9th Class Physics
Chapter-2: Kinematics
A. Multiple Choice Questions
Tick (✓) the correct option
2.1 The numerical ratio of displacement to distance is:
Explanation: Displacement is always less than or equal to distance (displacement ≤ distance). They are equal only for straight-line motion without direction change.
2.2 If a body does not change its position w.r.t. a fixed point, it is in:
Explanation: When a body's position remains unchanged relative to a fixed reference point, it is considered to be at rest.
2.3 A ball is dropped from a tower; distance in the 1ˢᵗ second is:
Explanation: Using h = ½ g t² = ½ × 10 × 1² = 5 m (taking g = 10 m/s²).
2.4 Body accelerates from rest to 144 km h⁻¹ (40 m s⁻¹) in 20 s. Distance covered:
Explanation: Using S = ½(v+u)t = ½ × 40 × 20 = 400 m (u = 0, v = 40 m/s).
2.5 Starting from rest with constant acceleration, if distance S is covered in 4 s, time to cover ¼ S is:
Explanation: For constant acceleration from rest, S ∝ t². If S/4, then t²/4, so t/2 = 2 s.
2.6 Two displacement-time graphs A & B are shown. True statement:
Explanation: Velocity is the slope of displacement-time graph. Graph A has smaller slope than B, so velocity of A is less than B.
2.7 The area under a speed-time graph equals:
Explanation: The area under a speed-time graph represents the total distance covered by the object.
2.8 The gradient (slope) of a speed-time graph equals:
Explanation: The slope of a speed-time graph gives the rate of change of speed, which is acceleration.
2.9 The gradient of a distance-time graph equals:
Explanation: The slope of a distance-time graph represents the rate of change of distance with time, which is speed.
2.10 A car accelerates uniformly from 80.5 km h⁻¹ to 113 km h⁻¹ in 9 s. Which graph best describes its motion?
Explanation: For uniform acceleration, the velocity-time graph is a straight line with positive slope.
B. Short Answer Questions
2.1 Define scalar and vector quantities.
Answer:
Scalar: A physical quantity completely described by magnitude only (no direction).
Vector: A quantity described by both magnitude and direction.
2.2 Give 5 examples each.
| Scalars | Vectors |
|---|---|
| mass | displacement |
| time | velocity |
| temperature | acceleration |
| speed | force |
| energy | momentum |
2.3 State the head-to-tail rule for adding vectors.
Answer:
- Draw the first vector to scale.
- Place the tail of the second vector at the head of the first.
- Continue for all vectors.
- The resultant is the single arrow from the tail of the first to the head of the last.
Head-to-tail method of vector addition
2.4 What are distance-time and speed-time graphs?
Answer:
Distance-time: Plots distance on y-axis vs time on x-axis; slope = speed.
Speed-time: Plots speed on y-axis vs time on x-axis; slope = acceleration, area under graph = distance covered.
2.5 Falling objects near Earth have the same constant acceleration. Does this imply a heavier object falls faster?
Answer: No.
In vacuum (no air resistance) a = g for all masses, so they fall equally fast. Differences observed are due to air-drag, not weight.
2.6 Vector quantities are sometimes written without bold face. How is direction indicated?
Answer: By a + or – sign along a stated reference direction (e.g., +5 m s⁻¹ right, –3 m s⁻¹ left). Or by an angle/compass bearing in diagrams.
2.7 A body moves with uniform speed. Will its velocity be uniform? Give reason.
Answer: Only if it moves in a straight line.
Velocity = speed + direction; if direction changes (circular path), velocity changes even though speed is constant.
2.8 Is it possible for a body to have acceleration when moving with:
(i) constant velocity? No – acceleration is zero by definition.
(ii) constant speed? Yes – in uniform circular motion speed is constant but direction changes continuously, producing centripetal acceleration.
C. Constructed Response Questions
2.1 Distance and displacement may or may not be equal in magnitude. Explain this statement.
Answer:
"Distance and displacement may or may not be equal in magnitude."
Equal: Straight-line motion without change of direction (e.g. 10 m east → distance = displacement = 10 m).
Unequal: Path involves turning back (e.g. walk 10 m east then 6 m west: distance = 16 m, displacement = 4 m).
Hence equality depends on whether the object reverses its motion.
Comparison of distance and displacement
2.2 When a bullet is fired, its velocity with which it leaves the barrel is called the muzzle velocity of the gun. The muzzle velocity of one gun with a longer barrel is lesser than that of another gun with a shorter barrel. In which gun is the acceleration of the bullet larger? Explain your answer.
Answer:
Which gun gives the bullet the larger acceleration?
Shorter-barrel gun attains higher muzzle velocity in less time.
Since a = (v – u)/t, smaller t for the same Δv ⇒ larger acceleration.
Conclusion: The shorter-barrel gun produces greater acceleration for the bullet.
2.3 For a complete trip, average velocity was calculated. Its value came out to be positive. Is it possible that its instantaneous velocity at any time during the trip had the negative value? Give justification of your answer.
Answer: Yes.
Average velocity only cares about net displacement (positive here).
During the trip the object could have moved backward (negative v) for some time, provided the final position lies ahead of the start.
Example: Drive 60 m east, then 20 m west; net displacement = +40 m (positive average velocity), yet while moving west instantaneous velocity was negative.
2.4 A ball is thrown vertically upward with velocity v. It returns to the ground in time T. Which of the following graphs correctly represents the motion? Explain your reasoning.
Answer: Correct choice: Graph (c) – a straight line with negative slope, first half in +y axis and second half in negative y axis.
Reasoning:
Upward journey: velocity decreases linearly (a = –g), reaches zero at the top.
Downward journey: velocity increases negatively, continuing the same straight line below the time axis.
Slope is constant (equal to –g) throughout; final velocity = –initial velocity.
Velocity-time graph for a ball thrown vertically upward - single continuous line
2.5 The figure given below shows the distance - time graph for the travel of a cyclist. Find the velocities for the segments a, b and c.
Answer:
| Segment | Δs (km) | Δt (min) | Velocity (km min⁻¹) | Velocity (km h⁻¹) |
|---|---|---|---|---|
| a | 2.0 – 0 = 2.0 | 6 – 0 = 6 | 2/6 = 0.33 | 0.33 × 60 = 20 km h⁻¹ |
| b | 2.0 – 2.0 = 0 | 10 – 6 = 4 | 0 (horizontal) | 0 km h⁻¹ (rest) |
| c | 0 – 2.0 = –2.0 | 20 – 10 = 10 | –2.0/10 = –0.2 | –0.2 × 60 = –12 km h⁻¹ (returning) |
Distance-time graph for a cyclist - distance drops to 0 km at 20 minutes
2.6 Is it possible that the velocity of an object is zero at an instant of time, but its acceleration is not zero? If yes, give an example of such a case.
Answer: Yes – at the turning point of any to-and-fro motion.
Example: A ball thrown vertically upward has zero velocity at the highest point, but acceleration = g (≈ 9.8 m s⁻² downward) continuously.
Likewise, a pendulum at the extreme position has zero speed but non-zero centripetal & tangential acceleration.
Ball at highest point: velocity = 0, acceleration = g
D. Numerical Problems
(a) A velocity of 400 m/s making an angle of 60° with x-axis.
(b) A force of 50 N making an angle of 120° with x-axis.
(a) Velocity Vector: 400 m/s at 60°
Scale: 1 cm ≡ 100 m/s → 4 cm line
(b) Force Vector: 50 N at 120°
Scale: 1 cm ≡ 10 N → 5 cm line
Given: Speed = 72 km/h, Distance = 360 km
Formula: Time = Distance ÷ Speed
Calculation: t = 360 km ÷ 72 km/h = 5 hours
Answer: The car will take 5 hours to cover 360 km.
Given: Initial velocity u = 0, Final velocity v = 90 km/h = 25 m/s, Time t = 50 s
Formula: Acceleration a = (v - u) ÷ t
Calculation: a = (25 - 0) ÷ 50 = 0.5 m/s²
Answer: The average acceleration is 0.5 m/s².
Given: Initial velocity u = 5 m/s, Acceleration a = 1.5 m/s², Time t = 5 s
Formula: Final velocity v = u + at
Calculation: v = 5 + (1.5 × 5) = 5 + 7.5 = 12.5 m/s
Answer: The velocity after 5 seconds is 12.5 m/s.
Given: Initial velocity u = 18 km/h = 5 m/s, Acceleration a = 2 m/s², Time t = 10 s
Formula: Distance S = ut + ½at²
Calculation: S = (5 × 10) + ½(2 × 100) = 50 + 100 = 150 m
Answer: The motorcycle will travel 150 meters.
Given: Initial velocity u = 54 km/h = 15 m/s, Final velocity v = 0, Distance S = 25 m
Formula: v² = u² + 2aS
Calculation: 0 = (15)² + 2a(25) ⇒ 0 = 225 + 50a ⇒ a = -4.5 m/s²
Answer: The acceleration is -4.5 m/s² (deceleration).
Given: Height h = 45 m, Initial velocity u = 0, Acceleration g = 10 m/s²
Time calculation: h = ½gt² ⇒ 45 = ½(10)t² ⇒ 45 = 5t² ⇒ t² = 9 ⇒ t = 3 s
Velocity calculation: v = gt = 10 × 3 = 30 m/s
Answer: Time = 3 seconds, Velocity = 30 m/s downward.
Leg 1: Distance S₁ = 10 km = 10000 m, Velocity v₁ = 20 m/s
Time t₁ = S₁/v₁ = 10000/20 = 500 s
Leg 2: Distance S₂ = 0.8 km = 800 m, Velocity v₂ = 4 m/s
Time t₂ = S₂/v₂ = 800/4 = 200 s
Total: Distance S_total = 10800 m, Time t_total = 700 s
Average velocity v_avg = S_total/t_total = 10800/700 ≈ 15.4 m/s
Answer: The average velocity for the total journey is 15.4 m/s.
Given: Time t = 5 s, Initial velocity u = 0, Acceleration g = 10 m/s²
Height calculation: h = ½gt² = ½(10)(25) = 125 m
Velocity calculation: v = gt = 10 × 5 = 50 m/s
Answer: Height of tower = 125 m, Impact velocity = 50 m/s downward.
Given: Time to highest point = 3 s, Initial height = 1 m, Acceleration g = 10 m/s²
Initial velocity: v = u - gt ⇒ 0 = u - (10 × 3) ⇒ u = 30 m/s
Height above hit-point: h = ut - ½gt² = (30 × 3) - ½(10 × 9) = 90 - 45 = 45 m
Height above ground: 45 m + 1 m = 46 m
Answer: Initial velocity = 30 m/s, Maximum height above ground = 46 m.