Kinematics — NEET Physics MCQs with answers

Kinematics is a Class 11 Physics chapter in the NEET (UG) syllabus. NEET720 has 1,009 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Motion under gravity · Graphical Analysis · Speed and Velocity · Non-uniform Acceleration · Reference Frames · Graphical analysis of motion · Scalars and vectors · Motion basics and frames of reference · Motion in a straight line with uniform acceleration · Position-time graphs · Motion graphs · Relative velocity in two dimensions · Relative velocity · Vectors · Projectile motion · Uniform circular motion (kinematics) · Equations of motion · Average speed and velocity · Kinematics with Variable Acceleration · Motion in a Straight Line · Uniform Circular Motion · Uniformly Accelerated Motion · Relative Motion · Motion in a Plane · Distance and Displacement · Graphical Motion · Kinematics · 1D Motion · Circular motion · Basic kinematics · Graphs · Circular motion + 1D braking · Projectile + accelerated horizontal wind · Circular release + projectile fall · Graphs + relative motion · Vectors + 1D component growth · Circular + tangential braking · Relative Velocity + Graphs · Projectile + Circular Motion · 1D Kinematics + Relative Motion

8 free Kinematics practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Motion basics and frames of reference

   The magnitude of a particle's average velocity over an interval equals its average speed over the same interval

   • A.for every one-dimensional motion
   • B.only when the acceleration is zero throughout
   • C.only when the particle moves along a straight line without reversing direction
   • D.never, if the particle is accelerating
   Show answer and explanation

   Answer: C. only when the particle moves along a straight line without reversing direction

   Equality needs distance = |displacement|, which happens exactly when the path is straight and never doubles back. Acceleration along that fixed direction is allowed.

   Average speed = distance/time and |average velocity| = |displacement|/time, so they are equal iff distance = |displacement|. That requires the path to coincide with the straight segment joining the endpoints, traversed one way only. A ball speeding up in free fall satisfies this even though it accelerates, ruling out B and D. A particle that goes forward then back in 1D violates it, ruling out A.

   Common mistake: Thinking all 1D motion qualifies, forgetting direction reversal.

2. Question 2 · easy · Motion basics and frames of reference

   A passenger walks toward the front of a train that is moving at a constant velocity relative to the ground. Which statement correctly describes the passenger's motion?

   • A.The passenger is in motion relative to the train and also in motion relative to the ground, and the two velocities are generally different
   • B.The passenger is at rest relative to the ground because the train's motion cancels out the walking
   • C.Motion is meaningless unless stated relative to the ground, since the ground is the only valid reference frame
   • D.The passenger's velocity is the same number in every frame, since velocity is a property of the object alone
   Show answer and explanation

   Answer: A. The passenger is in motion relative to the train and also in motion relative to the ground, and the two velocities are generally different

   Velocity is always relative to a chosen frame. The passenger has one velocity relative to the train (her walking speed) and a different velocity relative to the ground (train's velocity plus her walking velocity).

   Motion has no meaning without specifying a reference frame. Relative to the train, the passenger simply walks forward at her own walking speed. Relative to the ground, her velocity is the vector sum of the train's velocity and her velocity relative to the train, which is generally a different number from either individual value. Option B wrongly assumes the two motions cancel; option C wrongly privileges the ground frame as the only legitimate one — any inertial frame works, though the numerical value of velocity changes between frames; option D denies the very fact that motion is frame-dependent.

   Common mistake: Assuming a single 'true' velocity exists independent of the frame of reference.

3. Question 3 · medium · Motion basics and frames of reference

   A ball is released from rest by a passenger inside a train moving at constant velocity along a straight, level track. Which statement correctly compares the ball's trajectory as seen by the passenger inside the train and by a person standing still on the platform outside?

   • A.A straight vertical line in both frames, since the ball is simply falling under gravity
   • B.A straight horizontal line for the passenger and a straight vertical line for the platform observer
   • C.A curved path for the passenger and a straight vertical line for the platform observer
   • D.A straight vertical line for the passenger, but a curved (parabolic) path for the platform observer
   Show answer and explanation

   Answer: D. A straight vertical line for the passenger, but a curved (parabolic) path for the platform observer

   In the train's frame the ball has zero horizontal velocity, so it simply drops straight down; in the ground frame it retains the train's horizontal velocity while gaining downward speed, producing a curved path.

   At the instant of release, the ball shares the train's constant horizontal velocity. In the train's own frame (moving at that same horizontal velocity), the ball's horizontal velocity is zero, so it falls straight down under gravity — a vertical line. In the ground frame, the ball keeps moving horizontally at the train's speed while accelerating downward under gravity, giving a combination of uniform horizontal motion and uniformly accelerated vertical motion — a parabolic (curved) path. This illustrates that the shape of a trajectory is frame-dependent, even though the underlying physics (gravity acting downward) is the same in both frames.

   Common mistake: Assuming the trajectory's shape (straight vs curved) must be the same for every observer.

4. Question 4 · medium · Motion graphs

   A particle starts from the origin. Its velocity-time graph is positive for 0 < t < 4 s, crosses zero at t = 4 s, and is negative afterwards. Which statement about t = 4 s is correct?

   • A.The particle has returned to the origin
   • B.The particle is at its maximum distance from the origin
   • C.The particle's acceleration is zero at that instant
   • D.The particle remains at rest for all t > 4 s
   Show answer and explanation

   Answer: B. The particle is at its maximum distance from the origin

   Positive v means x increases until t = 4 s; negative v afterwards means x decreases. So x peaks exactly at the zero crossing — the turning point, farthest from the origin.

   Displacement accumulates as the area under v-t. For 0 < t < 4 s the area (and x) grows; after 4 s the negative velocity subtracts from x. Hence x is maximum at t = 4 s: the particle momentarily stops and reverses. It has not returned to the origin (that happens later, when the negative area cancels the positive area). The acceleration at t = 4 s equals the slope of the v-t graph there, which is negative, not zero.

   Common mistake: Thinking the particle is back at the origin when its v-t graph crosses the time axis.

5. Question 5 · medium · Motion graphs

   The position-time graphs of two cyclists riding along the same straight road are two straight lines of different slopes that intersect at time t₀. The intersection means that at t₀ the cyclists have the same

   • A.velocity
   • B.acceleration
   • C.speed for the entire journey
   • D.position — they are side by side at that instant
   Show answer and explanation

   Answer: D. position — they are side by side at that instant

   A point on an x-t graph is a (time, position) pair; two curves crossing share the same x at the same t — the riders meet or pass each other there.

   On a position-time plot, intersection means identical coordinates: same instant, same position — the faster cyclist catches or passes the slower one. Their velocities are the slopes, which remain different; that is exactly why the lines cross rather than run parallel. Students often transfer 'crossing' on a v-t graph (equal velocities) to an x-t graph, but each graph type must be read by its own axes.

   Common mistake: Claiming equal velocities at the crossing, importing v-t intuition into an x-t plot.

6. Question 6 · medium · Motion graphs

   A segment of a particle's velocity-time graph lies below the time axis and slopes downward (the velocity is becoming more negative). During this segment the particle is

   • A.slowing down, because the graph slopes downward
   • B.speeding up while moving in the negative direction
   • C.momentarily at rest
   • D.moving in the positive direction with decreasing speed
   Show answer and explanation

   Answer: B. speeding up while moving in the negative direction

   Below the axis, v < 0: motion is in the negative direction. Sloping further downward makes |v| larger — the particle speeds up.

   Two independent readings: the sign of v (below the axis → negative direction of motion) and the trend of |v| (v going from, say, -2 to -6 m/s means speed 2 → 6 m/s, increasing). A body speeds up whenever velocity and acceleration have the same sign — here both negative. 'Downward slope = slowing down' is only true above the axis; below it, the same slope direction means the opposite. This is the standard trap when reading braking versus reversing on v-t graphs.

   Common mistake: Reading any downward-sloping v-t segment as deceleration.

7. Question 7 · medium · Motion graphs

   An object's velocity-time graph consists of two straight segments: it rises linearly from 0 to 10 m/s during 0 to 2 s, and then stays constant at 10 m/s from 2 s to 5 s. The total distance covered in 5 s is

   • A.40 m
   • B.50 m
   • C.30 m
   • D.35 m
   Show answer and explanation

   Answer: A. 40 m

   Distance = area under v-t graph = triangle (0 to 2 s) + rectangle (2 to 5 s) = 10 + 30 = 40 m.

   Area of triangle for 0–2 s: (1/2)(2)(10) = 10 m. Area of rectangle for 2–5 s: (3)(10) = 30 m. Total distance = 10 + 30 = 40 m.

   Common mistake: Forgetting to add the triangular ramp-up portion separately

8. Question 8 · medium · Motion graphs

   A car's velocity-time graph is a straight line rising from 5 m/s at t = 0 to 25 m/s at t = 4 s. The displacement covered in these 4 s is

   • A.100 m
   • B.60 m
   • C.20 m
   • D.40 m
   Show answer and explanation

   Answer: B. 60 m

   Displacement equals the trapezoidal area under the v-t graph: (1/2)(5 + 25)(4) = 60 m.

   The v-t graph is a trapezoid with parallel sides 5 m/s and 25 m/s and height (time) 4 s. Area = (1/2)(sum of parallel sides)(height) = (1/2)(5+25)(4) = (1/2)(30)(4) = 60 m.

   Common mistake: Ignoring the initial velocity and computing only the triangular increase

Questions about Kinematics for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (1,009 active practice questions in this chapter today).