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.
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.
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.
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
Velocity–time graph
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.
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.
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.
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.
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.
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.
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.
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
Velocity–time graph (upward positive)
Choose a sign convention before solving
| Chosen positive direction | Upward velocity | Downward velocity | Acceleration due to gravity |
|---|---|---|---|
| Upward is positive | positive | negative | a = −g throughout |
| Downward is positive | negative | positive | a = +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.
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.
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.
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.
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.
