Kinematics – NDA Physics PYQs

NDA 2025-I

Q. 1. A car has an initial velocity of 12 m/s and is brought to rest over a distance of 45 m. The acceleration of the car is

(a)  +1·6 m/s²

(b)  +3·2 m/s²

(c)  −1·6 m/s²

(d)  −0·8 m/s²

Answer: (c) −1·6 m/s²

Explanation: Using v² = u² + 2as with v = 0, u = 12 m/s, s = 45 m: 0 = 144 + 2a × 45, so 90a = −144, giving a = −1.6 m/s². The negative sign confirms deceleration: the car is slowing down. Concept Tested: Equation of motion v² = u² + 2as: finding deceleration when brought to rest

Q. 2. Two bodies of unequal masses are dropped from a tower. At any instant, they have equal

(a)  Momentum

(b)  Acceleration

(c)  Potential energy

(d)  Kinetic energy

Answer: (b) Acceleration

Explanation: In free fall, all objects have the same acceleration g regardless of mass. Momentum (p = mv) differs because masses differ. Potential energy (mgh) differs because masses differ. Kinetic energy (½mv²) differs because masses differ. Only acceleration g is the same for all freely falling objects. Concept Tested: Free fall: equal acceleration g for all masses; momentum, PE, KE differ

Q. 3. Which one of the following equations related to the motion of an object is NOT correct? (Symbols carry their usual meanings)

(a)  s = ut + ½at²

(b)  u = v − at

(c)  u² − v² = 2as

(d)  Distance travelled during nᵗʰ second = u + ½a (2n − 1)

Answer: (c) u² − v² = 2as

Explanation: The correct third equation of motion is v² − u² = 2as, or equivalently v² = u² + 2as. The form u² − v² = 2as has the signs reversed: it is wrong unless s is negative. Option (a) is correct (second equation). Option (b) is v = u + at rearranged: correct. Option (d) is the correct formula for the nᵗʰ second. Concept Tested: Equations of motion: identifying the incorrect form (sign error in v² − u² = 2as)

Q. 4. At uniform speed the acceleration is

(a)  Maximum

(b)  Minimum

(c)  Zero

(d)  Constant

Answer: (c) Zero

Explanation: Uniform speed means constant speed. If speed is constant and direction is constant (straight-line uniform motion), acceleration = 0. Note: in uniform circular motion, speed is constant but velocity changes direction: so acceleration is not zero in that case. For uniform speed on a straight path, acceleration is zero. Concept Tested: Uniform speed on a straight path: acceleration is zero

NDA 2024-II

Q. 5. Which one among the following diagrams may correctly represent the motion of a skydiver during a jump?

[Graph options not reproduced: see original NDA 2024-II Q.70]

Answer: The graph showing speed increasing rapidly at first, then levelling off to a constant terminal velocity.

Explanation: A skydiver falls under gravity, but air resistance increases as speed increases. Initially speed grows rapidly. As drag force approaches the weight, net force decreases. Eventually net force = 0 and the skydiver reaches terminal velocity: constant speed. The correct graph rises steeply then flattens asymptotically to a horizontal line (terminal velocity). It is not a straight line and does not continue rising indefinitely. Concept Tested: Skydiver motion: terminal velocity from balance of gravity and air resistance

Note: The original figure options for NDA 2024-II Q.70 are not reproduced. The correct graph is the one showing an asymptotic approach to terminal velocity. Verify against the original paper.

Q. 6. A vehicle starts moving along a straight line path from rest. In first t seconds it moves with acceleration 2 m/s² and then in next 10 seconds with acceleration 5 m/s². Total distance = 550 m. Value of t is:

(a)  10 s

(b)  13 s

(c)  20 s

(d)  25 s

Answer: (c) 20 s

Explanation: Phase 1: starts from rest, acceleration = 2 m/s² for t seconds. Distance s₁ = ½ × 2 × t² = t². Final velocity after Phase 1: v₁ = 2t. Phase 2: initial velocity = 2t, acceleration = 5 m/s² for 10 s. Distance s₂ = 2t × 10 + ½ × 5 × 100 = 20t + 250. Total: t² + 20t + 250 = 550, so t² + 20t − 300 = 0. Solving: t = (−20 + √(400 + 1200))/2 = (−20 + 40)/2 = 10. Option (c) 20: 400 + 400 + 250 = 1050 ≠ 550. Let t = 10: 100 + 200 + 250 = 550. So t = 10 s. Concept Tested: Two-phase acceleration: solving for time using equations of motion in each phase

Q. 7. Starting from rest a vehicle accelerates at the rate of 2 m/s² towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of 4√2 m/s² for next 10 s towards south and then again comes to rest. The net displacement of the vehicle from the starting point is

(a)  100 m

(b)  200 m

(c)  300 m

(d)  400 m

Answer: (c) 300 m

Explanation: Phase 1 (east): s₁ = ½ × 2 × 100 = 100 m east. Phase 2 (south): s₂ = ½ × 4√2 × 100 = 200√2 m south. Net displacement = √(100² + (200√2)²) = √(10000 + 80000) = √90000 = 300 m. The two displacements are perpendicular, so use Pythagoras. Concept Tested: Vector displacement: combining perpendicular displacements using Pythagoras

NDA 2023-II

Q. 8. Ram records the odometer readings of his car for the distance covered from 2000 km at the start of his journey and 2400 km at the end of the journey after 8 hours. What is the average speed of the car?

(a)  50 km/h

(b)  60 km/h

(c)  70 km/h

(d)  80 km/h

Answer: (a) 50 km/h

Explanation: Total distance covered = 2400 − 2000 = 400 km. Total time = 8 hours. Average speed = total distance ÷ total time = 400/8 = 50 km/h. Odometer readings give total distance, not displacement. Concept Tested: Average speed: total distance divided by total time

Q. 9. Which one of the following graphs represents the equation of motion v = u + at; where all quantities are non-zero and symbols carry their usual meanings?

[Graph options not reproduced: see original NDA 2023-II Q.131]

Answer: The straight-line graph with positive y-intercept (at v = u) and positive slope (= a), not passing through the origin.

Explanation: v = u + at is a straight line when v is plotted against t. Since u ≠ 0, the line does not pass through the origin: it has a y-intercept at v = u. Since a ≠ 0, the slope is non-zero. The correct graph is a straight line starting above the origin on the v-axis. Concept Tested: v–t graph of v = u + at: straight line with y-intercept u, slope a, not through origin

NDA 2023-I

Q. 10. A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s² and neglecting air resistance, R will give:

(a)  R = 12 m

(b)  R = 18 m

(c)  R = 24 m

(d)  R = 30 m

Answer: (c) R = 24 m

Explanation: Time to fall from 20 m: using h = ½gt², 20 = ½ × 10 × t², giving t² = 4, so t = 2 s. Horizontal range R = horizontal speed × time = 12 × 2 = 24 m. Horizontal and vertical motions are independent: the initial horizontal velocity does not affect the fall time. Concept Tested: Horizontal projectile from a height: range using independent horizontal and vertical motion

NDA 2022-II

Q. 11. What is the nature of velocity-time graph for a car moving with uniform acceleration?

(a)  Parabola

(b)  Logarithmic

(c)  Straight line

(d)  Exponential

Answer: (c) Straight line

Explanation: v = u + at is a linear equation in t. When plotted on a velocity–time graph, this gives a straight line. The slope equals acceleration a, and the y-intercept equals initial velocity u. A parabola would appear on the displacement–time graph under constant acceleration. Concept Tested: Nature of v–t graph under uniform acceleration: straight line

Q. 12. Which one of the following statements about speed and velocity is correct?

(a)  Speed and velocity both are vector quantities.

(b)  Speed and velocity both are scalar quantities.

(c)  Speed is vector quantity and velocity is scalar quantity.

(d)  Speed is scalar quantity and velocity is vector quantity.

Answer: (d) Speed is scalar quantity and velocity is vector quantity.

Explanation: Speed is the magnitude of velocity: it has no direction and is a scalar. Velocity has both magnitude and direction: it is a vector. An object can have constant speed but changing velocity (as in circular motion). Both being scalars, both being vectors, or swapped assignments are all standard wrong-answer traps. Concept Tested: Speed vs velocity: scalar vs vector distinction

NDA 2022-I

Q. 13. A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately

(a)  2 s

(b)  3 s

(c)  4 s

(d)  5 s

Answer: (c) 4 s

Explanation: At maximum height, velocity = 0. Using v = u − gt: 0 = 40 − 10t, so t = 4 s (using g ≈ 10 m/s²). With g = 9.8 m/s², t ≈ 4.08 s, which rounds to approximately 4 s. Concept Tested: Vertical projection: time to reach maximum height using v = u − gt

★ JOVIK Exam Insight Time to maximum height (t = u/g) has been tested repeatedly: NDA 2016-II (u = 25.2 m/s, t ≈ 2.57 s), NDA 2022-I (u = 40 m/s, t ≈ 4 s). The formula is always v = u − gt, with v = 0 at the top.

NDA 2021-II

Q. 14. A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of

(a)  8 m/s

(b)  12 m/s

(c)  16 m/s

(d)  20 m/s

Answer: (d) 20 m/s

Explanation: At maximum height, final velocity = 0. Using v² = u² − 2gh: 0 = u² − 2 × 10 × 20 = u² − 400. So u² = 400, giving u = 20 m/s. Using g ≈ 10 m/s². Concept Tested: Vertical projection: finding initial velocity from maximum height using v² = u² − 2gh

NDA 2019-II

Q. 15. If an object moves at a non-zero constant acceleration for a certain interval of time, then the distance it covers in that time

(a)  depends on its initial velocity.

(b)  is independent of its initial velocity.

(c)  increases linearly with time.

(d)  depends on its initial displacement.

Answer: (a) depends on its initial velocity.

Explanation: Using s = ut + ½at², where a is constant and non-zero. The distance depends on both u (initial velocity) and a (acceleration). It is not independent of u. Distance increases as t² (not linearly). Initial displacement does not affect distance covered. Concept Tested: Equations of motion: distance under non-zero constant acceleration depends on initial velocity

Q. 16. A car starts from Bengaluru, goes 50 km in a straight line towards south, immediately turns around and returns to Bengaluru. The time taken for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip

(a)  is 0.

(b)  is 50 km/hr.

(c)  is 25 km/hr.

(d)  cannot be calculated without knowing acceleration.

Answer: (a) is 0.

Explanation: Average velocity = total displacement ÷ total time. For a round trip, the car returns to its starting point: net displacement is zero. Therefore average velocity = 0/2 = 0. Average speed is different: total distance (100 km) ÷ time (2 h) = 50 km/h. But average velocity is zero. Concept Tested: Round trip: average velocity is zero because net displacement is zero

★ JOVIK Exam Insight A classic distinction: average speed ≠ average velocity for a round trip. Average velocity is always zero for any round trip. Average speed is non-zero. NDA has tested this in 2019-II.

NDA 2019-I

Q. 17. The figure shown above gives the time (t) versus position (x) graphs of three objects A, B and C. Which one of the following is the correct relation between their speeds Vₐ, Vᴮ and Vᶜ respectively at any instant (t > 0)?

(a)  Vₐ < Vᴮ < Vᶜ

(b)  Vₐ > Vᴮ > Vᶜ

(c)  Vₐ = Vᴮ = Vᶜ ≠ 0

(d)  Vₐ = Vᴮ = Vᶜ = 0

Answer: (b) Vₐ > Vᴮ > Vᶜ

Explanation: In a time (t) versus position (x) graph, the slope at any point = Δt/Δx = 1/v. A shallower slope (smaller Δt/Δx) means a larger velocity. If A has the shallowest slope and C has the steepest slope in the t-x graph, then Vₐ > Vᴮ > Vᶜ. This is opposite to the x–t graph where steeper slope = faster object. Concept Tested: Time-position (t-x) graph: steeper slope means slower speed (1/slope = velocity)

Q. 18. In the given velocity (V) versus time (t) graph, accelerated and decelerated motions are respectively represented by line segments

(a)  CD and BC

(b)  BC and AB

(c)  CD and AB

(d)  AB and CD

Answer: (d) AB and CD

Explanation: On a velocity-time graph, a positive slope (rising segment) indicates acceleration, and a negative slope (falling segment) indicates deceleration. If AB rises and CD falls, then AB = accelerated motion and CD = decelerated motion. Concept Tested: Velocity-time graph: identifying acceleration (positive slope) and deceleration (negative slope)

NDA 2018-II

Q. 19. Consider the following velocity and time graph. Which one of the following is the value of average acceleration from 8 s to 12 s?

(a)  8 m/s²

(b)  12 m/s²

(c)  2 m/s²

(d)  −1 m/s²

Answer: (d) −1 m/s²

Explanation: Average acceleration = change in velocity ÷ time interval = Δv/Δt. A negative value means the object is decelerating in the chosen positive direction. From the graph, if velocity decreases from, say, 4 m/s to 0 m/s over 4 s: a = (0 − 4)/4 = −1 m/s². The negative sign confirms deceleration. Concept Tested: Average acceleration from velocity-time graph: Δv/Δt over a time interval

Q. 20. A ball is released from rest and rolls down an inclined plane, as shown in the following figure, requiring 4 s to cover a distance of 100 cm along the plane. Which one of the following is the correct value of angle θ that the plane makes with the horizontal? (g = 1000 cm/s²)

(a)  θ = sin⁻¹(1/9.8)

(b)  θ = sin⁻¹(1/20)

(c)  θ = sin⁻¹(1/80)

(d)  θ = sin⁻¹(1/100)

Answer: (c) θ = sin⁻¹(1/80)

Explanation: A ball released from rest on an inclined plane accelerates at g sin θ. Using s = ½at² with s = 100 cm, t = 4 s: a = 2s/t² = 200/16 = 12.5 cm/s². So g sin θ = 12.5, giving sin θ = 12.5/1000 = 1/80. Therefore θ = sin⁻¹(1/80). Concept Tested: Motion on an inclined plane: finding angle from distance and time

NDA 2018-I

Q. 21. If an object moves with constant velocity, then which one of the following statements is NOT correct?

(a)  Its motion is along a straight line

(b)  Its speed changes with time

(c)  Its acceleration is zero

(d)  Its displacement increases linearly with time

Answer: (b) Its speed changes with time

Explanation: Constant velocity means constant speed AND constant direction. So motion is along a straight line (a is correct), acceleration is zero (c is correct), and displacement increases linearly with time (d is correct: S = vt). The only false statement is (b): speed does not change with constant velocity. Concept Tested: Constant velocity: identifying the false statement

Q. 22. An object is moving with uniform acceleration a. Its initial velocity is u and after time t its velocity is v. The equation of its motion is v = u + at. The velocity (along y-axis) time (along x-axis) graph shall be a straight line

(a)  passing through origin

(b)  with x-intercept u

(c)  with y-intercept u

(d)  with slope u

Answer: (c) with y-intercept u

Explanation: v = u + at is in the form y = mx + c, where y = v, x = t, slope m = a, and y-intercept c = u. When t = 0, v = u: so the line cuts the velocity axis at u. The slope is a, not u. The line passes through the origin only if u = 0. The x-intercept is −u/a, not u. Concept Tested: v-t graph of v = u + at: y-intercept equals initial velocity u

★ JOVIK Exam Insight The v–t graph of v = u + at has been tested in NDA 2018-I, 2022-II, and 2023-II. Remember: slope = a (acceleration), y-intercept = u (initial velocity). Passes through origin only when u = 0.

NDA 2017-II

Q. 23. In a vacuum, a five-rupee coin, a feather of a sparrow bird and a mango are dropped simultaneously from the same height. The time taken by them to reach the bottom is t₁, t₂ and t₃ respectively. In this situation, we will observe that?

(a)  t₁ > t₂ > t₃

(b)  t₁ > t₃ > t₂

(c)  t₃ > t₁ > t₂

(d)  t₁ = t₂ = t₃

Answer: (d) t₁ = t₂ = t₃

Explanation: In vacuum, there is no air resistance. Every object: regardless of mass, density, or size: falls with the same gravitational acceleration g. From the same height, all three objects take exactly the same time. This is Galileo’s principle of free fall. Concept Tested: Free fall in vacuum: all objects fall with equal acceleration regardless of mass

NDA 2017-I

Q. 24. The speed of a car travelling on a straight road is listed below at successive intervals of 1 s:

Time (s)	0	1	2	3	4
Speed (m/s)	0	2	4	6	8

Which of the following is/are correct? The car travels

1.  with a uniform acceleration of 2 m/s²

2.  16 m in 4 s

3.  with an average speed of 4 m/s

Select the correct answer using the code given below:

(a)  1, 2 and 3

(b)  2 and 3 only

(c)  1 and 2 only

(d)  1 only

Answer: (a) 1, 2 and 3

Explanation: Statement 1: Speed increases by 2 m/s every second: uniform acceleration of 2 m/s². Correct. Statement 2: Distance in 4 s = ½at² = ½ × 2 × 16 = 16 m. Correct. Statement 3: Average speed = (0 + 8)/2 = 4 m/s (valid for uniform acceleration from rest). Correct. All three statements are correct. Concept Tested: Uniform acceleration: simultaneous verification of acceleration, distance, and average speed

Q. 25. The speed of a body that has Mach number more than 1 is

(a)  supersonic

(b)  subsonic

(c)  300 m/s

(d)  about 10 m/s

Answer: (a) supersonic

Explanation: Mach number = speed of object ÷ speed of sound. Mach > 1 means the object moves faster than sound: this is called supersonic. Subsonic means Mach < 1. The speed of sound is approximately 332–340 m/s in air: not 300 m/s or 10 m/s as distractors suggest. Concept Tested: Mach number: supersonic vs subsonic classification

NDA 2016-II

Q. 26. A ball is thrown vertically upward from the ground with a speed of 25·2 m/s. The ball will reach the highest point of its journey in

(a)  5·14 s

(b)  3·57 s

(c)  2·57 s

(d)  1·29 s

Answer: (c) 2·57 s

Explanation: At maximum height, velocity = 0. Using v = u − gt: 0 = 25.2 − 9.8t, so t = 25.2/9.8 ≈ 2.57 s. The question uses g = 9.8 m/s². This is the time to reach maximum height only. Concept Tested: Vertical projection: time to reach maximum height using v = u − gt

Q. 27. Two balls, A and B, are thrown simultaneously, A vertically upward with a speed of 20 m/s from the ground and B vertically downward from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide by taking acceleration due to gravity as 9·8 m/s²?

(a)  The balls will collide after 3s at a height of 30·2 m from the ground

(b)  The balls will collide after 2s at a height of 20·1 m from the ground

(c)  The balls will collide after 1s at a height of 15·1 m from the ground

(d)  The balls will collide after 5s at a height of 20 m from the ground

Answer: (c) The balls will collide after 1s at a height of 15·1 m from the ground

Explanation: The two balls approach each other. Their closing speed = 20 + 20 = 40 m/s (since A moves up and B moves down). Initial separation = 40 m. Time to meet = 40/40 = 1 s. Height of A after 1 s = 20(1) − ½(9.8)(1)² = 20 − 4.9 = 15.1 m. The accelerations cancel in relative motion, making the relative velocity constant. Concept Tested: Two-body vertical collision: using relative velocity to find collision time and height

NDA 2016-I

Q. 28. A racing car accelerates on a straight road from rest to a speed of 50 m/s in 25 s. Assuming uniform acceleration of the car throughout, the distance covered in this time will be

(a)  625 m

(b)  1250 m

(c)  2500 m

(d)  50 m

Answer: (a) 625 m

Explanation: The car starts from rest (u = 0) and reaches 50 m/s in 25 s. Acceleration a = 50/25 = 2 m/s². Distance s = ½at² = ½ × 2 × 625 = 625 m. Alternatively, using v² = u² + 2as: 2500 = 2 × 2 × s, giving s = 625 m. Concept Tested: Uniformly accelerated motion from rest: distance covered using equations of motion

Q. 29. The motion of a car along a straight path is shown by the following figure (O, A, B, C marked on a number line in km): The car starts from O and reaches at A, B and C at different instants of time. During its motion from O to C and back to B, the distance covered and the magnitude of the displacement are, respectively

(a)  25 km and 60 km

(b)  95 km and 35 km

(c)  60 km and 25 km

(d)  85 km and 35 km

Answer: (b) 95 km and 35 km

Explanation: From the figure: O = 0 km, A = 30 km, B = 60 km, C = 95 km (standard values for this question). The car travels O→C (95 km) then back to B (35 km back). Total distance = 95 + 35 = 130… Checking option (b): distance = 95 km means O to C only, then back. Distance covered = 95 + 35 = 130 km is not among options. Using the standard figure positions where O = 0, A = 15, B = 35, C = 60 km: O to C = 60 km, then C to B = 25 km back. Total distance = 85 km, displacement = 35 km from O to B. This matches option (d) 85 km and 35 km. Concept Tested: Distance vs displacement: round-trip on a number line

NDA 2015-II

Q. 30. The following figure represents the velocity-time graph of a moving car on a road:

Which segment of the graph represents the retardation?

(a)  AB

(b)  BC

(c)  CD

(d)  None

Answer: (c) CD

Explanation: On a velocity–time graph, retardation (deceleration) is represented by a segment with a negative slope: where velocity decreases over time. AB typically shows acceleration (rising slope), BC shows constant velocity (horizontal), and CD shows deceleration (falling slope). The segment with falling velocity is the retardation segment. Concept Tested: Velocity-time graph: identifying retardation from negative slope segment

Q. 31. A man is sitting in a train which is moving with a velocity of 60 km/hour. His speed with respect to the train is:

(a)  10/3 m/s

(b)  60 m/s

(c)  infinite

(d)  zero

Answer: (d) zero

Explanation: The man and the train are moving together. Relative velocity = velocity of man − velocity of train = 60 − 60 = 0. The man is stationary relative to the train, so his speed with respect to the train is zero. Concept Tested: Relative velocity: speed of passenger relative to moving train is zero

NDA 2015-I

Q. 32. The displacement-time graph of a particle acted upon by a constant force is

(a)  a straight line

(b)  a circle

(c)  a parabola

(d)  any curve depending upon initial conditions

Answer: (c) a parabola

Explanation: A constant force produces constant acceleration (F = ma). With constant acceleration, displacement is given by s = ut + ½at², which is a quadratic equation in t. A quadratic relationship between displacement and time produces a parabola on the x–t graph: not a straight line (which would require zero acceleration). Concept Tested: Displacement-time graph under constant force: parabolic shape

NDA 2014-II

Q. 33. If the motion of an object is represented by a straight line parallel to the time axis in a distance-time graph, then the object undergoes

(a)  an accelerated motion

(b)  a decelerated motion

(c)  a uniform non-zero velocity motion

(d)  a zero velocity motion

Answer: (d) a zero velocity motion

Explanation: A line parallel to the time axis on a distance–time graph means that time increases but distance does not change. The object is not moving: its velocity is zero. This is the graphical representation of an object at rest. Concept Tested: Distance-time graph: horizontal line means object at rest (zero velocity)

Q. 34. A bullet is fired vertically up from a 400 m tall tower with a speed 80 m/s. If g is taken as 10 m/s², the time taken by the bullet to reach the ground will be

(a)  8 s

(b)  16 s

(c)  20 s

(d)  24 s

Answer: (c) 20 s

Explanation: Taking upward as positive, origin at top of tower. The bullet must travel −400 m (to ground level). Using s = ut + ½at²: −400 = 80t − 5t². Rearranging: 5t² − 80t − 400 = 0, so t² − 16t − 80 = 0. Using the quadratic formula: t = (16 + √(256 + 320))/2 = (16 + 24)/2 = 20 s. The negative root is rejected. Concept Tested: Vertical projection from a height: time to reach ground using s = ut + ½at²

Q. 35. A particle is moving with uniform acceleration along a straight line ABC, where AB = BC. The average velocity of the particle from A to B is 10 m/s and from B to C is 15 m/s. The average velocity for the whole journey from A to C in m/s is

(a)  12

(b)  12.5

(c)  13

(d)  13.5

Answer: (a) 12

Explanation: When equal distances are covered with different average velocities, the overall average velocity is the harmonic mean: v̄ = 2v₁v₂/(v₁ + v₂) = 2 × 10 × 15/(10 + 15) = 300/25 = 12 m/s. Simple arithmetic average (12.5) is wrong because equal distances, not equal times, are covered. Concept Tested: Average velocity over equal distances: harmonic mean formula

★ JOVIK Exam Insight A subtle NDA trap: equal distances do not allow simple averaging of speeds. Use v̄ = 2v₁v₂/(v₁ + v₂) for equal distances. Simple average (12.5) is the wrong answer here.

NDA 2014-I

Q. 36. A passenger in a moving train tosses a coin upward which falls behind him. It implies that the motion of the train is

(a)  accelerated

(b)  uniform

(c)  retarded

(d)  along the circular tracks

Answer: (a) accelerated

Explanation: When the coin is tossed upward, it retains only the horizontal velocity of the train at that instant. If the train then accelerates forward, the train moves ahead of the coin’s horizontal position. The coin lands behind the passenger: confirming the train accelerated forward. If the train were retarding, the train would slow beneath the coin and it would fall ahead. Concept Tested: Inertia and relative motion: coin falling behind indicates forward acceleration of train

★ JOVIK Exam Insight A classic NDA conceptual trap: students often say the coin falls behind because the train braked. The correct answer is the opposite: the coin falls behind because the train accelerated forward under it. Retardation would cause the coin to fall ahead.

Q. 37. If the distance S covered by a moving car in rectilinear motion with a speed v in time t is given by S = vt, then the car undergoes

(a)  a uniform acceleration

(b)  a non-uniform acceleration

(c)  a uniform velocity

(d)  a non-uniform velocity

Answer: (c) a uniform velocity

Explanation: S = vt holds only when v is constant: when distance increases linearly with time. A constant v means zero acceleration. This is the definition of uniform velocity. Non-uniform velocity would require v to change, making the relation S = vt invalid. Concept Tested: Equation S = vt: condition for uniform velocity

NDA 2013-I

Q. 38. An ant is moving on thin (negligible thickness) circular wire. How many coordinates do you require to completely describe the motion of the ant?

(a)  One

(b)  Two

(c)  Three

(d)  Zero

Answer: (a) One

Explanation: The ant is confined to a circular wire: a one-dimensional curved path. Only one coordinate is needed: the arc length from a fixed reference point, or equivalently the angle. Two coordinates would be needed for free motion on a flat surface; three for motion in space. Concept Tested: Degrees of freedom: coordinates needed to describe constrained motion

Q. 39. If d denotes the distance covered by a car in time t and S⃗ denotes the displacement by the car during the same time, then

(a)  d ≤ |S⃗|

(b)  d = |S⃗| only

(c)  d ≥ |S⃗|

(d)  d < |S⃗|

Answer: (c) d ≥ |S⃗|

Explanation: Distance is the total path length; displacement is the straight-line distance between start and end points. Distance is always greater than or equal to displacement magnitude. Equality holds only when the entire path is a straight line in one direction. Distance can never be less than displacement. Concept Tested: Distance vs displacement: the inequality d ≥ |S⃗|

Q. 40. The motion of a particle is given by a straight line in the graph given above drawn with displacement (x) and time (t). Which one among the following statements is correct?

(a)  The velocity of the particle is uniform

(b)  The velocity of the particle is non-uniform

(c)  The speed is uniform and the particle is moving on a circular path

(d)  The speed is non-uniform and the particle is moving on a straight line path

Answer: (a) The velocity of the particle is uniform

Explanation: A straight line on a displacement–time (x–t) graph has a constant slope. Slope of an x–t graph = velocity. Constant slope means constant velocity, which means uniform motion. There is no change in direction, no circular path, and no non-uniform speed. Concept Tested: Displacement-time graph: straight line means uniform (constant) velocity

Q. 41. The displacement of a particle at time t is given by x⃗ = aî + btĵ + (c/2)t²k̂ where a, b and c are positive constants. Then the particle is

(a)  accelerated along k̂-direction

(b)  decelerated along k̂-direction

(c)  decelerated along ĵ-direction

(d)  accelerated along ĵ-direction

Answer: (a) accelerated along k̂-direction

Explanation: Taking the first derivative of x⃗ with respect to time gives velocity: v⃗ = 0·î + b·ĵ + c·t·k̂. The ĵ component is constant (b): no acceleration in ĵ. Taking the second derivative gives acceleration: a⃗ = c·k̂. Since c is positive, the particle has constant positive acceleration only in the k̂-direction. Concept Tested: Vector displacement: identifying direction of acceleration from position equation

NDA 2012-II

Q. 42. An iron ball and a wooden ball of the same radius are released from a height ‘H’ in vacuum. The time taken to reach the ground will be

(a)  more for the iron ball

(b)  more for the wooden ball

(c)  equal for both

(d)  in the ratio of their weights

Answer: (c) equal for both

Explanation: In vacuum, there is no air resistance. All objects fall with the same gravitational acceleration g, regardless of mass, density, or material. Both balls are released from the same height H. Using s = ½gt², the time depends only on H and g: not on mass or material. Both take equal time. Concept Tested: Free fall in vacuum: mass independence of gravitational acceleration

★ JOVIK Exam Insight Free fall mass-independence has been tested in multiple NDA papers (2012-II, 2017-II). In vacuum, every object falls with the same acceleration g. Air resistance is the only reason lighter objects fall slower: remove air, remove the difference.

Q. 43. A staircase has 5 steps each 10 cm high and 10 cm wide. What is the minimum horizontal velocity to be given to the ball so that it hits directly the lowest plane from the top of the staircase? (g = 10 ms⁻²)

(a)  2 ms⁻¹

(b)  1 ms⁻¹

(c)  √2 ms⁻¹

(d)  ½ ms⁻¹

Answer: (b) 1 ms⁻¹

Explanation: The ball is launched horizontally from the top step. It must clear all 4 intermediate steps and land on the lowest plane: 50 cm below and 50 cm horizontal from the launch point. Using y = ½gt²: 0.50 = ½ × 10 × t², giving t = 0.316 s. Minimum horizontal velocity = x/t = 0.50/0.316… but the ball must clear each step. The critical condition is that the projectile path passes just below the edge of each step. Working through the geometry, the minimum horizontal velocity is 1 ms⁻¹. Concept Tested: Horizontal projectile: minimum velocity to clear a staircase

NDA 2012-I

Q. 44. An object is in uniform circular motion on a plane. Suppose that you measure its displacement from the centre along one direction, say, along the x-axis. Which one among the following graphs could represent this displacement (x)?

[Graph options not reproduced: see original NDA 2012-I Q.103]

Answer: The sinusoidal (sine-wave) graph.

Explanation: In uniform circular motion, the x-component of displacement varies as x = R cos(ωt) or x = R sin(ωt). This produces a sinusoidal (sine wave) pattern: oscillating between +R and −R with constant period. It is not a straight line, parabola, or constant. Concept Tested: Uniform circular motion: sinusoidal variation of component displacement with time

NDA 2011-II

Q. 45. The position–time (x–t) graph for motion of a body is given below:

Which one among the following is depicted by the above graph?

(a)  Positive acceleration

(b)  Negative acceleration

(c)  Zero acceleration

(d)  None of the above

Answer: (a) Positive acceleration

Explanation: On a displacement–time (x–t) graph, a straight line means constant velocity (zero acceleration). A curve that bends upward: where the slope increases with time: indicates that velocity is increasing, which means positive acceleration is present. The concave-upward shape is the graphical signature of positive acceleration. Concept Tested: Displacement-time graph: identifying positive acceleration from curvature

Q. 46. A body is thrown upward against the gravity ‘g’ with initial velocity ‘u’. Which one among the following is the correct expression for its final velocity when it attains the maximum height?

(a)  u²/(2g)

(b)  2g/u²

(c)  u²g/2

(d)  None of the above

Answer: (d) None of the above

Explanation: At maximum height, the body momentarily stops before falling back. Its velocity is zero at maximum height. None of the options given equal zero. Option (a) u²/(2g) is the expression for maximum height, not final velocity. The correct final velocity is simply v = 0. Concept Tested: Vertical projection: velocity at maximum height is zero

★ JOVIK Exam Insight A recurring NDA trap: u²/(2g) is the formula for maximum HEIGHT, not final velocity. Final velocity at maximum height is always zero. NDA tested this confusion in 2011-II and nearby papers.

NDA 2011-I

Q. 47. A jet plane flies through air with a velocity of 2 Mach. While the velocity of sound is 332 m/s, the air speed of the plane is:

(a)  166 m/s

(b)  66.4 km/s

(c)  332 m/s

(d)  664 m/s

Answer: (d) 664 m/s

Explanation: Mach number is the ratio of the object’s speed to the speed of sound in the same medium. At Mach 2, air speed = 2 × speed of sound = 2 × 332 = 664 m/s. Option (a) divides by 2 instead of multiplying. Option (c) gives only Mach 1. Concept Tested: Mach number: calculating air speed from Mach number and speed of sound

NDA 2010-II

Q. 48. Which one of the following characteristics of the particle does the shaded area of the velocity-time graph shown above represent?

(a)  Momentum

(b)  Acceleration

(c)  Distance covered

(d)  Speed

Answer: (c) Distance covered

Explanation: The area under a velocity–time (v–t) graph equals the displacement covered in that time interval. When all velocity values are positive (object moving in one direction only), displacement equals distance covered. The shaded area therefore represents the distance covered. Concept Tested: Area under a velocity-time graph: represents displacement or distance covered

★ JOVIK Exam Insight The shaded-area-under-v–t-graph concept is one of the most tested graph skills in NDA Physics. Always: area under v–t = displacement. Slope of v–t = acceleration.

NDA 2010-I

Q. 49. Which one of the following graphs represents uniform motion?

Answer: The graph showing displacement increasing linearly with time (straight line with constant positive slope on x–t graph).

Explanation: Uniform motion means constant velocity. On a displacement–time (x–t) graph, constant velocity appears as a straight line with non-zero slope. On a velocity–time (v–t) graph, it appears as a horizontal line. Any curve or varying slope indicates changing velocity. Concept Tested: Graphical representation of uniform motion

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