Motion in a Plane NEET PYQs (2005–2026) with Solutions

10 solved Motion in a Plane PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2025)

   NEET 2025HardResolution & Relative Velocity

   Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of service and the speed (assumed constant) of the buses.

   1. AOption A: 9 min, 40 km/h
   2. BOption B: 25 min, 100 km/h
   3. COption C: 10 min, 90 km/h
   4. DOption D: 15 min, 120 km/h

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   Correct answer: (D) 15 min, 120 km/h

   Explanation

   Let bus speed be v and interval between buses be T hours. Same direction condition: v − 60 = distance between buses / 0.5 h. Opposite direction condition: v + 60 = distance between buses / (1/6) h. Solving gives v = 120 km/h and T = 15 min.

2. Question 2 (NEET 2024)

   NEET 2024EasyUniform Circular Motion

   A particle moving with uniform speed in a circular path maintains:

   1. AOption A: constant velocity but varying acceleration.
   2. BOption B: varying velocity and varying acceleration.
   3. COption C: constant velocity.
   4. DOption D: constant acceleration.

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   Correct answer: (B) varying velocity and varying acceleration.

   Explanation

   In uniform circular motion, speed remains constant but the direction of velocity changes continuously. Therefore, velocity varies. The centripetal acceleration also changes direction continuously, so acceleration is varying. Hence, option (B) is correct.

3. Question 3 (NEET 2023)

   NEET 2023MediumProjectile Motion

   A bullet is fired from a gun at the speed of 280 m s⁻¹ in the direction 30° above the horizontal. The maximum height attained by the bullet is : (g = 9.8 m s⁻², sin 30° = 0.5)

   1. AOption A: 2800 m
   2. BOption B: 2000 m
   3. COption C: 1000 m
   4. DOption D: 3000 m

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   Correct answer: (C) 1000 m

   Explanation

   Vertical component of velocity: uy = 280 × sin30° = 280 × 0.5 = 140 m/s Maximum height: H = uy²/(2g) = 140² / (2 × 9.8) = 19600 / 19.6 = 1000 m Hence, option (C) is correct.

4. Question 4 (NEET 2022)

   NEET 2022EasyProjectile Motion

   A ball is projected with a velocity 10 ms⁻¹, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be:

   1. AOption A: 10 ms⁻¹
   2. BOption B: Zero
   3. COption C: 5√3 ms⁻¹
   4. DOption D: 5 ms⁻¹

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   Correct answer: (C) 5√3 ms⁻¹

   Explanation

   Angle with horizontal: θ = 30° At highest point, vertical velocity = 0. Remaining horizontal velocity: v = u cos30° = 10 × √3/2 = 5√3 ms⁻¹ Hence option (C) is correct.

5. Question 5 (NEET 2021)

   NEET 2021MediumProjectile Motion

   A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s²)

   1. AOption A: 20√2 m/s, 10 m/s²
   2. BOption B: 20 m/s, 5 m/s²
   3. COption C: 20 m/s, 0
   4. DOption D: 20√2 m/s, 0

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   Correct answer: (A) 20√2 m/s, 10 m/s²

   Explanation

   Velocity of car at t=4s: v = u + at = 0 + 5×4 = 20 m/s. Ball retains horizontal velocity = 20 m/s. After 2 s, vertical velocity = gt = 20 m/s. Resultant velocity = √(20²+20²)=20√2 m/s. Acceleration = g = 10 m/s².

6. Question 6 (NEET 2021)

   NEET 2021MediumProjectile Motion

   A particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it equals $4R$. The angle of projection, $\theta$, is given by:

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (A) Option 1

   Explanation

   For circular motion, speed of the particle is: $$v = \frac{2\pi R}{T}$$ For projectile motion, maximum height: $$H = \frac{v^2 \sin^2\theta}{2g}$$ Given: $$H = 4R$$ Substituting: $$\frac{v^2 \sin^2\theta}{2g} = 4R$$ $$\frac{\left(\frac{2\pi R}{T}\right)^2 \sin^2\theta}{2g} = 4R$$ $$\frac{4\pi^2 R^2 \sin^2\theta}{2gT^2} = 4R$$ $$\frac{2\pi^2 R \sin^2\theta}{gT^2} = 4$$ $$\sin^2\theta = \frac{2gT^2}{\pi^2 R}$$ $$\theta = \sin^{-1}\left(\sqrt{\frac{2gT^2}{\pi^2 R}}\right)$$ Hence, option (A) is correct.

7. Question 7 (NEET 2019)

   NEET 2019MediumUniform Circular Motion

   A particle starting from rest, moves in a circle of radius r. It attains a velocity of V₀ m/s in the nᵗʰ round. Its angular acceleration will be:

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 14

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   Correct answer: (D) Option 14

   Explanation

   Initial speed u = 0 Final speed v = V₀ Tangential acceleration a = αr Distance covered in n rounds: s = 2πrn Using equation of motion: v² = u² + 2as V₀² = 0 + 2(αr)(2πrn) V₀² = 4πnαr² Therefore, α = V₀²/(4πnr²)

8. Question 8 (NEET 2019)

   NEET 2019MediumProjectile Motion

   Two bullets are fired horizontally and simultaneously towards each other from rooftops of two buildings $100\ \text{m}$ apart and of same height $200\ \text{m}$, with the same velocity of $25\ \text{m/s}$ $(g = 10\ \text{m/s}^2)$. When and where will the two bullets collide?

   1. AOption A: They will not collide
   2. BOption B: After $2\ \text{s}$ at a height of $180\ \text{m}$
   3. COption C: After $2\ \text{s}$ at a height of $20\ \text{m}$
   4. DOption D: After $4\ \text{s}$ at a height of $120\ \text{m}$

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   Correct answer: (B) After $2\ \text{s}$ at a height of $180\ \text{m}$

   Explanation

   Let the bullets collide after time $t$. Since both bullets move horizontally towards each other with equal speeds: $$x_1 + x_2 = 100$$ $$25t + 25t = 100$$ $$50t = 100$$ $$t = 2\ \text{s}$$ In this time, the vertical distance fallen by each bullet is: $$y = \frac{1}{2}gt^2$$ $$y = \frac{1}{2} \times 10 \times (2)^2$$ $$y = 20\ \text{m}$$ Since the initial height is $200\ \text{m}$, the height at collision is:$$h = 200 - 20 = 180\ \text{m}$$Hence, the bullets collide after$2\ \text{s}$at a height of$180\ \text{m}$ above the ground.

9. Question 9 (NEET 2017)

   NEET 2017MediumMotion in a Plane

   The $x$ and $y$ coordinates of the particle at any time are $x = 5t - 2t^2$ and $y = 10t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2\ \text{s}$ is

   1. AOption A: $-8\ \text{m/s}^{2}$
   2. BOption B: $0$
   3. COption C: $5\ \text{m/s}^{2}$
   4. DOption D: $-4\ \text{m/s}^{2}$

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   Correct answer: (D) $-4\ \text{m/s}^{2}$

   Explanation

   Given: $$x = 5t - 2t^2, \qquad y = 10t$$ Velocity components are: $$v_x = \frac{dx}{dt} = 5 - 4t$$ $$v_y = \frac{dy}{dt} = 10$$ Acceleration components are: $$a_x = \frac{dv_x}{dt} = -4$$ $$a_y = \frac{dv_y}{dt} = 0$$ Hence, acceleration of the particle at $t = 2\ \text{s}$is:$$-4\ \text{m s}^{-2}$$

10. Question 10 (AIPMT 2015)

    AIPMT 2015MediumVector Operations

    If vectors $$\vec{A}=\cos\omega t\,\hat{i}+\sin\omega t\,\hat{j}$$ $$\vec{B}=\cos\left(\frac{\omega t}{2}\right)\hat{i}+\sin\left(\frac{\omega t}{2}\right)\hat{j}$$ are functions of time, then the value of $t$ at which they are orthogonal to each other is:

    1. AOption A: $t=0$
    2. BOption B: $t=\dfrac{\pi}{4\omega}$
    3. COption C: $t=\dfrac{\pi}{2\omega}$
    4. DOption D: $t=\dfrac{\pi}{\omega}$

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    Correct answer: (D) $t=\dfrac{\pi}{\omega}$

    Explanation

    For orthogonal vectors: $$\vec{A}\cdot\vec{B}=0$$ Therefore, $$\cos(\omega t)\cos\left(\frac{\omega t}{2}\right)+\sin(\omega t)\sin\left(\frac{\omega t}{2}\right)=0$$ Using: $$\cos x\cos y+\sin x\sin y=\cos(x-y)$$ $$\cos\left(\omega t-\frac{\omega t}{2}\right)=0$$ $$\cos\left(\frac{\omega t}{2}\right)=0$$ The least positive solution is: $$\frac{\omega t}{2}=\frac{\pi}{2}$$ Hence, $$t=\frac{\pi}{\omega}$$ Therefore, option (D) is correct.