9th Class Physics · Punjab Board · Complete learning guide
Chapter 2Kinematics
Every topic, every MCQ and every exercise question, explained step by step in simple language. Written for the Punjab Board new-syllabus Physics book, English medium.
The branch of mechanics that studies the motion of objects without asking what causes the motion (the causes, forces, come in Chapter 3).
Definition
Rest
A body is at rest if it does not change its position with respect to its surroundings.
Definition
Motion
A body is in motion if it keeps changing its position with respect to its surroundings.
Rest and motion are relative: they depend on what you compare with. A passenger sitting in a moving bus is at rest with respect to the bus and the other passengers, but in motion with respect to the trees by the road.
Remember
There is no absolute rest. Even a "still" building is moving with the Earth, which spins on its axis and goes round the Sun.
Topic 2
Types of motion
Translatory: the whole body moves from one place to another. Rotatory: it spins about an axis. Vibratory: it moves to and fro about a mean position.
Definition
Translatory motion
Motion in which every particle of a body moves the same distance in the same direction, so the whole body shifts from one place to another.
Kind of translatory motion
Path
Examples
Linear motion
A straight line
A car on a straight road, a stone falling straight down
Circular motion
A circle
A stone whirled on a string, a car on a round track
Random motion
Irregular, no fixed pattern
A flying bee, dust particles in air, gas molecules
Definition
Rotatory motion
Motion of a body spinning about an axis that passes through the body itself. Examples: a ceiling fan, a spinning top, a wheel, the Earth spinning on its axis.
Definition
Vibratory motion
To-and-fro motion of a body about its mean position. Examples: a swing, a simple pendulum, a plucked guitar string.
Exam tip
Circular motion and rotatory motion are different. In circular motion the body goes round a point outside itself (a stone on a string). In rotatory motion it spins about an axis through itself (a fan).
Topic 3
Scalars and vectors
Definition
Scalar quantity
A physical quantity that is fully described by its magnitude (a number with a unit) alone.
Definition
Vector quantity
A physical quantity that needs both magnitude and direction to be fully described.
Scalars (magnitude only)
Vectors (magnitude and direction)
Distance
Displacement
Speed
Velocity
Mass
Weight
Time
Acceleration
Energy, work, power
Force, momentum, torque
Volume, temperature
Electric field
In print, a vector is written in bold (F) or with an arrow over it. When a vector is written in ordinary type, its direction is given in words ("20 m s⁻¹ towards the north"), as an angle ("at 30° to the x-axis"), or by a + or − sign along a line.
Showing a vector on paper
A vector is drawn as a straight line with an arrowhead. The length of the line, on a chosen scale, shows the magnitude, and the arrowhead shows the direction.
Solved Example 2.2: a 350 N force at 60° to the x-axis, on a scale of 1 cm = 100 N, is a 3.5 cm line at 60°.
Worked exampleRepresent a velocity of 300 m s⁻¹ at 60° east of north (Solved Example 2.1)
Choose a scale: 1 cm = 100 m s⁻¹.
Length of the line = 300 ÷ 100 = 3 cm.
Draw north (up) and east (right) lines. From the origin, draw a 3 cm line turned 60° from north towards east, and put an arrowhead at its end.
Answera 3 cm arrow at 60° east of north
Topic 4
Adding vectors: the head-to-tail rule
Definition
Resultant vector
The single vector that has the same effect as two or more vectors acting together.
Head-to-tail rule
Draw the first vector to scale in its direction.
Draw the second vector to the same scale, starting from the head (arrow end) of the first one.
Repeat for any more vectors.
The resultant is the vector from the tail of the first vector to the head of the last one. Measure its length and convert back with the scale.
The resultant R joins the tail of A to the head of B.
Worked exampleA girl walks 3 km west and then 4 km south. Find her displacement.
Draw 3 km west, then 4 km south from its head: the two lines make a right angle.
The resultant joins the start to the end: R = √(3² + 4²) = √25 = 5 km.
The distance she walked is 3 + 4 = 7 km, but her displacement is only 5 km, towards the south-west.
Answer5 km towards the south-west
Exam tip
Vectors along one line add like numbers with signs: 5 N east + 3 N east = 8 N east, but 5 N east + 3 N west = 2 N east.
Topic 5
Distance and displacement
Definition
Distance
The total length of the path a body actually travels, whatever its direction. It is a scalar; SI unit metre (m).
Definition
Displacement
The shortest straight-line distance from the starting point to the final point, in the direction from start to end. It is a vector; SI unit metre (m).
The curved path is the distance; the straight arrow from start to end is the displacement.
Distance
Displacement
Measures
Length of the actual path
Straight line from start to end
Type
Scalar
Vector
Can it be zero for a moving body?
No
Yes, after a round trip
Example: one lap of a 400 m track
400 m
0 m (back at the start)
Remember
Displacement ≤ distance. They are equal only when the body moves in a straight line in one direction, so displacement ÷ distance is 1 or less.
Topic 6
Speed and velocity
Definition
Speed
The distance covered by a body in unit time. It is a scalar; SI unit m s⁻¹.
speed v = distance S ÷ time t
Definition
Uniform speed
A body has uniform speed if it covers equal distances in equal intervals of time, however short the intervals are.
Definition
Average speed
Total distance travelled ÷ total time taken, for a body whose speed changes during the trip.
Definition
Velocity
The rate of change of displacement: displacement covered in unit time, in a stated direction. It is a vector; SI unit m s⁻¹.
Definition
Uniform velocity
A body has uniform velocity if it covers equal displacements in equal intervals of time in the same direction, so both its speed and direction stay the same.
Speed
Velocity
Defined using
Distance
Displacement
Type
Scalar
Vector
On a circular track at constant pace
Stays the same
Changes all the time (direction changes)
Worked exampleAn eagle flies 300 m at 60 m s⁻¹. How long does it take? (Solved Example 2.3)
Given: S = 300 m, v = 60 m s⁻¹.
From v = S ÷ t, the time t = S ÷ v.
t = 300 ÷ 60.
Answer5 s
1 km h⁻¹ = 1000 m ÷ 3600 s = 1/3.6 m s⁻¹
km h⁻¹ → m s⁻¹: divide by 3.6. m s⁻¹ → km h⁻¹: multiply by 3.6.
Object
Typical speed
Person walking
about 1.5 m s⁻¹
Car on a motorway (120 km h⁻¹)
about 33 m s⁻¹
Sound in air
about 340 m s⁻¹
Light
3 × 10⁸ m s⁻¹
Did you know?
The universal speed limit: nothing can travel faster than light, about 3 × 10⁸ m s⁻¹. Einstein's special theory of relativity (1905) showed that no object with mass can even reach it.
Topic 7
Acceleration
Definition
Acceleration
The rate of change of velocity of a body. It is a vector; SI unit m s⁻².
a = (vf − vᵢ) ÷ t
vᵢ = initial velocity, vf = final velocity, t = time taken
Definition
Uniform acceleration
A body has uniform acceleration if its velocity changes by equal amounts in equal intervals of time.
Definition
Deceleration (retardation)
The acceleration of a body whose velocity is decreasing. It comes out negative and points opposite to the motion.
Acceleration is…
Effect
Daily-life example
In the direction of motion
Speeds up
An apple falling from a tree
Opposite to the motion
Slows down
A ball thrown upward; a car braking
Zero
Velocity stays the same
A car moving straight at a steady 60 km h⁻¹
Worked exampleAn aeroplane starts from rest and reaches 252 km h⁻¹ in 20 s. Find its average acceleration. (Solved Example 2.4)
Change the unit: vf = 252 ÷ 3.6 = 70 m s⁻¹; vᵢ = 0.
a = (vf − vᵢ) ÷ t = (70 − 0) ÷ 20.
Answer3.5 m s⁻²
Remember
A body moving at constant speed can still accelerate if its direction changes, as in circular motion. Constant velocity, though, always means zero acceleration.
Topic 8
Distance-time graphs
A distance-time graph has time on the x-axis and distance on the y-axis. Its shape tells you at a glance how the body is moving.
Distance = area of the triangle = ½ × base × height = ½ × 10 × 20 = 100 m.
Answer2 m s⁻² and 100 m
Topic 10
Equations of motion
For a body moving in a straight line with uniform acceleration, five quantities are linked: initial velocity vᵢ, final velocity vf, acceleration a, time t and distance S. Three equations connect them.
1st equation: vf = vᵢ + at
2nd equation: S = vᵢt + ½at²
3rd equation: 2aS = vf² − vᵢ²
Worked exampleDeriving the 1st equation (from the speed-time graph)
The graph is a straight line from vᵢ at time 0 to vf at time t.
Its gradient is the acceleration: a = (vf − vᵢ) ÷ t.
Rearranging: vf = vᵢ + at.
Worked exampleDeriving the 2nd equation
Distance = area under the graph = rectangle (vᵢ × t) + triangle (½ × t × (vf − vᵢ)).
From the 1st equation, vf − vᵢ = at, so the triangle is ½ × t × at = ½at².
So S = vᵢt + ½at².
Worked exampleDeriving the 3rd equation
Distance = average velocity × time: S = ½(vf + vᵢ) × t.
From the 1st equation, t = (vf − vᵢ) ÷ a.
So S = (vf + vᵢ)(vf − vᵢ) ÷ 2a = (vf² − vᵢ²) ÷ 2a, which gives 2aS = vf² − vᵢ².
Worked exampleA car moving at 10 m s⁻¹ accelerates at 2 m s⁻² for 4 s. Find its final velocity and the distance covered.
vf = vᵢ + at = 10 + 2 × 4 = 18 m s⁻¹.
S = vᵢt + ½at² = 10 × 4 + ½ × 2 × 4² = 40 + 16 = 56 m.
Check with the 3rd equation: 2aS = 2 × 2 × 56 = 224, and vf² − vᵢ² = 324 − 100 = 224.
Answer18 m s⁻¹ and 56 m
Exam tip
Write the given values first, change km h⁻¹ to m s⁻¹, then pick the equation that holds the unknown you want and nothing else unknown. "Starts from rest" means vᵢ = 0; "comes to rest" means vf = 0.
Topic 11
Motion under gravity (free fall)
Definition
Free fall
The motion of a body falling under gravity alone, with air resistance neglected.
Definition
Acceleration due to gravity (g)
The acceleration of a freely falling body near the Earth's surface: about 9.8 m s⁻², taken as 10 m s⁻² in this book.
Remember
Without air resistance, all bodies fall with the same acceleration, whatever their mass. A heavy stone and a light coin dropped together reach the ground together; a feather falls slowly only because of the air.
The equations of motion work for free fall with a replaced by g and S written as the height h:
vf = vᵢ + gt h = vᵢt + ½gt² 2gh = vf² − vᵢ²
Case
Starting values
Sign of g
Body dropped (falling down)
vᵢ = 0
+10 m s⁻² (speeds up)
Body thrown straight up
vf = 0 at the highest point
−10 m s⁻² (slows down)
Worked exampleAn iron bob falls freely for 4 s. Find the height it falls and its speed on hitting the ground. (Solved Example 2.5)
Given: vᵢ = 0, t = 4 s, g = 10 m s⁻².
h = vᵢt + ½gt² = 0 + ½ × 10 × 4² = 80 m.
vf = vᵢ + gt = 0 + 10 × 4 = 40 m s⁻¹.
Answer80 m and 40 m s⁻¹
Worked exampleAn arrow is shot straight up at 30 m s⁻¹. Find the time to reach the top and the maximum height. (Solved Example 2.6)
Given: vᵢ = 30 m s⁻¹, vf = 0 at the top, g = −10 m s⁻².
Time: vf = vᵢ + gt gives 0 = 30 − 10t, so t = 3 s.
Height: 2gh = vf² − vᵢ² gives 2(−10)h = 0 − 30², so h = 900 ÷ 20 = 45 m.
Answer3 s and 45 m
For a ball thrown up and caught again, velocity falls in a straight line from +v to −v; the slope is the constant −g.
Exam tip
Going up and coming down take the same time, and the ball comes back to your hand at the speed it was thrown. At the top its velocity is zero but its acceleration is still g downward.
The wording follows published solved exercises and every answer was re-checked by hand, so the wording or the order of options may differ slightly from your print.
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Set A: Textbook exercise 10 MCQs
The multiple-choice questions at the end of Chapter 2 in your book, in the book's option order.
Q2.1Distance, Displacement, Speed and Velocity
The ratio of displacement to distance is:
Answer(d) equal to or less than one
Step by step
Distance is the length of the whole path; displacement is the straight line from start to end.
A straight line is the shortest way between two points, so displacement ≤ distance.
They are equal only for straight-line motion in one direction, so the ratio is equal to or less than one.
Why the others are wrong
(a) for straight-line motion in one direction the two are equal, so the ratio can be exactly 1.
(b) on a curved path or a round trip, displacement is smaller than distance.
(c) displacement can never be greater than distance.
Remember
Displacement ≤ distance, always.
Q2.2Rest and Types of Motion
If an object keeps the same position with respect to a fixed point, it is said to be:
Answer(a) at rest
Step by step
Rest means no change of position with respect to the surroundings (a fixed point).
Motion, uniform or variable, needs the position to change.
Why the others are wrong
(b) motion needs a change of position.
(c) uniform motion is still motion: the position changes.
(d) variable motion is still motion: the position changes.
Remember
No change in position = rest.
Q2.3Motion under Gravity
A ball is dropped from the top of a tower. The distance it covers in the first second is (take g = 10 m s⁻²):
Answer(a) 5 m
Step by step
Dropped from rest: vᵢ = 0, a = g = 10 m s⁻², t = 1 s.
S = vᵢt + ½gt² = 0 + ½ × 10 × 1².
S = 5 m.
Why the others are wrong
(b) 10 m forgets the ½ in S = ½gt²; 10 m s⁻¹ is the speed after 1 s, not the distance.
(c) far too large for 1 s of falling from rest.
(d) far too large for 1 s of falling from rest.
Remember
From rest, a falling body covers 5 m in the first second.
Q2.4Equations of Motion
A body starts from rest and accelerates uniformly to 144 km h⁻¹ in 20 s. The distance it covers is:
Answer(b) 400 m
Step by step
Change the unit: 144 km h⁻¹ ÷ 3.6 = 40 m s⁻¹.
Acceleration a = (40 − 0) ÷ 20 = 2 m s⁻².
S = vᵢt + ½at² = 0 + ½ × 2 × 20² = 400 m.
Check: average speed (0 + 40) ÷ 2 = 20 m s⁻¹, and 20 × 20 = 400 m.
Why the others are wrong
(a) 100 m does not come from these numbers.
(c) 1400 m comes from mixing km h⁻¹ and seconds.
(d) 1440 m uses 144 km h⁻¹ as if it were m s⁻¹ and halves it (72 × 20); the unit must be changed first.
Remember
Always turn km h⁻¹ into m s⁻¹ first: divide by 3.6.
Q2.5Equations of Motion
A body starting from rest moves with constant acceleration and covers a distance S in 4 s. How long does it take to cover the first S/4?
Answer(b) 2 s
Step by step
From rest: S = ½a × 4² = 8a.
For S/4 = 2a: 2a = ½at², so t² = 4.
t = 2 s.
Why the others are wrong
(a) 1 s would cover only S/16.
(c) 4 s covers the whole distance S.
(d) 16 s would cover 16 S.
Remember
From rest, distance ∝ time²: ¼ of the distance takes ½ the time.
Q2.6Distance-Time and Speed-Time Graphs
On a displacement-time graph, the line for object A slopes upward while the line for object B is parallel to the time axis. Which statement is true?
Answer(a) The velocity of A is greater than that of B
Step by step
Gradient of a displacement-time graph = velocity.
B's line is parallel to the time axis, so its gradient is zero: B is at rest.
A's line slopes upward, so A has a positive velocity, greater than B's.
Why the others are wrong
(b) B is at rest, so nothing is slower than B here.
(c) the slopes are different, so the velocities are different.
(d) the slope of the graph is exactly the velocity.
Remember
Steeper line = faster; flat line = at rest.
Q2.7Distance-Time and Speed-Time Graphs
The area under a speed-time graph is equal to:
Answer(d) distance covered
Step by step
The area under the line has units of (m s⁻¹) × s = m.
For constant speed it is a rectangle v × t, which is the distance.
So the area under a speed-time graph is the distance covered.
Why the others are wrong
(a) the value read off the y-axis is the speed; the area is not.
(b) the area has units of metres, not m s⁻¹.
(c) acceleration is the gradient (slope), not the area.
Remember
Slope gives acceleration; area gives distance.
Q2.8Distance-Time and Speed-Time Graphs
The gradient (slope) of a speed-time graph is equal to:
Answer(c) acceleration
Step by step
Gradient = rise ÷ run = Δv ÷ Δt.
Change in speed per unit time is acceleration (m s⁻²).
Why the others are wrong
(a) speed is read straight off the y-axis.
(b) velocity is the slope of a displacement-time graph.
(d) distance is the area under the graph.
Remember
Speed-time: slope = acceleration, area = distance.
Q2.9Distance-Time and Speed-Time Graphs
The gradient (slope) of a distance-time graph is equal to:
Answer(a) speed
Step by step
Gradient = ΔS ÷ Δt, where S here is distance.
Distance per unit time is speed, a scalar.
Velocity would need displacement on the y-axis, not distance.
Why the others are wrong
(b) velocity is the slope of a displacement-time graph. Some answer keys mark (b) here; the textbook's own definition gives speed.
(c) distance is read off the y-axis.
(d) acceleration is the slope of a speed-time graph.
A scalar quantity is fully described by its magnitude alone, for example mass, time and speed.
A vector quantity needs both magnitude and direction to be fully described, for example displacement, velocity and force.
Short Q 2.2
Give 5 examples each of scalar and vector quantities.
Answer
Scalars
Vectors
Distance
Displacement
Speed
Velocity
Mass
Weight
Time
Acceleration
Energy
Force
Short Q 2.3
State the head-to-tail rule for the addition of vectors.
Answer
Draw the vectors to scale, joining the tail of each vector to the head of the one before it. The resultant is the vector drawn from the tail of the first vector to the head of the last vector.
R = A + B, from the tail of A to the head of B.
Short Q 2.4
What are a distance-time graph and a speed-time graph?
Answer
A distance-time graph shows how the distance covered by a body changes with time (time on the x-axis, distance on the y-axis). Its gradient gives the speed.
A speed-time graph shows how the speed of a body changes with time (time on the x-axis, speed on the y-axis). Its gradient gives the acceleration and the area under it gives the distance.
Short Q 2.5
Falling objects near the Earth have the same constant acceleration. Does this mean that a heavy object will fall faster than a light one?
Answer
No. Since every object falls with the same acceleration g (about 10 m s⁻²), whatever its mass, a heavy and a light object dropped together from the same height reach the ground together.
In everyday life a feather or a sheet of paper falls more slowly only because air resistance affects it much more; in a vacuum it would fall as fast as a stone.
Short Q 2.6
Vector quantities are sometimes written in scalar notation (not bold). How is the direction shown then?
Answer
The direction is stated separately: in words (for example "20 m s⁻¹ towards the north"), as an angle (for example "at 30° to the x-axis"), or with a + or − sign for the two directions along a straight line.
Short Q 2.7
A body is moving with uniform speed. Will its velocity be uniform? Give a reason.
Answer
Not necessarily. Velocity has both magnitude and direction. If the body moves in a straight line, its velocity is uniform too.
But if its direction changes, for example a car going round a circular track at a steady 40 km h⁻¹, its velocity changes all the time even though its speed is uniform.
Short Q 2.8
Is it possible for a body to have acceleration when moving with (i) constant velocity (ii) constant speed?
Answer
(i) No. Acceleration is the rate of change of velocity. If the velocity is constant, it does not change, so the acceleration is zero.
(ii) Yes. If the speed is constant but the direction keeps changing, as in circular motion, the velocity changes, so the body is accelerating.
These test whether you really understand. Give a reason every time.
CRQ 2.1
Explain why distance and displacement may or may not be equal in magnitude.
Answer
Distance is the length of the actual path, while displacement is the straight line from the start to the end.
Equal: when a body moves along a straight line in one direction, its path is itself that straight line, so the two are equal.
Not equal: when the path bends or the body turns back, the path is longer than the straight line, so distance is greater than displacement. After a full round trip the displacement is zero.
The curved path is longer than the straight arrow.
CRQ 2.2
Two guns are identical except that one has a longer barrel, and the longer gun gives its bullet a lower muzzle velocity. Which gun gives the bullet the greater acceleration? Explain.
Answer
The shorter gun. Inside the barrel the bullet starts from rest and speeds up until it leaves the muzzle.
From 2aS = vf² − vᵢ² with vᵢ = 0, the acceleration is a = vf² ÷ 2S, where S is the barrel length and vf the muzzle velocity.
The shorter gun has the larger vf and the smaller S, so vf² ÷ 2S is larger: its bullet gains more velocity in less distance, which means a greater acceleration.
Key idea
Acceleration is about how quickly velocity changes, not how far the bullet travels.
CRQ 2.3
The average velocity of a body for a complete trip is positive. Is it possible for its instantaneous velocity to be negative at some time during the trip?
Answer
Yes. Average velocity depends only on the overall displacement from start to end divided by the total time. The body could move backwards (negative velocity) for part of the trip and still end up ahead of its starting point.
Example: a car goes 10 km forward, reverses 2 km, then goes 5 km forward. Its overall displacement is +13 km, so its average velocity is positive, but while reversing its velocity was negative.
CRQ 2.4
A ball is thrown straight up with speed v and comes back to the thrower's hand after time T. Which velocity-time graph describes its motion correctly? Give a reason.
Answer
The correct graph: one straight line from +v to −v, crossing zero at T/2.
Throughout the flight the only acceleration is g, downward and constant. A constant acceleration gives a straight line of constant negative slope on a velocity-time graph.
The velocity starts at +v, falls to zero at the top (at time T/2), becomes negative as the ball comes down, and reaches −v at time T.
Check your book
Your book shows four graphs. Choose the one that is a single straight line falling from +v through zero to −v; reject any that bend, stop at zero, or jump.
CRQ 2.5
The distance-time graph of a cyclist has three parts, a, b and c. Find the velocity in each part.
Answer
a: 0 to 2 km in 6 min. b: stays at 2 km from 6 to 10 min. c: back to 0 km at 20 min.
Part
Working
Velocity
a
2 km ÷ 6 min
0.33 km min⁻¹ (= 20 km h⁻¹)
b
No change in distance from 6 to 10 min
0 (the cyclist is at rest)
c
(0 − 2) km ÷ (20 − 10) min
−0.2 km min⁻¹ (= −12 km h⁻¹)
The negative sign in part c means the cyclist is coming back towards the starting point.
Check your book
Read the values off the graph in your book; if its numbers differ, use the same method: velocity = change in distance ÷ change in time for each part.
CRQ 2.6
Can an object have zero velocity and still be accelerating? Give an example.
Answer
Yes. A ball thrown straight up stops for an instant at its highest point, so its velocity is zero there.
But gravity still acts on it, so its acceleration at that moment is g = 10 m s⁻² downward. That is why it starts falling straight away.
Set A: Textbook comprehensive questions 7 questions
The long questions at the end of the chapter in your book.
Long Q 2.1
How can a vector be represented graphically? Explain.
Answer
Vectors need direction
A vector quantity has both magnitude and direction, so it cannot be shown by a number alone. On paper it is drawn as a straight line with an arrowhead.
What the drawing shows
The length of the line, on a chosen scale, shows the magnitude.
The arrowhead shows the direction.
The angle of the line, measured from a reference line (x-axis or north), fixes the direction exactly.
Steps
Choose a suitable scale, for example 1 cm = 100 N.
Draw the reference directions (x-y axes, or north and east).
Work out the length: magnitude ÷ scale.
From the origin, draw a line of that length at the given angle.
Put an arrowhead at the end.
Worked exampleRepresent a force of 350 N at 60° to the x-axis
Scale: 1 cm = 100 N.
Length = 350 ÷ 100 = 3.5 cm.
Draw a 3.5 cm line at 60° to the x-axis and add an arrowhead.
The 350 N force drawn to scale.
Notation
In print, a vector is shown in bold (F) or with an arrow over the letter.
Long Q 2.2
Differentiate between: (i) rest and motion (ii) speed and velocity.
Answer
(i) Rest and motion
Rest
Motion
A body does not change its position with respect to its surroundings
A body changes its position with respect to its surroundings
Example: a book lying on a table
Example: a car moving on a road
Relative: a passenger is at rest in a moving bus
Relative: the same passenger is moving with respect to the road
Remember
Rest and motion are relative: they depend on what the body is compared with.
(ii) Speed and velocity
Speed
Velocity
Distance covered in unit time
Displacement covered in unit time (rate of change of displacement)
Scalar: magnitude only
Vector: magnitude and direction
v = distance ÷ time
v = displacement ÷ time
Never negative
Can be negative (opposite direction)
Same on a circular track at a steady pace
Changes on a circular track, because the direction changes
SI unit m s⁻¹
SI unit m s⁻¹
Long Q 2.3
Describe the different types of motion, with examples.
Answer
Introduction
Bodies move in different ways. Motion is divided into three types.
The three types of motion.
1. Translatory motion
Definition
Translatory motion
Motion in which every particle of the body moves the same distance in the same direction, so the whole body shifts from one place to another.
Kind
Path
Examples
Linear
Straight line
A car on a straight road; a stone falling
Circular
A circle around a point
A stone whirled on a string; a car on a round track
Random
Irregular, no pattern
A flying bee; dust particles in air
2. Rotatory motion
Definition
Rotatory motion
The body spins about an axis passing through itself.
Examples: a ceiling fan, a spinning top, a wheel, the Earth spinning on its axis.
3. Vibratory motion
Definition
Vibratory motion
The body moves to and fro about a mean position.
Examples: a swing, a simple pendulum, the string of a guitar.
Exam tip
Circular motion is around a point outside the body; rotatory motion is about an axis through the body.
Long Q 2.4
Explain the difference between distance and displacement.
Answer
Distance
Definition
Distance
The total length of the path actually travelled, whatever its direction. It is a scalar; SI unit metre.
Example: the road from Lahore to Multan is about 320 km long, so a car driving it covers a distance of about 320 km.
Displacement
Definition
Displacement
The shortest straight-line distance from the starting point to the final point, in the direction from start to end. It is a vector; SI unit metre.
The path (distance) and the straight arrow (displacement).
Differences
Distance
Displacement
Scalar
Vector
Depends on the path taken
Depends only on the start and end points
Always positive
Can be zero after a round trip
Always ≥ displacement
Always ≤ distance
Worked exampleOne lap of a 400 m track
The runner covers the whole track: distance = 400 m.
She ends where she started: displacement = 0.
Remember
They are equal only for straight-line motion in one direction.
Long Q 2.5
What do the gradients of a distance-time graph and a speed-time graph represent? Explain by drawing diagrams.
Answer
Gradient (slope)
The gradient of a straight line is how steeply it rises: rise ÷ run. To find it, pick two points on the line and draw a right-angled triangle under it.
Gradient of a distance-time graph = speed
Rise ΔS over run Δt.
gradient = ΔS ÷ Δt = speed
A steeper line means a greater speed. A flat (horizontal) line means the distance is not changing: the body is at rest. A curve getting steeper means the speed is increasing.
Gradient of a speed-time graph = acceleration
Flat, rising and falling speed-time graphs.
gradient = Δv ÷ Δt = acceleration
A rising straight line means uniform acceleration, a falling one uniform deceleration, and a horizontal line zero acceleration (constant speed).
Worked exampleExample
A speed-time line rises from 0 to 20 m s⁻¹ in 10 s.
Gradient = 20 ÷ 10.
Answeracceleration = 2 m s⁻²
Long Q 2.6
Prove that the area under a speed-time graph is equal to the distance covered by an object.
Answer
Idea
Distance = speed × time. On a speed-time graph, speed is the height and time is the width, so speed × time is an area.
The two cases.
Case 1: constant speed
Worked exampleA body moves at constant speed v for time t
The graph is a horizontal line at height v, so the area under it is a rectangle.
Area = length × breadth = t × v.
Distance = speed × time = v × t.
Answerarea = distance
Case 2: uniform acceleration from rest
Worked exampleA body speeds up uniformly from 0 to v in time t
The graph is a straight line from 0 to v, so the area under it is a triangle.
Area = ½ × base × height = ½ × t × v.
Average speed = (0 + v) ÷ 2 = v/2, so distance = (v/2) × t = ½vt.
Answerarea = distance
Conclusion
In every case the area between a speed-time graph and the time axis equals the distance covered. For more complicated graphs, split the area into rectangles and triangles and add them.
Long Q 2.7
How can the equations of motion be applied to bodies moving under the action of gravity?
Answer
Free fall
Near the Earth, a freely falling body (air resistance ignored) has a uniform acceleration g ≈ 10 m s⁻², the same for every body. So the equations of motion apply, with a replaced by g and S by height h.
vf = vᵢ + gt h = vᵢt + ½gt² 2gh = vf² − vᵢ²
Sign rules
Case
Known values
g taken as
Body dropped (falling down)
vᵢ = 0
+10 m s⁻² (it speeds up)
Body thrown up (going up)
vf = 0 at the top
−10 m s⁻² (it slows down)
Example 1: a body falling
Worked exampleA stone is dropped from a height of 80 m. How long does it take to fall?
h = vᵢt + ½gt² with vᵢ = 0: 80 = ½ × 10 × t².
t² = 16.
Answert = 4 s
Example 2: a body thrown up
Worked exampleA ball is thrown up at 20 m s⁻¹. How high does it go?
At the top vf = 0; g = −10 m s⁻².
2gh = vf² − vᵢ²: 2(−10)h = 0 − 400.
Answerh = 20 m
Velocity of a ball thrown up: a straight line of slope −g.
Remember
The time going up equals the time coming down, and at the top the velocity is zero but the acceleration is still g downward.
Set B: More important long questions 6 questions
Long questions often set in board papers from this chapter, answered in the same full style.
Imp. LQ 1
Define speed, velocity and acceleration. Give their units and explain their types.
Answer
Speed
Definition
Speed
Distance covered in unit time: v = S ÷ t. Scalar; SI unit m s⁻¹.
Uniform speed: equal distances in equal intervals of time.
Variable speed: unequal distances in equal intervals.
Average speed: total distance ÷ total time.
Velocity
Definition
Velocity
Rate of change of displacement, in a stated direction. Vector; SI unit m s⁻¹.
Uniform velocity: equal displacements in equal intervals, in the same direction.
Variable velocity: the speed or the direction (or both) changes.
Average velocity: total displacement ÷ total time.
Acceleration
Definition
Acceleration
Rate of change of velocity: a = (vf − vᵢ) ÷ t. Vector; SI unit m s⁻².
Uniform acceleration: velocity changes by equal amounts in equal intervals.
Variable acceleration: the change is not equal in equal intervals.
Deceleration (retardation): velocity decreases; a is negative.
Worked exampleAn aeroplane reaches 252 km h⁻¹ from rest in 20 s
vf = 252 ÷ 3.6 = 70 m s⁻¹.
a = (70 − 0) ÷ 20.
Answer3.5 m s⁻²
Imp. LQ 2
Derive the three equations of motion by the graphical method.
Answer
Set-up
A body moves in a straight line with uniform acceleration a. Its velocity is vᵢ at time 0 and vf at time t, and it covers distance S. Its speed-time graph is a straight line from vᵢ up to vf.
The area under the line splits into a rectangle and a triangle.
1st equation: vf = vᵢ + at
Worked exampleFrom the gradient
Gradient of the speed-time line = acceleration: a = (vf − vᵢ) ÷ t.
Multiply both sides by t: at = vf − vᵢ.
Answervf = vᵢ + at
2nd equation: S = vᵢt + ½at²
Worked exampleFrom the area
Distance = area under the line = rectangle + triangle.
Rectangle = base × height = t × vᵢ.
Triangle = ½ × base × height = ½ × t × (vf − vᵢ) = ½ × t × at = ½at².
AnswerS = vᵢt + ½at²
3rd equation: 2aS = vf² − vᵢ²
Worked exampleFrom the trapezium area
The whole area is a trapezium: S = ½(vᵢ + vf) × t.
From the 1st equation, t = (vf − vᵢ) ÷ a.
S = ½(vf + vᵢ)(vf − vᵢ) ÷ a = (vf² − vᵢ²) ÷ 2a.
Answer2aS = vf² − vᵢ²
Remember
These equations hold only for uniform acceleration in a straight line.
Imp. LQ 3
Explain distance-time graphs for a body (i) at rest (ii) moving with uniform speed (iii) moving with variable speed. How is speed found from such a graph?
Answer
What the graph shows
Time is on the x-axis and distance on the y-axis. The shape of the line shows how the body moves.
Three distance-time graphs.
(i) Body at rest
The distance does not change as time passes, so the graph is a horizontal straight line parallel to the time axis. Its gradient is zero: speed = 0.
(ii) Uniform speed
Equal distances are covered in equal times, so the graph is a straight line sloping upward. A steeper line means a greater speed.
(iii) Variable speed
The graph is a curve. If it gets steeper, the speed is increasing; if it gets flatter, the speed is decreasing.
Finding the speed
Gradient = ΔS ÷ Δt.
speed = gradient = ΔS ÷ Δt
Worked exampleA straight distance-time line rises from 20 m to 80 m between t = 2 s and t = 8 s
ΔS = 80 − 20 = 60 m; Δt = 8 − 2 = 6 s.
Speed = 60 ÷ 6.
Answer10 m s⁻¹
Imp. LQ 4
Explain speed-time graphs for a body moving with (i) constant speed (ii) uniform acceleration (iii) uniform deceleration. How are acceleration and distance found from them?
Answer
What the graph shows
Time is on the x-axis and speed on the y-axis.
Three speed-time graphs.
(i) Constant speed
The speed does not change: a horizontal line. Acceleration = 0.
(ii) Uniform acceleration
The speed rises by equal amounts in equal times: a straight line sloping upward.
(iii) Uniform deceleration
The speed falls by equal amounts in equal times: a straight line sloping downward.
Finding acceleration and distance
acceleration = gradient = Δv ÷ Δt
distance = area under the graph
Worked exampleA car speeds up uniformly from 10 m s⁻¹ to 30 m s⁻¹ in 4 s
Acceleration = (30 − 10) ÷ 4 = 5 m s⁻².
Distance = area of the trapezium = ½ × (10 + 30) × 4 = 80 m.
Answer5 m s⁻² and 80 m
Imp. LQ 5
What are scalars and vectors? Explain the head-to-tail rule for adding vectors with an example.
Answer
Scalars and vectors
Definition
Scalar
A quantity fully described by its magnitude alone: mass, time, distance, speed, energy.
Definition
Vector
A quantity that needs magnitude and direction: displacement, velocity, acceleration, force, weight.
Resultant vector
Definition
Resultant vector
The single vector that has the same effect as two or more vectors together.
Head-to-tail rule
Draw the first vector to scale in its direction.
Draw the second vector to the same scale, its tail starting at the head of the first.
Repeat for any more vectors.
Join the tail of the first to the head of the last: this is the resultant. Measure its length and angle.
R = A + B.
Worked exampleA girl walks 3 km west and then 4 km south
Draw 3 km west, then 4 km south from its head (scale 1 cm = 1 km).
The resultant joins start to end; the two walks are at right angles, so R = √(3² + 4²) = 5 km.
Answerdisplacement = 5 km towards the south-west (the distance walked is 7 km)
Imp. LQ 6
What is free fall? Describe the motion of a ball thrown vertically upward.
Answer
Free fall
Definition
Free fall
The motion of a body under gravity alone, with air resistance neglected. Every freely falling body has the same acceleration g ≈ 10 m s⁻² downward.
Going up
The ball leaves the hand with velocity vᵢ upward. Gravity acts downward, opposite to the motion, so the ball slows down by 10 m s⁻¹ every second (g = −10 m s⁻²).
At the highest point
The velocity becomes zero for an instant. The acceleration is still g downward, so the ball does not stay there.
Coming down
Now gravity acts in the direction of motion, so the ball speeds up by 10 m s⁻¹ every second, and reaches the hand at the speed it was thrown.
Velocity falls in a straight line from +v through zero to −v.
Represent (a) a velocity of 400 m s⁻¹ at 60° to the x-axis and (b) a force of 50 N at 120° to the x-axis.
Given Data
Scale for (a): 1 cm = 100 m s⁻¹
Scale for (b): 1 cm = 10 N
To Find
Length of each arrow, and the drawing
Formula
length = magnitude ÷ scale
Calculation
(a) Length = 400 ÷ 100 = 4 cm. Draw a 4 cm line at 60° from the positive x-axis and add an arrowhead.
(b) Length = 50 ÷ 10 = 5 cm. Draw a 5 cm line at 120° from the positive x-axis (it points up and to the left) and add an arrowhead.
Result
(a) a 4 cm arrow at 60° (b) a 5 cm arrow at 120°
Note: Any sensible scale is fine, as long as you state it and the length matches it.
Numerical 2.2
A car is moving at an average speed of 72 km h⁻¹. How much time will it take to cover a distance of 360 km?
Given Data
v = 72 km h⁻¹
S = 360 km
To Find
Time t
Formula
v = S ÷ t, so t = S ÷ v
Calculation
From v = S ÷ t, the time t = S ÷ v.
t = 360 km ÷ 72 km h⁻¹ = 5 h.
Here both values are already in km and hours, so no conversion is needed.
Result
t = 5 hours
Numerical 2.3
A truck starts from rest and reaches a velocity of 90 km h⁻¹ in 50 s. Find its average acceleration.
Given Data
vᵢ = 0
vf = 90 km h⁻¹ = 90 ÷ 3.6 = 25 m s⁻¹
t = 50 s
To Find
Average acceleration a
Formula
a = (vf − vᵢ) ÷ t
Calculation
a = (vf − vᵢ) ÷ t.
a = (25 − 0) ÷ 50.
Result
a = 0.5 m s⁻²
Numerical 2.4
A car is moving at 5 m s⁻¹ and speeds up with an acceleration of 1.5 m s⁻². What is its velocity after 5 s?
Given Data
vᵢ = 5 m s⁻¹
a = 1.5 m s⁻²
t = 5 s
To Find
Final velocity vf
Formula
vf = vᵢ + at
Calculation
vf = vᵢ + at.
vf = 5 + 1.5 × 5 = 5 + 7.5.
Result
vf = 12.5 m s⁻¹
Numerical 2.5
A motorcycle moving at 18 km h⁻¹ accelerates at 2 m s⁻². How far will it travel in 10 s?
Given Data
vᵢ = 18 km h⁻¹ = 18 ÷ 3.6 = 5 m s⁻¹
a = 2 m s⁻²
t = 10 s
To Find
Distance S
Formula
S = vᵢt + ½at²
Calculation
S = vᵢt + ½at².
S = 5 × 10 + ½ × 2 × 10².
S = 50 + 100.
Result
S = 150 m
Numerical 2.6
A wagon moving at 54 km h⁻¹ is brought to rest within 25 m. Find its acceleration.
Given Data
vᵢ = 54 km h⁻¹ = 54 ÷ 3.6 = 15 m s⁻¹
vf = 0 (comes to rest)
S = 25 m
To Find
Acceleration a
Formula
2aS = vf² − vᵢ²
Calculation
Time is not given, so use 2aS = vf² − vᵢ².
2 × a × 25 = 0² − 15² = −225.
a = −225 ÷ 50.
Result
a = −4.5 m s⁻² (a deceleration of 4.5 m s⁻²)
Note: The minus sign shows the acceleration is opposite to the motion: the wagon is slowing down.
Numerical 2.7
A stone is dropped from a height of 45 m. How long will it take to reach the ground, and with what velocity will it hit the ground?
Given Data
vᵢ = 0
h = 45 m
g = 10 m s⁻²
To Find
Time t to reach the ground
Velocity vf on hitting the ground
Formulas
h = vᵢt + ½gt²
vf = vᵢ + gt
Calculation
Time: h = vᵢt + ½gt² gives 45 = ½ × 10 × t², so t² = 9 and t = 3 s.
Velocity: vf = vᵢ + gt = 0 + 10 × 3 = 30 m s⁻¹.
Check: 2gh = 2 × 10 × 45 = 900 = 30², so vf = 30 m s⁻¹.
Result
t = 3 s, vf = 30 m s⁻¹
Numerical 2.8
A car travels 10 km at an average speed of 20 m s⁻¹ and then 0.8 km at an average speed of 4 m s⁻¹. Find its average speed for the whole journey.
Given Data
S₁ = 10 km = 10 000 m at v₁ = 20 m s⁻¹
S₂ = 0.8 km = 800 m at v₂ = 4 m s⁻¹
To Find
Average speed for the whole journey
Formulas
t = S ÷ v
average speed = total distance ÷ total time
Calculation
Time for the first part: t₁ = 10 000 ÷ 20 = 500 s.
Time for the second part: t₂ = 800 ÷ 4 = 200 s.
Average speed = total distance ÷ total time = (10 000 + 800) ÷ (500 + 200) = 10 800 ÷ 700.
Result
Average speed ≈ 15.4 m s⁻¹
Note: The average is not (20 + 4) ÷ 2 = 12 m s⁻¹: the car spends different times at each speed, so always divide total distance by total time.
Numerical 2.9
A ball is dropped from the top of a building and takes 5 s to reach the ground. Find the height of the building and the velocity with which the ball hits the ground.
Given Data
vᵢ = 0
t = 5 s
g = 10 m s⁻²
To Find
Height h of the building
Velocity vf on hitting the ground
Formulas
h = vᵢt + ½gt²
vf = vᵢ + gt
Calculation
Height: h = vᵢt + ½gt² = 0 + ½ × 10 × 5² = 125 m.
Velocity: vf = vᵢ + gt = 0 + 10 × 5 = 50 m s⁻¹.
Result
h = 125 m, vf = 50 m s⁻¹
Numerical 2.10
A ball is thrown straight up and takes 3 s to reach its highest point. Find its initial velocity and the maximum height it reaches above the ground, if it was released 1 m above the ground.
The branch of mechanics that studies the motion of objects without asking what causes the motion (the causes, forces, come in Chapter 3).
Rest
A body is at rest if it does not change its position with respect to its surroundings.
Motion
A body is in motion if it keeps changing its position with respect to its surroundings.
Translatory motion
Motion in which every particle of a body moves the same distance in the same direction, so the whole body shifts from one place to another.
Rotatory motion
Motion of a body spinning about an axis that passes through the body itself. Examples: a ceiling fan, a spinning top, a wheel, the Earth spinning on its axis.
Vibratory motion
To-and-fro motion of a body about its mean position. Examples: a swing, a simple pendulum, a plucked guitar string.
Scalar quantity
A physical quantity that is fully described by its magnitude (a number with a unit) alone.
Vector quantity
A physical quantity that needs both magnitude and direction to be fully described.
Resultant vector
The single vector that has the same effect as two or more vectors acting together.
Distance
The total length of the path a body actually travels, whatever its direction. It is a scalar; SI unit metre (m).
Displacement
The shortest straight-line distance from the starting point to the final point, in the direction from start to end. It is a vector; SI unit metre (m).
Speed
The distance covered by a body in unit time. It is a scalar; SI unit m s⁻¹.
Uniform speed
A body has uniform speed if it covers equal distances in equal intervals of time, however short the intervals are.
Average speed
Total distance travelled ÷ total time taken, for a body whose speed changes during the trip.
Velocity
The rate of change of displacement: displacement covered in unit time, in a stated direction. It is a vector; SI unit m s⁻¹.
Uniform velocity
A body has uniform velocity if it covers equal displacements in equal intervals of time in the same direction, so both its speed and direction stay the same.
Acceleration
The rate of change of velocity of a body. It is a vector; SI unit m s⁻².
Uniform acceleration
A body has uniform acceleration if its velocity changes by equal amounts in equal intervals of time.
Deceleration (retardation)
The acceleration of a body whose velocity is decreasing. It comes out negative and points opposite to the motion.
Free fall
The motion of a body falling under gravity alone, with air resistance neglected.
Acceleration due to gravity (g)
The acceleration of a freely falling body near the Earth's surface: about 9.8 m s⁻², taken as 10 m s⁻² in this book.
Common mistakes to avoid
Using km h⁻¹ in an equation without changing it to m s⁻¹ first.
Adding distance and displacement as if they were the same: displacement is the straight line, with direction.
Taking the average of two speeds instead of total distance ÷ total time.
Forgetting the ½ in S = vᵢt + ½at².
Using the final velocity instead of the change in velocity to find acceleration.
Thinking acceleration is zero at the top of a vertical throw: only the velocity is zero there.
Mixing up the graphs: distance-time slope = speed, speed-time slope = acceleration.
Leaving out the minus sign for a deceleration or for g when a body goes up.
Extra practice tap a question to see its answer
Convert 36 km h⁻¹ into m s⁻¹.Show answer
36 ÷ 3.6 = 10 m s⁻¹
Convert 25 m s⁻¹ into km h⁻¹.Show answer
25 × 3.6 = 90 km h⁻¹
A runner completes one lap of a 400 m track. Give the distance and the displacement.Show answer
Distance 400 m; displacement 0 m
A boy walks 6 m east and then 8 m north. Find his displacement.Show answer
√(6² + 8²) = 10 m towards the north-east
A bus covers 150 km in 3 h. Find its average speed in m s⁻¹.Show answer
150 ÷ 3 = 50 km h⁻¹ = 50 ÷ 3.6 ≈ 13.9 m s⁻¹
A car goes from 20 m s⁻¹ to 5 m s⁻¹ in 5 s. Find its acceleration.Show answer
(5 − 20) ÷ 5 = −3 m s⁻² (a deceleration)
A body starts from rest with a = 4 m s⁻². Find its velocity after 6 s.Show answer
vf = 0 + 4 × 6 = 24 m s⁻¹
For the body above, find the distance covered in those 6 s.Show answer
S = ½ × 4 × 6² = 72 m
A car at 30 m s⁻¹ brakes at −6 m s⁻². How far does it go before stopping?Show answer
2(−6)S = 0 − 30², so S = 900 ÷ 12 = 75 m
A speed-time graph rises in a straight line from 0 to 12 m s⁻¹ in 4 s. Find the distance covered.Show answer
Area = ½ × 4 × 12 = 24 m
A stone is dropped from rest. How far does it fall in 2 s?Show answer
h = ½ × 10 × 2² = 20 m
A ball is thrown up at 20 m s⁻¹. How high does it go?Show answer
h = vᵢ² ÷ 2g = 400 ÷ 20 = 20 m
How long does the ball above take to come back to the thrower?Show answer
2 s up + 2 s down = 4 s
A distance-time graph is a straight line rising 60 m in 12 s. Find the speed.Show answer
60 ÷ 12 = 5 m s⁻¹
Final self-test
Final self-test: all 24 MCQs, no answers
Give yourself 25 minutes. Pick one option for each question, then submit to see your score and which questions to revise. Any question you get wrong tells you which topic to read again.
