Saturday, August 15, 2026

Kinematics Chapter 2 – Complete Definitions, Formulas, Derivations and Graphs

Kinematics Chapter 2: Master Revision Guide

Definitions · formulas · derivations · graphs · worked examples · common mistakes

How to use this page: Read the quick formula sheet first. For revision, try to recall each definition or derivation before opening its detailed explanation.

1. Symbols and SI Units

SymbolMeaningSI unitType
ddistancemetre (m)scalar
sdisplacementmetre (m)vector
uinitial velocitym/svector
vfinal velocitym/svector
aaccelerationm/s²vector
ttimesecond (s)scalar
gmagnitude of gravitational acceleration9.8 m/s² (approximately 10)downward

2. Quick Formula Sheet

Speed
speed = distance/time
Velocity
velocity = displacement/time
Acceleration
a = (v − u)/t
First equation
v = u + at
Second equation
s = ut + ½at²
Third equation
v² = u² + 2as
Average velocity
v̄ = total displacement/total time
nth-second distance
Sₙ = u + (a/2)(2n − 1)
Vertical projection
t↑ = u/g; H = u²/(2g); T = 2u/g
Dropped body
v = gt; h = ½gt²; v² = 2gh

3. Definitions

Motion, rest and reference point

Motion: a body is in motion when its position changes with time relative to a chosen reference point. Rest: its position does not change relative to that reference. Motion is relative: a seated passenger is at rest relative to the bus but moving relative to the road.

Scalar and vector quantities

A scalar has magnitude only: distance, speed, time and mass. A vector has magnitude and direction: displacement, velocity and acceleration. “5 m” is a scalar distance; “5 m east” is a vector displacement.

Distance and displacement

Distance is the total path length travelled. It is scalar and never negative. Displacement is the straight-line change from initial to final position, with direction. Its magnitude is always ≤ distance.

Example: walk 3 m east and 3 m west. Distance = 6 m; displacement = 0.
Speed, velocity and their averages

Speed is distance covered per unit time. Velocity is displacement per unit time in a specified direction. Average speed = total distance/total time. Average velocity = total displacement/total time. Instantaneous speed or velocity is the value at one particular instant.

Uniform and non-uniform motion

Uniform velocity: equal displacements in equal time intervals in the same direction. Non-uniform velocity: unequal displacements in equal times, or a changing direction. Uniform speed in a circle is not uniform velocity because direction continuously changes.

Acceleration and retardation

Acceleration is rate of change of velocity: a = Δv/Δt. It occurs when speed changes, direction changes, or both. Retardation/deceleration means acceleration opposite to the velocity, causing speed to decrease. Negative acceleration does not automatically mean slowing; the relative directions matter.

Types of motion
  • Translational: the whole body changes position; rectilinear along a straight line or curvilinear along a curve.
  • Rotational: body turns about an axis, such as a wheel.
  • Circular: object follows a circular path.
  • Oscillatory: repeated to-and-fro motion about a mean position, such as a pendulum.
  • Random: irregular path, such as a dust particle in air.
Projectile, trajectory and free fall

A projectile is an object that, after being projected, moves under gravity alone (air resistance ignored). Its path is the trajectory. An oblique projectile is launched at an angle to the horizontal. Free fall is motion under gravity alone. Near Earth, all freely falling bodies have the same acceleration g if air resistance is neglected.

4. Equations of Uniformly Accelerated Motion

Condition: These equations apply only when acceleration is constant and motion is along one straight line. Choose a positive direction first and keep the signs consistent.
Derivation 1: v = u + at

Acceleration is change in velocity per unit time:

a = (v − u)/t

at = v − u

v = u + at

Example: u = 4 m/s, a = 3 m/s², t = 5 s. Then v = 4 + 3(5) = 19 m/s.
Derivation 2: s = ut + ½at² (average-velocity method)

For constant acceleration, average velocity = (u + v)/2.

s = [(u + v)/2]t

Using v = u + at:

s = [(u + u + at)/2]t = [u + ½at]t

s = ut + ½at²

Example: u = 5 m/s, a = 2 m/s², t = 4 s. s = 5(4) + ½(2)(4²) = 36 m.
Graphical derivation: s = ut + ½at²
tvutrectangle = uttriangle = ½(v − u)t

Displacement equals area under the velocity-time graph:

s = rectangle + triangle = ut + ½(v − u)t. Since v − u = at, s = ut + ½at².

Derivation 3: v² = u² + 2as

Start with s = [(u + v)/2]t and use t = (v − u)/a:

s = [(u + v)/2][(v − u)/a]

2as = (u + v)(v − u) = v² − u²

v² = u² + 2as

Example: u = 3 m/s, a = 2 m/s², s = 10 m. v² = 9 + 40 = 49, so v = 7 m/s.
Distance travelled in the nth second

Total displacement after n seconds: sₙ = un + ½an². After n − 1 seconds: sₙ₋₁ = u(n − 1) + ½a(n − 1)².

Sₙ = sₙ − sₙ₋₁

Sₙ = u + (a/2)(2n − 1)

Example: u = 3 m/s, a = 2 m/s², n = 4. S₄ = 3 + (2/2)(7) = 10 m.

5. Motion Graphs

Displacement–timeslope = velocitytimes Velocity–timeslope = accelerationarea = displacementtimev Acceleration–timearea = change in vtimea At rest: s–thorizontal: position fixed Constant velocity: v–thorizontal: a = 0
  • s–t graph: slope gives velocity; horizontal means rest; straight slope means uniform velocity; curved line means changing velocity.
  • v–t graph: slope gives acceleration; signed area gives displacement; horizontal means constant velocity.
  • a–t graph: signed area gives change in velocity; line on the time axis means zero acceleration.

6. Gravity and Vertical Motion

Sign convention for g

The sign depends on the chosen positive direction, not on whether the object is moving upward or downward.

  • If upward is positive: a = −g throughout.
  • If downward is positive: a = +g throughout.

At maximum height v = 0, but acceleration remains g downward.

Body dropped from rest

Put u = 0 and a = g in the motion equations:

v = gt; h = ½gt²; v² = 2gh.

Example: dropped for 3 s with g = 9.8 m/s². v = 29.4 m/s downward and h = 44.1 m.
Vertical projection: time of ascent, maximum height and total time

Choose upward positive, so a = −g. At the top, v = 0.

From v = u − gt: t↑ = u/g.

From v² = u² − 2gH: H = u²/(2g).

If it returns to the same level, descent time equals ascent time: T = 2u/g.

Eliminating u using u = gT/2 gives H = gT²/8.

Example: u = 19.6 m/s. t↑ = 2 s, H = 19.6 m and total time T = 4 s.

7. Average Speed and Velocity

Equal time intervals

If velocities v₁ and v₂ act for equal times, average velocity = (v₁ + v₂)/2.

10 m/s for 4 s and 18 m/s for 4 s gives average = 14 m/s.
Equal distances

For equal distances travelled at speeds v₁ and v₂:

Let each distance be d. Total distance = 2d; total time = d/v₁ + d/v₂.

average speed = 2v₁v₂/(v₁ + v₂)

Equal distances at 40 and 60 km/h: average speed = 2(40)(60)/(100) = 48 km/h. If the trip returns to its start, average velocity = 0.
Three equal distances

For equal distances at speeds v₁, v₂ and v₃:

average speed = 3/(1/v₁ + 1/v₂ + 1/v₃).

At 10, 20 and 30 km/h: average = 3/(1/10 + 1/20 + 1/30) = 180/11 ≈ 16.36 km/h.

8. Projectile Motion Essentials

uu cos θu sin θmaximum heightrange R

Ignoring air resistance, horizontal velocity remains constant while vertical velocity changes due to gravity. Components: uₓ = u cosθ, uᵧ = u sinθ. For launch and landing at the same level:

  • time of flight: T = 2u sinθ/g
  • maximum height: H = u²sin²θ/(2g)
  • range: R = u²sin2θ/g

Range is maximum at 45° because sin90° = 1. Complementary angles such as 30° and 60° have equal range because sin60° = sin120°, although 60° goes higher and stays airborne longer.

9. Common Mistakes Checklist

  • Do not confuse distance with displacement or speed with velocity.
  • Convert units before substituting: 1 km/h = 5/18 m/s.
  • Choose a positive direction and keep signs consistent.
  • At maximum height, velocity is zero but acceleration is not.
  • The standard three equations require constant acceleration.
  • Area below a velocity-time axis is negative displacement.
  • For average speed, use total distance/total time—not usually the simple arithmetic mean.
  • In circular motion, constant speed can still mean changing velocity.

10. Mixed Worked Revision Example

Problem: A car begins at 5 m/s and accelerates uniformly at 2 m/s² for 6 s. Find final velocity, displacement and distance in the sixth second.

Final velocity: v = u + at = 5 + 2(6) = 17 m/s.
Displacement: s = ut + ½at² = 5(6) + ½(2)(36) = 66 m.
Sixth-second distance: S₆ = u + (a/2)(2×6 − 1) = 5 + 1(11) = 16 m.
Check: distance in first 5 s = 5(5) + ½(2)(25) = 50 m; 66 − 50 = 16 m.
Final memory map: slope of s–t → velocity; slope of v–t → acceleration; area under v–t → displacement; area under a–t → change in velocity.

Back to top

Kinematics Chapter 2 – Questions 45–50 and Concept Application 1–25

Kinematics Revision: Questions 45–50 and Concept Application 1–25

Class 9 Physics · Chapter 2 · Checked answers with step-by-step corrections

How to revise: Attempt each question first, then open the answer. Green indicates correct work, amber means partially correct, and red identifies an incorrect or unanswered item.

Result: Test Your Concepts — 4 correct, 1 partial, 1 unanswered. Concept Application — 21 correct, 2 partial, 2 incorrect/unanswered.

Sections: Test Your Concepts 45–50 · Concept Application 1–25

Test Your Concepts

Question 45: Give the equations of motion of a body falling under gravity after being dropped from a certain height.

Status: Correct
Show answer and explanation
Because the body is dropped, initial velocity u = 0. Taking downward as positive and replacing a by g: v = gt, h = ½gt², and v² = 2gh. These are the usual equations of uniformly accelerated motion specialized for free fall.

Question 46: Explain the characteristics of displacement–time, velocity–time and acceleration–time graphs.

Status: Partially correct
Show answer and explanation
Displacement–time: slope gives velocity; a horizontal line means rest; a straight sloping line means uniform velocity; changing slope means non-uniform velocity.
Velocity–time: slope gives acceleration and signed area gives displacement; a horizontal line means constant velocity and zero acceleration.
Acceleration–time: signed area gives change in velocity; a line on the time axis means zero acceleration. The written answer had the main rules but needed these cases stated clearly.

Question 47: Derive the distance travelled in the nth second: Sₙ = u + (a/2)(2n − 1).

Status: Correct
Show answer and explanation
Total displacement after n seconds is sₙ = un + ½an². After (n − 1) seconds it is sₙ₋₁ = u(n − 1) + ½a(n − 1)². Therefore Sₙ = sₙ − sₙ₋₁ = u + ½a[n² − (n − 1)²] = u + (a/2)(2n − 1).

Question 48: Derive s = ut + ½at².

Status: Correct
Show answer and explanation
For uniform acceleration, average velocity = (u + v)/2. Hence s = [(u + v)/2]t. Using v = u + at gives s = [(2u + at)/2]t = ut + ½at².

Question 49: Obtain s = ut + ½at² by the graphical method.

Status: Not answered
Show answer, graph and detailed explanation

What the graph represents

Draw a velocity–time graph for a body whose velocity increases uniformly from initial velocity u at time 0 to final velocity v at time t. Uniform acceleration produces a straight sloping line. The displacement s equals the complete area under this line.

timevelocity uvt rectangle = ut triangle = ½(v − u)t
Blue rectangle: initial-velocity displacement   +   orange triangle: extra displacement caused by acceleration

Step 1 — Find displacement from graph area

s = area of rectangle + area of triangle

s = ut + ½ × t × (v − u)

Step 2 — Use the acceleration relation

a = (v − u)/t, therefore v − u = at.

Step 3 — Substitute

s = ut + ½ × t × at

s = ut + ½at²

Worked numerical example
Let u = 5 m/s, a = 2 m/s² and t = 4 s.
Final velocity: v = u + at = 5 + 2(4) = 13 m/s.
Rectangle area = ut = 5 × 4 = 20 m.
Triangle area = ½(v − u)t = ½(13 − 5)(4) = 16 m.
Total displacement = 20 + 16 = 36 m.
Formula check: s = ut + ½at² = 5(4) + ½(2)(4²) = 20 + 16 = 36 m.

Remember: this graphical derivation assumes constant acceleration, so the velocity–time graph is a straight line.

Question 50: Derive v² − u² = 2as.

Status: Correct
Show answer and explanation
For uniform acceleration, s = [(u + v)/2]t and t = (v − u)/a. Substitution gives s = [(u + v)(v − u)]/(2a) = (v² − u²)/(2a). Therefore v² − u² = 2as.

Concept Application

Question 1: A body moves with retardation when projected vertically upward. True or false?

Status: Correct
Show answer and explanation
True. During upward motion, velocity is upward but gravity acts downward. The speed therefore decreases at the constant rate g until the body reaches maximum height.

Question 2: At maximum height, the velocity of a vertically projected body becomes zero. True or false?

Status: Correct
Show answer and explanation
True. Its instantaneous velocity is zero at the turning point, but its acceleration is still downward and equal to g.

Question 3: A velocity–time graph can be used to find displacement. True or false?

Status: Correct
Show answer and explanation
True. The signed area between the velocity–time graph and the time axis equals displacement.

Question 4: For constant velocity, displacement depends on the square of time. True or false?

Status: Correct
Show answer and explanation
False. With constant velocity, s = vt, so displacement is directly proportional to time, not time squared.

Question 5: Two balls of different masses are thrown vertically upward with the same speed. The heavier ball rises higher. True or false?

Status: Correct
Show answer and explanation
False. Ignoring air resistance, H = u²/(2g). Mass is absent from the formula, so both balls reach the same height.

Question 6: The equations of motion are applicable only to uniform velocity. True or false?

Status: Correct
Show answer and explanation
False. The standard equations v = u + at, s = ut + ½at² and v² − u² = 2as apply when acceleration is uniform.

Question 7: A freely falling body travels the same distance in every successive second. True or false?

Status: Correct
Show answer and explanation
False. Its speed increases under gravity. From rest, the distances in successive seconds are proportional to 1 : 3 : 5 : 7 …

Question 8: Find the ratio of the velocities of a freely falling body after 1 s and 2 s.

Status: Correct
Show answer and explanation
Using v = gt: v₁ : v₂ = g(1) : g(2) = 1 : 2.

Question 9: A stone is thrown vertically upward and is caught at the same level after total time t. Find its maximum height.

Status: Not answered
Show answer, diagrams and detailed explanation

Answer: H = gt²/8

The stone starts and finishes at the same height. If air resistance is ignored, upward and downward motion are symmetric. Therefore the total flight time t is divided equally:

time of ascent = time of descent = t/2

Motion of the stone
launch and catch level maximum height: v = 0 up: t/2 down: t/2 H Gravity g acts downward throughout
Velocity–time graph
velocitytime ut/2t upwarddownward slope = −g
Positive triangular area = upward displacement H. The equal negative area represents the return to the starting level.

Method 1 — Use time of ascent

Take upward as positive. During ascent, acceleration is −g. At maximum height, final velocity v = 0 and ascent time = t/2.

Using v = u + at:

0 = u − g(t/2)

u = gt/2

Now use v² = u² + 2as for the upward journey, where s = H:

0 = u² − 2gH

H = u²/(2g)

Substitute u = gt/2:

H = (gt/2)²/(2g) = g²t²/(8g)

H = gt²/8

Method 2 — Read the velocity–time graph

The upward displacement is the area of the blue triangle from 0 to t/2:

H = ½ × base × height = ½ × (t/2) × u = ut/4

Using u = gt/2:

H = (gt/2)t/4 = gt²/8

Worked numerical example
Suppose the stone returns to the hand after t = 6 s. Take g = 9.8 m/s².
Time of ascent = t/2 = 6/2 = 3 s.
Initial velocity u = g(t/2) = 9.8 × 3 = 29.4 m/s.
Maximum height H = gt²/8 = 9.8 × 6²/8 = 9.8 × 36/8 = 44.1 m.
Check: H = u²/(2g) = 29.4²/(19.6) = 44.1 m.
Why “caught at the same level” matters
This tells us that the upward and downward displacements are equal in magnitude, so the ascent and descent times are equal. If the stone were caught at a different height, we could not simply use t/2 as the ascent time.

Common mistake: Do not put the complete time t directly into H = ½gt². The stone spends only t/2 reaching its maximum height; the remaining t/2 is the descent.

Assumptions: air resistance is neglected, g is constant, and the stone is caught exactly at its launch level.

Question 10: What does the area under a velocity–time graph represent?

Status: Correct
Show answer and explanation
It represents displacement. Area above the time axis is positive; area below it is negative. Adding the signed areas gives net displacement.

Question 11: What is a body projected at an angle other than 90° to the horizontal called?

Status: Partially correct
Show answer and explanation
It is specifically called an oblique projectile. “Projectile” is broadly correct, but “oblique” identifies that it has both horizontal and vertical velocity components.

Question 12: What can be said about the ratio |average velocity| / average speed?

Status: Correct
Show answer and explanation
The ratio is always less than or equal to 1, because displacement cannot exceed distance. It equals 1 only when motion is along a straight line without reversing direction.

Question 13: Compare the directions of displacement and average velocity.

Status: Correct
Show answer and explanation
They are the same, because average velocity = displacement/time and time is a positive scalar.

Question 14: What is produced when velocity changes?

Status: Correct
Show answer and explanation
A change in velocity produces acceleration. Velocity may change because its magnitude changes, its direction changes, or both.

Question 15: Match the types of motion with graphs a–h.

Status: Partially correct
Show answer and explanation
Complete matching: A–c, B–d, C–f, D–g, E–b, F–h, G–a, H–e.
A: uniform velocity; B: uniform acceleration with initial velocity; C: uniform acceleration; D: increasing acceleration; E: uniform retardation; F: acceleration decreasing steadily; G: uniform acceleration with initial displacement; H: body at rest with initial displacement. A–c and B–d were correct; the remaining matches were not attempted.

Question 16: The ratio of displacement magnitude to distance is: greater than 1, equal to 1, less than 1, or less than/equal to 1?

Status: Correct
Show answer and explanation
Correct option: 4 — less than or equal to 1. The magnitude of displacement is the shortest straight-line separation and therefore cannot exceed the actual path length.

Question 17: Two bodies fall freely through heights in the ratio 3 : 5. Find the ratio of their final velocities.

Status: Correct
Show answer and explanation
From v² = 2gh, v ∝ √h. Therefore v₁ : v₂ = √3 : √5, which is option 2.

Question 18: Find the displacement from the given velocity–time graph during 0–4 s.

Status: Correct
Show answer and explanation
Add the areas under the graph: first triangle = 10 m, rectangle = 20 m, trapezium = 15 m, and final rectangle = 10 m. Total displacement = 10 + 20 + 15 + 10 = 55 m, option 4.

Question 19: A person crosses an 8 m by 6 m rectangular hall diagonally. Find the displacement.

Status: Correct
Show answer, diagram and detailed explanation

Answer: displacement = 10 m

Displacement is the shortest straight-line distance from the starting point to the finishing point, together with its direction. Because the person crosses the rectangular hall diagonally, the displacement is the hall’s diagonal.

A C 8 m 6 m displacement d = 10 m Start at AFinish at C
The 8 m side and 6 m side form the perpendicular legs of a right triangle. The diagonal is its hypotenuse.

Step 1 — Identify the right triangle

The hall’s length and width are perpendicular:

  • horizontal side = 8 m
  • vertical side = 6 m
  • diagonal displacement = d

Step 2 — Apply the Pythagorean theorem

d² = 8² + 6²

d² = 64 + 36 = 100

d = √100 = 10 m

Why the answer is not 14 m
Adding 8 m + 6 m gives 14 m. That would be the distance travelled if the person first walked along one wall and then along the adjoining wall. But the person moves directly across the diagonal, so the straight-line displacement is 10 m.

Direction of displacement

If direction is required, let θ be the angle above the 8 m side:

tan θ = opposite/adjacent = 6/8 = 0.75

θ ≈ 36.9°

Thus the displacement vector is 10 m at about 36.9° to the 8 m side. If the multiple-choice question asks only for magnitude, the answer is simply 10 m.

Second sample
A rectangular field is 12 m long and 5 m wide. A student crosses it diagonally.
d = √(12² + 5²) = √(144 + 25) = √169 = 13 m.
This is another Pythagorean triple: 5–12–13, just as the hall uses 6–8–10.

Quick recognition: 6–8–10 is twice the familiar 3–4–5 right triangle, so its diagonal can also be recognized immediately as 10 m.

Remember: distance describes the path travelled; displacement connects the initial and final positions by the shortest straight line.

Question 20: The displacement–time graph shows straight lines for P and Q, with P steeper. Interpret the motion.

Status: Correct
Show answer and explanation
Both straight lines have constant slopes, so both bodies move with uniform speeds. P has the larger slope, so P moves faster than Q, option 3.

Question 21: A vehicle slows from 40 m/s to 10 m/s in 10 s. Find acceleration.

Status: Correct
Show answer and explanation
a = (v − u)/t = (10 − 40)/10 = −3 m/s², option 1. The negative sign indicates retardation when forward is positive.

Question 22: For a body starting from rest with uniform acceleration, how does velocity depend on time?

Status: Correct
Show answer and explanation
From v = u + at and u = 0, v = at. Thus velocity is directly proportional to time, option 1.

Question 23: At maximum height, which statement about a vertically projected body is correct?

Status: Incorrect
Show answer, diagrams and detailed explanation

Correct option: 1 — velocity is zero, but acceleration due to gravity is not zero.

At maximum height, the ball stops only for an instant before reversing direction. Its instantaneous velocity is v = 0, but Earth is still pulling it downward. Therefore its acceleration remains g downward (approximately 9.8 m/s² downward).

Most important rule
The sign of g does not change because the ball changes from moving upward to moving downward. The sign is determined entirely by which direction we choose as positive.
Gravity during the complete journey
moving upv = 0moving down g downwardg downwardg downward Gravity points downward everywhere
Velocity–time graph (upward positive)
velocity vtime v positive: moving up v negative: moving down maximum height v = 0 constant slope = a = −g

Choose a sign convention before solving

Chosen positive directionUpward velocityDownward velocityAcceleration due to gravity
Upward is positivepositivenegativea = −g throughout
Downward is positivenegativepositivea = +g throughout

So, is g positive while going up and negative while going down?

No. If upward is selected as positive, gravity is −g both while the ball goes up and while it comes down. If downward is selected as positive, gravity is +g throughout both parts. What changes at the top is the sign of velocity, not the direction of gravitational acceleration.

Worked example: upward chosen as positive
A ball is thrown upward with u = +19.6 m/s. Take g = 9.8 m/s², so a = −9.8 m/s².
Using v = u + at:
At 1 s: v = 19.6 − 9.8(1) = +9.8 m/s — moving upward.
At 2 s: v = 19.6 − 9.8(2) = 0 — maximum height.
At 3 s: v = 19.6 − 9.8(3) = −9.8 m/s — moving downward.
In all three cases, acceleration remains −9.8 m/s².
The same motion with downward positive
Initial upward velocity u = −19.6 m/s and acceleration a = +9.8 m/s².
At 1 s, v = −9.8 m/s; at 2 s, v = 0; at 3 s, v = +9.8 m/s.
The numerical signs change, but the physical motion is identical.

At maximum height:

  • velocity v = 0 for an instant;
  • acceleration has magnitude g and points downward;
  • the ball is not in equilibrium because the gravitational force mg still acts;
  • immediately afterward, it begins moving downward.

Common mistake: “The ball stops at the top, therefore acceleration is zero.” This is incorrect. Zero velocity at one instant does not imply zero acceleration. Acceleration tells us how velocity is changing; at the top, gravity changes the velocity from upward to downward.

Question 24: A body starts from rest with uniform acceleration. If the times are in the ratio 1 : 2, find the ratio of distances.

Status: Correct
Show answer and explanation
From s = ½at², distance is proportional to t². Therefore s₁ : s₂ = 1² : 2² = 1 : 4, option 2.

Question 25: A body has u = 3 m/s and a = 2 m/s². Find the distance travelled in the fourth second.

Status: Correct
Show answer and explanation
Sₙ = u + (a/2)(2n − 1). For n = 4: S₄ = 3 + (2/2)(7) = 10 m, option 1.
Key revision targets: graphical derivation of s = ut + ½at²; acceleration at maximum height; precise term “oblique projectile”; and interpreting motion graphs.

Kinematics Revision: Questions 45–50 and Concept Application 1–25

Class 9 Physics · Chapter 2 · Checked answers with step-by-step corrections

How to revise: Attempt each question first, then open the answer. Green indicates correct work, amber means partially correct, and red identifies an incorrect or unanswered item.

Result: Test Your Concepts — 4 correct, 1 partial, 1 unanswered. Concept Application — 21 correct, 2 partial, 2 incorrect/unanswered.

Sections: Test Your Concepts 45–50 · Concept Application 1–25

Test Your Concepts

Question 45: Give the equations of motion of a body falling under gravity after being dropped from a certain height.

Status: Correct
Show answer and explanation
Because the body is dropped, initial velocity u = 0. Taking downward as positive and replacing a by g: v = gt, h = ½gt², and v² = 2gh. These are the usual equations of uniformly accelerated motion specialized for free fall.

Question 46: Explain the characteristics of displacement–time, velocity–time and acceleration–time graphs.

Status: Partially correct
Show answer and explanation
Displacement–time: slope gives velocity; a horizontal line means rest; a straight sloping line means uniform velocity; changing slope means non-uniform velocity.
Velocity–time: slope gives acceleration and signed area gives displacement; a horizontal line means constant velocity and zero acceleration.
Acceleration–time: signed area gives change in velocity; a line on the time axis means zero acceleration. The written answer had the main rules but needed these cases stated clearly.

Question 47: Derive the distance travelled in the nth second: Sₙ = u + (a/2)(2n − 1).

Status: Correct
Show answer and explanation
Total displacement after n seconds is sₙ = un + ½an². After (n − 1) seconds it is sₙ₋₁ = u(n − 1) + ½a(n − 1)². Therefore Sₙ = sₙ − sₙ₋₁ = u + ½a[n² − (n − 1)²] = u + (a/2)(2n − 1).

Question 48: Derive s = ut + ½at².

Status: Correct
Show answer and explanation
For uniform acceleration, average velocity = (u + v)/2. Hence s = [(u + v)/2]t. Using v = u + at gives s = [(2u + at)/2]t = ut + ½at².

Question 49: Obtain s = ut + ½at² by the graphical method.

Status: Not answered
Show answer and explanation
On a velocity–time graph, displacement equals the area under the graph. The area is a rectangle plus a triangle: s = ut + ½(v − u)t. Since v − u = at, s = ut + ½at².

Question 50: Derive v² − u² = 2as.

Status: Correct
Show answer and explanation
For uniform acceleration, s = [(u + v)/2]t and t = (v − u)/a. Substitution gives s = [(u + v)(v − u)]/(2a) = (v² − u²)/(2a). Therefore v² − u² = 2as.

Concept Application

Question 1: A body moves with retardation when projected vertically upward. True or false?

Status: Correct
Show answer and explanation
True. During upward motion, velocity is upward but gravity acts downward. The speed therefore decreases at the constant rate g until the body reaches maximum height.

Question 2: At maximum height, the velocity of a vertically projected body becomes zero. True or false?

Status: Correct
Show answer and explanation
True. Its instantaneous velocity is zero at the turning point, but its acceleration is still downward and equal to g.

Question 3: A velocity–time graph can be used to find displacement. True or false?

Status: Correct
Show answer and explanation
True. The signed area between the velocity–time graph and the time axis equals displacement.

Question 4: For constant velocity, displacement depends on the square of time. True or false?

Status: Correct
Show answer and explanation
False. With constant velocity, s = vt, so displacement is directly proportional to time, not time squared.

Question 5: Two balls of different masses are thrown vertically upward with the same speed. The heavier ball rises higher. True or false?

Status: Correct
Show answer and explanation
False. Ignoring air resistance, H = u²/(2g). Mass is absent from the formula, so both balls reach the same height.

Question 6: The equations of motion are applicable only to uniform velocity. True or false?

Status: Correct
Show answer and explanation
False. The standard equations v = u + at, s = ut + ½at² and v² − u² = 2as apply when acceleration is uniform.

Question 7: A freely falling body travels the same distance in every successive second. True or false?

Status: Correct
Show answer and explanation
False. Its speed increases under gravity. From rest, the distances in successive seconds are proportional to 1 : 3 : 5 : 7 …

Question 8: Find the ratio of the velocities of a freely falling body after 1 s and 2 s.

Status: Correct
Show answer and explanation
Using v = gt: v₁ : v₂ = g(1) : g(2) = 1 : 2.

Question 9: A stone is thrown vertically upward and is caught at the same level after total time t. Find its maximum height.

Status: Not answered
Show answer and explanation
Time of ascent = t/2. Since 0 = u − g(t/2), u = gt/2. Thus H = u²/(2g) = (g²t²/4)/(2g) = gt²/8.

Question 10: What does the area under a velocity–time graph represent?

Status: Correct
Show answer and explanation
It represents displacement. Area above the time axis is positive; area below it is negative. Adding the signed areas gives net displacement.

Question 11: What is a body projected at an angle other than 90° to the horizontal called?

Status: Partially correct
Show answer and explanation
It is specifically called an oblique projectile. “Projectile” is broadly correct, but “oblique” identifies that it has both horizontal and vertical velocity components.

Question 12: What can be said about the ratio |average velocity| / average speed?

Status: Correct
Show answer and explanation
The ratio is always less than or equal to 1, because displacement cannot exceed distance. It equals 1 only when motion is along a straight line without reversing direction.

Question 13: Compare the directions of displacement and average velocity.

Status: Correct
Show answer and explanation
They are the same, because average velocity = displacement/time and time is a positive scalar.

Question 14: What is produced when velocity changes?

Status: Correct
Show answer and explanation
A change in velocity produces acceleration. Velocity may change because its magnitude changes, its direction changes, or both.

Question 15: Match the types of motion with graphs a–h.

Status: Partially correct
Show answer and explanation
Complete matching: A–c, B–d, C–f, D–g, E–b, F–h, G–a, H–e.
A: uniform velocity; B: uniform acceleration with initial velocity; C: uniform acceleration; D: increasing acceleration; E: uniform retardation; F: acceleration decreasing steadily; G: uniform acceleration with initial displacement; H: body at rest with initial displacement. A–c and B–d were correct; the remaining matches were not attempted.

Question 16: The ratio of displacement magnitude to distance is: greater than 1, equal to 1, less than 1, or less than/equal to 1?

Status: Correct
Show answer and explanation
Correct option: 4 — less than or equal to 1. The magnitude of displacement is the shortest straight-line separation and therefore cannot exceed the actual path length.

Question 17: Two bodies fall freely through heights in the ratio 3 : 5. Find the ratio of their final velocities.

Status: Correct
Show answer and explanation
From v² = 2gh, v ∝ √h. Therefore v₁ : v₂ = √3 : √5, which is option 2.

Question 18: Find the displacement from the given velocity–time graph during 0–4 s.

Status: Correct
Show answer and explanation
Add the areas under the graph: first triangle = 10 m, rectangle = 20 m, trapezium = 15 m, and final rectangle = 10 m. Total displacement = 10 + 20 + 15 + 10 = 55 m, option 4.

Question 19: A person crosses an 8 m by 6 m rectangular hall diagonally. Find the displacement.

Status: Correct
Show answer and explanation
The diagonal is √(8² + 6²) = √100 = 10 m, option 2. Displacement uses the direct start-to-finish separation.

Question 20: The displacement–time graph shows straight lines for P and Q, with P steeper. Interpret the motion.

Status: Correct
Show answer and explanation
Both straight lines have constant slopes, so both bodies move with uniform speeds. P has the larger slope, so P moves faster than Q, option 3.

Question 21: A vehicle slows from 40 m/s to 10 m/s in 10 s. Find acceleration.

Status: Correct
Show answer and explanation
a = (v − u)/t = (10 − 40)/10 = −3 m/s², option 1. The negative sign indicates retardation when forward is positive.

Question 22: For a body starting from rest with uniform acceleration, how does velocity depend on time?

Status: Correct
Show answer and explanation
From v = u + at and u = 0, v = at. Thus velocity is directly proportional to time, option 1.

Question 23: At maximum height, which statement about a vertically projected body is correct?

Status: Incorrect
Show answer and explanation
Correct option: 1. At the highest point, velocity is momentarily zero, but acceleration is not zero: it remains g downward. Gravity does not switch off at the turning point.

Question 24: A body starts from rest with uniform acceleration. If the times are in the ratio 1 : 2, find the ratio of distances.

Status: Correct
Show answer and explanation
From s = ½at², distance is proportional to t². Therefore s₁ : s₂ = 1² : 2² = 1 : 4, option 2.

Question 25: A body has u = 3 m/s and a = 2 m/s². Find the distance travelled in the fourth second.

Status: Correct
Show answer and explanation
Sₙ = u + (a/2)(2n − 1). For n = 4: S₄ = 3 + (2/2)(7) = 10 m, option 1.
Key revision targets: graphical derivation of s = ut + ½at²; acceleration at maximum height; precise term “oblique projectile”; and interpreting motion graphs.

Kinematics Chapter 2

 Class 9 Physics · Chapter 2

Kinematics Revision Guide

Corrected answers and targeted practice for questions that were incorrect or only partially correct in the original exercise.

How to revise: Attempt each question first. Open the model answer only after writing your own response. Then complete the practice check and compare it with the answer key.

Question 6Explain the motion of a body undergoing circular motion, with an example.

Show model answer

Model answer

Circular motion is motion along a circular path around a fixed centre or axis. In uniform circular motion, the speed is constant but the velocity is not, because the direction of motion changes continuously. This change in velocity means that the body has an acceleration directed towards the centre (centripetal acceleration).

Remember

A stone tied to a string and whirled in a circle; a satellite orbiting Earth; or a point on a rotating fan blade.

What to fix

The earlier response mentioned non-uniform velocity and a merry-go-round, but did not clearly explain the circular path or the continuous change in direction.

Practice check

A car travels around a circular track at constant speed. Is its velocity constant? Explain.

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Question 8Define time of ascent and time of descent.

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Model answer

Time of ascent is the time taken by a body projected upward to travel from its point of projection to its maximum height. Time of descent is the time taken to travel from the maximum height back to the ground or the original level.

Remember

For a body that returns to its launch level without air resistance: time of ascent = time of descent = u/g.

What to fix

The definition of descent incorrectly described reaching maximum height; that is ascent.

Practice check

A ball takes 3 s to reach maximum height. How long will it take to return to the same level?

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Question 10A body projected at an angle other than 90 degrees to the horizontal is known as what?

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Model answer

It is called an oblique projectile. Its initial velocity has both horizontal and vertical components, and its trajectory is parabolic when air resistance is ignored.

Remember

The word 'projectile' is too general; 'oblique projectile' is the precise term required.

What to fix

The earlier answer omitted the word 'oblique'.

Practice check

Classify a ball projected at 35 degrees above the horizontal.

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Question 11What are scalar and vector quantities? Give examples.

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Model answer

A scalar quantity has magnitude only. A vector quantity has both magnitude and direction.

Remember

Scalars: distance, speed, mass, time. Vectors: displacement, velocity, acceleration, force.

What to fix

Saying only that scalars do not require direction is incomplete unless magnitude is also stated.

Practice check

Classify each as scalar or vector: 12 m, 12 m east, 5 m/s, 5 m/s north.

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Question 13Give the CGS and MKS units of distance and displacement.

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Model answer

Both distance and displacement use centimetre (cm) in the CGS system and metre (m) in the MKS system.

Remember

CGS = centimetre-gram-second; MKS = metre-kilogram-second.

What to fix

The MKS unit was left uncertain. It is metre (m).

Practice check

Convert a displacement of 2.4 m into centimetres.

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Question 16Why is distance scalar while displacement is vector?

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Model answer

Distance is the total length of the actual path travelled. It has magnitude but no direction, so it is scalar. Displacement is the straight-line change from the initial position to the final position and includes direction, so it is vector.

Remember

After one complete lap of a 400 m track: distance = 400 m, displacement = 0 m.

What to fix

The earlier response reversed the ideas of actual path and shortest straight-line separation.

Practice check

A student walks 3 m east and then 3 m west. Find distance and displacement.

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Question 22What are acceleration and retardation?

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Model answer

Acceleration is the rate of change of velocity with time: a = (v - u)/t. Retardation (deceleration) is acceleration that acts opposite to the velocity and reduces speed.

Remember

The sign of acceleration depends on the selected positive direction. Acceleration is not always positive, and retardation is not defined merely as negative acceleration.

What to fix

The earlier response treated positive and negative signs as absolute definitions.

Practice check

A car slows from 20 m/s to 8 m/s in 4 s. Calculate its acceleration.

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Question 26A particle moves P to Q at velocity v1 and Q to P at velocity v2. What is its average velocity for the complete trip?

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Model answer

Average velocity = total displacement / total time. The particle returns to P, so its total displacement is zero. Therefore its average velocity is 0, regardless of v1 and v2.

Remember

Do not confuse average velocity with average speed. The average speed would generally be non-zero.

What to fix

This question was left blank.

Practice check

A runner completes one full lap and returns to the starting point. What is the average velocity?

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Question 28A body travels at v1 for the first half of the total time and at v2 for the remaining half. Find its average velocity.

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Model answer

For equal time intervals, average velocity is the arithmetic mean: average velocity = (v1 + v2)/2, provided both velocities are along the stated straight-line direction.

Remember

If the question instead specified equal distances, the result would be 2v1v2/(v1 + v2).

What to fix

This question was left blank.

Practice check

A body moves at 10 m/s for 4 s and 18 m/s for the next 4 s. Find its average velocity.

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Question 30What is a trajectory?

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Model answer

A trajectory is the path followed by a moving body or particle through space.

Remember

For an oblique projectile under gravity, ignoring air resistance, the trajectory is a parabola. However, not every trajectory is parabolic.

What to fix

The earlier response described a condition for a parabolic projectile path instead of defining trajectory generally.

Practice check

State the trajectory of (a) an oblique projectile and (b) a satellite in a circular orbit.

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Question 31Derive the maximum height reached by a body projected vertically upward.

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Model answer

Take upward as positive. For upward motion, a = -g. Use v^2 = u^2 + 2as. At maximum height, v = 0 and s = H. Therefore 0 = u^2 - 2gH, so H = u^2/(2g).

Remember

The result must be accompanied by the equation used, the sign of acceleration, and the condition v = 0 at maximum height.

What to fix

Only the final formula was supplied; the requested derivation was missing.

Practice check

Find the maximum height for u = 14 m/s and g = 9.8 m/s^2.

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Question 33What are the different types of motion? Give an example of each.

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Model answer

Translational motion: the body moves from one position to another. Rectilinear example: a train on a straight track. Curvilinear example: a thrown ball. Rotational motion: a body turns about an axis, such as a ceiling fan. Oscillatory motion: repeated to-and-fro motion about a mean position, such as a pendulum.

Remember

Uniform and non-uniform describe how motion changes; they are not substitutes for the main path/behaviour categories requested here.

What to fix

The earlier list contained some valid terms but no examples and mixed categories.

Practice check

Name the type of motion: a lift moving vertically, a spinning wheel, and a playground swing.

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Question 34Derive the time taken by a vertically projected body to reach maximum height.

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Model answer

For upward motion, v = u - gt. At maximum height, v = 0. Hence 0 = u - gt, giving time of ascent t = u/g.

Remember

The maximum-height formula H = u^2/(2g) answers a different question.

What to fix

The earlier answer gave maximum height rather than time of ascent.

Practice check

A ball is projected upward at 19.6 m/s. Find its time of ascent.

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Question 35A runner covers 200 cm in the 2nd second and 220 cm in the 4th second. Find initial velocity and uniform acceleration.

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Model answer

Use distance in the nth second: S_n = u + (a/2)(2n - 1). Convert 200 cm = 2.00 m and 220 cm = 2.20 m. For n=2: 2.00 = u + 1.5a. For n=4: 2.20 = u + 3.5a. Subtraction gives 0.20 = 2a, so a = 0.10 m/s^2. Then u = 2.00 - 1.5(0.10) = 1.85 m/s.

Remember

Final answer: initial velocity = 1.85 m/s; acceleration = 0.10 m/s^2.

What to fix

The method and initial velocity were correct, but acceleration was written as 0.01 m/s^2.

Practice check

Check the result by substituting u and a into the expression for the 4th second.

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Question 42A train travels between two stations at 40 km/h and returns at 60 km/h. Find average speed and average velocity.

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Model answer

For equal one-way distances, average speed = 2v1v2/(v1 + v2) = 2(40)(60)/(40+60) = 48 km/h. For the complete return journey, total displacement is zero, so average velocity = 0 km/h.

Remember

Do not take (40+60)/2 because the train spends unequal times at the two speeds.

What to fix

Average velocity was correct, but average speed was written as 50 km/h.

Practice check

A car travels equal distances at 30 km/h and 50 km/h. Find its average speed.

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Question 43Use the velocity-time graph to find (i) velocity of projection and (ii) maximum height.

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Model answer

The graph starts at +20 m/s, so the velocity of projection is 20 m/s. Maximum height equals displacement during ascent, which is the positive area under the velocity-time graph from 0 to 2 s: H = 1/2 x 2 s x 20 m/s = 20 m.

Remember

Area under a velocity-time graph gives displacement. Only the portion above the time axis represents upward displacement.

What to fix

Part (i) was correct. Part (ii) identified the method but did not finish the calculation.

Practice check

If the same graph crossed the time axis at 3 s with initial velocity 24 m/s, find the maximum height.

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Question 44Determine acceleration: (a) constant speed 20 m/s for 12 s; (b) velocity changes from 0 m/h to 360 m/min in 4.2 s.

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Model answer

(a) Constant velocity gives a = 0 m/s^2. (b) Convert 360 m/min to 6 m/s. Then a = (v-u)/t = (6-0)/4.2 = 1.43 m/s^2 (approximately).

Remember

Always convert velocities to consistent units before applying a = (v-u)/t.

What to fix

Part (a) and the conversion in part (b) were correct, but the final division and acceleration unit were not completed.

Practice check

A bicycle speeds up from 2 m/s to 11 m/s in 6 s. Find its acceleration.

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Practice Answer Key

  1. Q6. No. Its speed is constant, but its direction changes, so velocity changes.
  2. Q8. 3 s.
  3. Q10. An oblique projectile.
  4. Q11. Scalar, vector, scalar, vector.
  5. Q13. 240 cm.
  6. Q16. Distance = 6 m; displacement = 0 m.
  7. Q22. -3 m/s^2 (taking forward as positive).
  8. Q26. 0.
  9. Q28. 14 m/s.
  10. Q30. Parabolic; circular.
  11. Q31. 10 m.
  12. Q33. Rectilinear translational; rotational; oscillatory.
  13. Q34. 2 s.
  14. Q35. 2.20 = 1.85 + 3.5(0.10) = 2.20 m.
  15. Q42. 37.5 km/h.
  16. Q43. 36 m.
  17. Q44. 1.5 m/s^2.