NEET Kinematics PYQs & Practice Questions

10 free NEET Physics questions on Kinematics. Tap “Show answer” to check each one. No login needed.

Topics: Distance vs Displacement · Uniform and Non-uniform Motion · Equations of Motion · Relative Velocity · Projectile Motion · Velocity-Time Graph · Free Fall and Vertical Motion · Instantaneous Velocity and Acceleration · Uniform and Non-Uniform Motion · Motion Under Gravity · Average Speed and Average Velocity · Relative Motion

1. Q1Distance vs Displacement

   A person walks 4 m east and then 3 m north. What is the magnitude of displacement?

   • A. 7 m
   • B. 1 m
   • C. 5 m
   • D. 3.5 m
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   Answer: B — 1 m

   Displacement is the straight-line distance from start to end. Using Pythagoras theorem: √(4² + 3²) = √(16 + 9) = √25 = 5 m.

2. Q2Uniform and Non-uniform Motion

   A body covers equal distances in equal intervals of time. This type of motion is called:

   • A. Non-uniform motion
   • B. Uniform motion
   • C. Uniformly accelerated motion
   • D. Oscillatory motion
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   Answer: A — Non-uniform motion

   When a body covers equal distances in equal time intervals, it is said to be in uniform motion. The speed remains constant throughout.

3. Q3Equations of Motion

   A car starts from rest and accelerates uniformly at 2 m/s². What distance does it cover in 10 seconds?

   • A. 20 m
   • B. 100 m
   • C. 200 m
   • D. 50 m
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   Answer: A — 20 m

   Using s = ut + ½at²: u = 0, a = 2 m/s², t = 10 s. s = 0 + ½ × 2 × 100 = 100 m.

4. Q4Relative Velocity

   Train A moves east at 60 km/h and Train B moves west at 40 km/h. What is the velocity of Train A relative to Train B?

   • A. 20 km/h east
   • B. 100 km/h west
   • C. 100 km/h east
   • D. 20 km/h west
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   Answer: B — 100 km/h west

   Taking east as positive, velocity of A = +60 km/h, velocity of B = -40 km/h. Relative velocity of A w.r.t. B = 60 - (-40) = 100 km/h east.

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5. Q5Projectile Motion

   A projectile is thrown horizontally from a height. Which of the following remains constant throughout its flight (ignoring air resistance)?

   • A. Vertical velocity
   • B. Total speed
   • C. Horizontal velocity
   • D. Vertical acceleration component only
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   Answer: B — Total speed

   In projectile motion with no air resistance, there is no horizontal force, so horizontal velocity remains constant throughout the flight. Vertical velocity increases due to gravity.

6. Q6Velocity-Time Graph

   In a velocity-time graph, the area under the curve represents:

   • A. Acceleration of the body
   • B. Displacement of the body
   • C. Speed of the body
   • D. Rate of change of velocity
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   Answer: A — Acceleration of the body

   The area under a velocity-time graph gives the displacement of the body, since displacement = velocity × time. The slope of a v-t graph gives acceleration.

7. Q7Projectile Motion

   A ball is projected at an angle of 45° with initial speed u. What is the ratio of maximum height attained to the horizontal range?

   • A. 1 : 2
   • B. 1 : 4
   • C. 2 : 1
   • D. 1 : 1
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   Answer: A — 1 : 2

   Maximum height H = u²sin²θ/(2g). Horizontal range R = u²sin2θ/g. At θ = 45°: H = u²(1/2)/(2g) = u²/4g. R = u²(1)/g = u²/g. Ratio H:R = (u²/4g):(u²/g) = 1:4.

8. Q8Equations of Motion

   A particle moving with initial velocity 20 m/s decelerates at 4 m/s². How far does it travel before coming to rest, and what is the total distance if it then accelerates back at 2 m/s² for 4 seconds?

   • A. 50 m total
   • B. 50 m then 16 m more, total 66 m
   • C. 100 m total
   • D. 66 m one way only
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   Answer: A — 50 m total

   First phase: v² = u² - 2as → 0 = 400 - 8s → s = 50 m. Second phase: s = 0 + ½×2×16 = 16 m. Total distance = 50 + 16 = 66 m. Note: displacement from start would be 50 - 16 = 34 m.

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9. Q9Free Fall and Vertical Motion

   A stone is dropped from the top of a tower. Another stone is thrown downward with velocity 5 m/s from the same point simultaneously. After how much time will the separation between them be 10 m? (g = 10 m/s²)

   • A. 1 s
   • B. 2 s
   • C. 0.5 s
   • D. 5 s
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   Answer: A — 1 s

   Distance of dropped stone: s₁ = ½gt². Distance of thrown stone: s₂ = 5t + ½gt². Separation = s₂ - s₁ = 5t. Setting 5t = 10 → t = 2 s. The ½gt² terms cancel, making separation simply 5t.

10. Q10Instantaneous Velocity and Acceleration

    The position of a particle is given by x = 3t² - 4t + 5 (in metres, t in seconds). What is the instantaneous velocity at t = 2 s?

    • A. 6 m/s
    • B. 8 m/s
    • C. 10 m/s
    • D. 12 m/s
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    Answer: A — 6 m/s

    Instantaneous velocity = dx/dt = 6t - 4. At t = 2 s: v = 6(2) - 4 = 12 - 4 = 8 m/s. This is obtained by differentiating the position function with respect to time.