All 24 MCQs of 9th Class Physics Chapter 2, Kinematics, with answers and the reason each option is right or wrong: the 10 textbook exercise MCQs plus 14 more board-style MCQs. Punjab Board new syllabus.
Try each one first: tap an option to check it, then read the reasoning and why the others are wrong.
A note about the questions
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.
Answered 0 of 24 · Correct 0
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.
