance from the bottom of the tower will the shell hit the ground?

Answer: 1000 m.

Horizontal and vertical motions are independent. For vertical motion: u = 0, h = 490 m, g = 9.8 m/s². From \(h=ut+\tfrac12gt^2\): \(490=\tfrac12\times9.8\,t^2\), so \(t^2=\dfrac{490}{4.9}=100\) and t = 10 s.

Horizontal distance = horizontal velocity × time = 100 × 10 = 1000 m. The shell strikes the ground 1000 m from the bottom of the tower.

(The source wrote 100 m in the last line; the correct value is 1000 m.)

VARIOUS GRAPHS RELATED TO MOTION

Displacement–time (s–t) graph

• A straight line inclined to the time axis represents constant velocity.
• A straight line inclined at an angle greater than 90° to the time axis shows negative velocity.
• A curve bending upwards (getting steeper) represents accelerated motion; a curve getting flatter represents decelerated motion.

Velocity–time (v–t) graph

• A line parallel to the time axis: constant velocity (zero acceleration).
• A straight line sloping down, not starting from zero velocity: constant retardation with non-zero initial velocity.
• A straight line rising from the origin: accelerated motion with zero initial velocity.
• A line sloping downwards: decelerated motion.

Acceleration–time (a–t) graph

• A line parallel to the time axis: acceleration is constant.
• A rising line: acceleration is increasing and positive.
• A falling line: acceleration is decreasing and negative.

CIRCULAR MOTION

When a body moves in such a way that its distance from a fixed point always remains constant, its motion is called circular motion.

Uniform circular motion. If the radius vector sweeps out equal angles in equal times, the motion is uniform circular motion.

• The speed remains constant.
• The linear velocity is a vector, so its direction changes continuously.
• The direction of velocity is along the tangent at every point.

Angular velocity \(\omega=\dfrac{\theta}{t}\) is a vector quantity whose direction is perpendicular to the plane of rotation. Its unit is radian per second.

• If the particle revolves clockwise, the direction of angular velocity is perpendicular to the plane downwards; for anticlockwise rotation it is upwards.
• In uniform circular motion the angular velocity is along the axis of rotation and is constant in magnitude and direction.
• \(v=r\omega\), where r is the radius of the circle.

Centripetal acceleration. In uniform circular motion the particle has an acceleration \(a_c=\dfrac{v^2}{r}=r\omega^2\), directed along the radius towards the centre.

Centripetal force always acts towards the centre and is required to move a particle in a circle. Since it is always perpendicular to the velocity (displacement), the work done by this force is always zero.

• Circular motion in a horizontal plane is usually uniform circular motion.
• The equations of motion are not applicable to circular motion.

Time period (T) is the time taken to complete one revolution. In one revolution the angle subtended is 2π, so \(\omega=\dfrac{2\pi}{T}\) and \(T=\dfrac{2\pi}{\omega}\).

Frequency (n) is the number of revolutions per second: \(n=\dfrac1T=\dfrac{\omega}{2\pi}\).

Example 16

A particle moves in a circle of radius 2 m and completes 5 revolutions in 10 seconds. Calculate (i) the angular velocity and (ii) the linear velocity.

Answer: (i) π rad/s; (ii) 2π m/s ≈ 6.28 m/s.

Time period \(T=\dfrac{10}{5}=2\) s. (i) \(\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{2}=\pi\) rad/s. (ii) \(v=r\omega=2\times\pi=2\pi\) m/s ≈ 6.28 m/s.

(The source gave the linear velocity as “2 m/s” with the π lost; it is 2π m/s.)

Example 17

The length of the second's needle in a watch is 1.2 cm. Calculate (i) the angular velocity and (ii) the linear velocity of the tip of the needle.

Answer: (i) π/30 rad/s ≈ 0.105 rad/s; (ii) ≈ 0.126 cm/s.

The second's needle completes one revolution in 60 s, so T = 60 s. (i) \(\omega=\dfrac{2\pi}{60}=\dfrac{\pi}{30}\) rad/s. (ii) The radius is 1.2 cm, so \(v=r\omega=1.2\times\dfrac{\pi}{30}=0.04\pi\approx1.26\times10^{-1}\) cm/s.

Example 18

The earth revolves around the sun in 365 days. Calculate its angular velocity.

Answer: 1.99 × 10⁻⁷ rad/s.

T = 365 × 24 × 60 × 60 s. \(\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{365\times24\times60\times60}=1.99\times10^{-7}\) rad/s.

 TYPE OF MOTION

1.According to Directions

(ii) To calculate the resultant displacement, we choose a convenient scale, where 1 cm represents 1 m. We draw a 5 cm long line AB towards east and then 4 cm long line BC towards north. Finally,

a 2 cm long line CD towards west. The resultant displacement is calculated by joining the initial position A to the final position D. We measure AB = 5 cm.

Since 1 cm = 1 m

\ 5 cm = 5 m

Hence, the displacement of the person

R = 32 + 42

= 5m towards AD.

♦  Speed = dis tance / time

♦ Velocity = displaceme nt / time

In C.G.S. system = cm/s

♦  If time distance graph is given then speed can be given by the slope of the line, at given time

average speed = total distance / total time

Average velocity = displaceme nt total time

♦  For a moving body average speed can never be –ve or zero (unless t ® ¥), while average velocity can be i.e. vav > 0 while vav > = or < 0

♦  If a particle travels distances L1, L2, L3 at speeds v1, v2, v3 etc respectively, then Ds

♦  If a particle travels at speeds v1, v2 etc for intervals t1, t2 etc respectively, then

Sol. (i) Distance from A to B = 100 m Distance from B to A = 100 m Thus, total distance = 200 m