Why this chapter decides more marks than it appears to
Kinematics is worth preparing carefully for a reason that has nothing to do with how many questions come from it directly. It is the chapter every other mechanics chapter is built on. Laws of Motion asks you to find acceleration and then use it; Work, Energy and Power asks you to relate speeds at two positions; Rotational Motion repeats the entire structure with angular quantities. A student who is slow at setting up displacement, velocity and acceleration pays that cost three more times.
It is also the chapter where the gap between recognising a formula and knowing when it applies is widest. Almost every student can recite v = u + at. The questions that separate scores are the ones where that equation is not valid, and the paper does not announce which those are.
How JEE actually asks Kinematics
NTA does not publish chapter-wise weightage for JEE Main, so every numeric breakdown you find online is a coaching estimate reverse-engineered from past papers. Treat those figures as indicative, and check the official information bulletin on the NTA website for the syllabus and exam pattern that actually applies to your session.
What is observable from past papers is the style rather than the count. Recent Kinematics questions lean toward graph interpretation, non-uniform acceleration requiring calculus, and relative motion - and away from direct substitution into the equations of motion. Projectile questions still appear regularly, but often combined with another idea, such as asking for the velocity at a particular instant rather than the range.
Key concepts, compressed
- ✓Distance is the total path length and is never negative. Displacement is the straight-line change in position and can be zero for a non-zero distance.
- ✓Average velocity is displacement over total time. Average speed is distance over total time. They are equal only for motion in a single direction without reversal.
- ✓Acceleration is the rate of change of velocity, so a body can have zero velocity and non-zero acceleration - a ball at the top of its flight is the standard example.
- ✓For constant acceleration, the three equations of motion apply. For variable acceleration, use a = dv/dt, v = dx/dt, and a = v(dv/dx).
- ✓In two dimensions, motion along perpendicular axes is independent. This is what makes projectile motion tractable.
- ✓Relative velocity of A with respect to B is v(A) − v(B), a vector subtraction. In one dimension, same direction subtracts and opposite directions adds.
Formulas you need before attempting the questions
| Quantity | Formula | Valid when |
|---|---|---|
| Velocity | v = u + at | constant acceleration |
| Displacement | s = ut + ½at² | constant acceleration |
| Velocity-displacement | v² = u² + 2as | constant acceleration |
| Distance in the nth second | sₙ = u + a(2n − 1)/2 | constant acceleration |
| Variable acceleration | a = dv/dt = v(dv/dx) | always |
| Projectile time of flight | T = 2u sinθ / g | level ground, no air resistance |
| Projectile maximum height | H = u² sin²θ / 2g | level ground, no air resistance |
| Projectile range | R = u² sin2θ / g | level ground, no air resistance |
| Relative velocity | v(A rel B) = v(A) − v(B) | always |
The five mistakes that cost the most marks
- ✓Using v = u + at when acceleration is not constant. If the question gives a as a function of t or x, the equations of motion do not apply and you must integrate.
- ✓Reading 'in the 5th second' as 'in the first 5 seconds'. For a body starting from rest with a = 2 m/s², these are 9 m and 25 m respectively - a difference large enough that no option will save you.
- ✓Taking the slope of a v-t graph as displacement. Slope is acceleration; area under the curve is displacement, and area below the time axis is negative displacement.
- ✓Treating the two components of projectile motion as connected. Horizontal velocity never changes; vertical motion is free fall. Solve them separately and combine only at the end.
- ✓Switching reference frames mid-problem. In river-boat and rain-man questions, decide whether you are working from the ground or from the moving object, write it down, and stay there.
Practice set 1: motion in a straight line
Eight questions on one-dimensional motion. Work each one before reading the solution - the value of a practice set collapses if you read the answer first.
1. A body starts from rest with a constant acceleration of 2 m/s². What distance does it cover in the 5th second?
9 m. Use sₙ = u + a(2n − 1)/2 with u = 0, a = 2 and n = 5: sₙ = 0 + 2(2 × 5 − 1)/2 = 2 × 9/2 = 9 m. Note the trap directly: the distance in the first 5 seconds is s = ut + ½at² = ½ × 2 × 25 = 25 m. The question asked for the 5th second alone, which is the interval from t = 4 s to t = 5 s.
2. A car moving at 20 m/s decelerates uniformly and comes to rest in 5 s. How far does it travel while stopping?
50 m. First find the acceleration: v = u + at gives 0 = 20 + a(5), so a = −4 m/s². Then v² = u² + 2as gives 0 = 400 + 2(−4)s, so s = 50 m. You could equally use the average-velocity shortcut: for constant acceleration the average velocity is (20 + 0)/2 = 10 m/s, and 10 × 5 = 50 m.
3. A particle moves such that x = 3t² − 2t + 1, with x in metres and t in seconds. What is its velocity at t = 2 s?
10 m/s. Velocity is dx/dt = 6t − 2. At t = 2, v = 12 − 2 = 10 m/s. Do not substitute t = 2 into the position equation and divide by 2 - that would give the average velocity from the origin of time, which is a different quantity and a common error.
4. For the same particle, what is the acceleration at t = 2 s?
6 m/s², and it is the same at every instant. a = dv/dt = d(6t − 2)/dt = 6. Because the position is quadratic in t, the acceleration is constant - which means the equations of motion would have worked here too. Recognising that saves time.
5. A ball is dropped from rest at a height of 80 m. How long does it take to reach the ground? (g = 10 m/s²)
4 s. With u = 0, s = ut + ½at² becomes 80 = ½ × 10 × t², so t² = 16 and t = 4 s. The negative root is discarded as unphysical. Note that the mass of the ball is irrelevant and is often supplied in the question purely as a distractor.
6. With what speed does that ball strike the ground?
40 m/s. Either v = u + at = 0 + 10 × 4 = 40 m/s, or v² = u² + 2as = 0 + 2 × 10 × 80 = 1600, giving v = 40 m/s. The second route is faster when the question does not ask for time, and avoids carrying an intermediate result.
7. A particle travels 4 m east and then 3 m north. What are the distance travelled and the magnitude of the displacement?
Distance 7 m, displacement 5 m. Distance is the total path length, 4 + 3 = 7 m. Displacement is the straight line from start to finish, and since the two legs are perpendicular its magnitude is √(4² + 3²) = 5 m. Questions asking for average speed and average velocity in the same stem are testing exactly this distinction.
8. A body covers the first half of a journey at 20 m/s and the second half at 30 m/s. What is its average speed for the whole journey?
24 m/s. Average speed is total distance over total time, not the arithmetic mean of the speeds. Let each half be d. Time taken is d/20 + d/30 = 5d/60 = d/12. Total distance is 2d, so average speed = 2d ÷ (d/12) = 24 m/s. The answer 25 m/s comes from averaging the speeds, and it is the option the paper will offer you. Equal distances give the harmonic mean 2v₁v₂/(v₁ + v₂); equal times would give the arithmetic mean.
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Practice set 2: motion graphs
Graph questions are increasingly common and are answered by knowing what slope and area mean on each type of graph - not by calculation.
9. On a velocity-time graph, what does the slope represent?
Acceleration. Slope is the change in the vertical quantity over the change in the horizontal one, which here is Δv/Δt - the definition of acceleration. A straight line therefore means constant acceleration, and a horizontal line means zero acceleration.
10. On a velocity-time graph, what does the area between the curve and the time axis represent?
Displacement. Area is velocity multiplied by time, which is displacement. Crucially, area below the time axis counts as negative because the velocity is negative there. That is why a body can have a large total area in magnitude and zero net displacement - and why questions distinguish carefully between the two.
11. On a position-time graph, what does the slope represent?
Velocity. Slope is Δx/Δt. A curved position-time graph therefore means the velocity is changing, so the body is accelerating. A horizontal line means the body is at rest, and a steeper line means a greater speed.
12. A ball is thrown vertically upward. At the highest point of its flight, what are its velocity and acceleration?
Velocity is zero; acceleration is g downward, which is 10 m/s². This is the single most-tested conceptual point in the chapter. Zero velocity does not imply zero acceleration - if the acceleration were zero at that instant, the ball would stay there. Gravity acts continuously throughout the flight, including at the top.
13. A velocity-time graph is a horizontal straight line above the time axis. Describe the motion.
Uniform velocity in the positive direction, with zero acceleration. Since the slope is zero, the acceleration is zero, and since the line sits above the axis the velocity is positive and constant. The corresponding position-time graph would be a straight line with a positive slope, and the acceleration-time graph would lie along the time axis.
Practice set 3: projectile motion
Every question here uses the same method: split into horizontal and vertical components, treat them independently, and recombine only if the question asks for a resultant. Take g = 10 m/s².
14. A projectile is launched at 20 m/s at 30° to the horizontal. What is its time of flight on level ground?
2 s. T = 2u sinθ/g = 2 × 20 × sin30° / 10. Since sin30° = 0.5, this is 2 × 20 × 0.5 / 10 = 20/10 = 2 s. A useful check: the time to reach the top is half of this, 1 s, which is also u sinθ/g = 20 × 0.5/10 = 1 s.
15. For the same projectile, what is the maximum height reached?
5 m. H = u² sin²θ / 2g = (20)² × (0.5)² / (2 × 10) = 400 × 0.25 / 20 = 100/20 = 5 m. Note that sin²θ means (sinθ)², so you square the 0.5 before multiplying - squaring the angle instead is a frequent slip.
16. For the same projectile, what is the horizontal range?
About 34.6 m. R = u² sin2θ / g = 400 × sin60° / 10. With sin60° ≈ 0.866, R ≈ 400 × 0.866/10 ≈ 34.6 m. Note the 2θ: the range formula uses double the launch angle, so a 30° launch involves sin60°, not sin30°. Substituting sinθ here is one of the most common errors in the chapter.
17. At what launch angle is the range maximum, for a fixed launch speed?
45°. R = u² sin2θ/g is greatest when sin2θ = 1, which requires 2θ = 90° and so θ = 45°. This assumes level ground and no air resistance - for a projectile launched from a height, the optimum angle is less than 45°, which advanced questions exploit.
18. Two projectiles are launched at the same speed with angles of 30° and 60°. How do their ranges compare?
They are equal. R depends on sin2θ, and sin60° = sin120°, so both give the same range. In general, complementary angles θ and (90° − θ) produce equal ranges at the same speed. Their times of flight and maximum heights differ, however - the 60° launch goes higher and stays up longer, which is what the follow-up question usually asks.
19. A projectile is launched at 20 m/s at 30°. What is its speed at the highest point?
About 17.3 m/s. At the highest point the vertical component of velocity is zero, so the speed equals the horizontal component, which never changes: u cosθ = 20 × cos30° ≈ 20 × 0.866 ≈ 17.3 m/s. Answering 'zero' is the classic error - that is true for a body thrown straight up, not for a projectile with a horizontal component.
20. During projectile motion with no air resistance, which quantity remains constant throughout?
The horizontal component of velocity. Gravity acts vertically, so it changes only the vertical component. The horizontal component is unchanged for the entire flight, which is exactly why the horizontal motion can be treated as uniform velocity while the vertical motion is treated as free fall.
21. A ball is thrown horizontally at 15 m/s from a height of 20 m. How long does it take to land, and how far from the base does it land?
2 s and 30 m. Vertically it starts with zero vertical velocity, so 20 = ½ × 10 × t², giving t² = 4 and t = 2 s. Horizontally the velocity is constant, so the distance is 15 × 2 = 30 m. Notice that the horizontal launch speed has no effect on the time to fall - a ball dropped from the same height lands at the same instant.
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Practice set 4: relative velocity
The recurring instruction for this set: name your reference frame before you calculate anything.
22. Two cars travel in the same direction at 60 km/h and 40 km/h. What is the velocity of the first relative to the second?
20 km/h in the direction of motion. Relative velocity is a vector subtraction: v(1 rel 2) = 60 − 40 = 20 km/h. Physically, the faster car pulls ahead at 20 km/h, which is why an overtaking manoeuvre feels slow even at high absolute speeds.
23. The same two cars now travel toward each other. What is their relative velocity?
100 km/h. Taking one direction as positive, the second car's velocity is −40 km/h, so v(1 rel 2) = 60 − (−40) = 100 km/h. Same subtraction, different signs - which is the whole reason to define a positive direction before starting rather than reasoning about 'adding when opposite'.
24. A river 100 m wide flows at 5 m/s. A boat that can travel at 10 m/s in still water heads straight across, perpendicular to the bank. How long does it take to cross, and how far downstream does it land?
10 s and 50 m downstream. The crossing time depends only on the component perpendicular to the bank, which is 10 m/s: t = 100/10 = 10 s. The current does not affect crossing time at all, because it acts along the bank. The drift is then the current speed times that time: 5 × 10 = 50 m. This independence of perpendicular components is the same principle as projectile motion.
25. Rain falls vertically at 10 m/s. A man walks horizontally at 10 m/s. At what angle to the vertical should he tilt his umbrella?
45°, tilted forward into his direction of motion. In the man's frame, the rain has a vertical component of 10 m/s downward and a horizontal component of 10 m/s directed backwards relative to him. So tanθ = 10/10 = 1 and θ = 45° from the vertical. Students routinely tilt it the wrong way - the rain appears to come from ahead, so the umbrella leans forward.
26. To cross a flowing river in the minimum possible time, in which direction should a boat head?
Perpendicular to the bank. Crossing time depends only on the velocity component across the river, and heading straight across maximises that component. The boat will be carried downstream, but the question asked for minimum time, not minimum drift. To land directly opposite instead, the boat must head partly upstream, which takes longer - and distinguishing these two questions is exactly what the exam tests.
Practice set 5: variable acceleration
These are the questions where reaching for v = u + at produces a confident wrong answer. The tell is that acceleration or velocity is given as a function.
27. A particle starts from rest with acceleration a = 6t m/s². What is its velocity at t = 2 s?
12 m/s. Acceleration is not constant, so the equations of motion do not apply. Integrate instead: v = ∫a dt = ∫6t dt = 3t² + C. Since the particle starts from rest, v = 0 at t = 0, so C = 0 and v = 3t². At t = 2, v = 3 × 4 = 12 m/s. Using v = u + at with a = 6 × 2 = 12 would give 24 m/s, which is the distractor.
28. For the same particle, what is the displacement in the first 2 seconds?
8 m. Integrate the velocity: x = ∫v dt = ∫3t² dt = t³ + C, and x = 0 at t = 0, so x = t³. At t = 2, x = 8 m. Each integration introduces a constant fixed by the initial conditions, which is why the question always tells you the body starts from rest or from the origin.
29. A particle's velocity is v = 4 − 2t m/s. At what time does it reverse direction, and what is its displacement in the first 4 seconds?
It reverses at t = 2 s, and the displacement in 4 s is zero. Velocity is zero when 4 − 2t = 0, so t = 2 s, after which v is negative. Displacement is ∫(4 − 2t)dt from 0 to 4 = [4t − t²] from 0 to 4 = (16 − 16) − 0 = 0. The distance travelled, however, is not zero: the particle goes 4 m forward in the first 2 s and 4 m back in the next 2 s, so the distance is 8 m. Expect both to be asked.
30. A particle's velocity is v = 3t² − 6t m/s. What is its acceleration at t = 1 s?
0 m/s². Differentiate: a = dv/dt = 6t − 6. At t = 1, a = 6 − 6 = 0. Note what this does not mean - the velocity at t = 1 is 3 − 6 = −3 m/s, so the particle is moving while momentarily not accelerating. Zero acceleration with non-zero velocity is the mirror image of the ball-at-the-top case in question 12, and papers like to set them together.
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How to study this chapter efficiently
- ✓Before writing anything, ask one question: is the acceleration constant? That single check routes you to either the equations of motion or to calculus, and getting it wrong invalidates everything after.
- ✓Draw the graph even when the question does not supply one. A rough v-t sketch makes signs, reversals and areas obvious in a way algebra does not.
- ✓For projectiles, write two columns - horizontal and vertical - and fill them in before touching a formula. It forces the independence of components and prevents mixing components mid-calculation.
- ✓Practise questions where the given data is deliberately excessive. JEE supplies masses and irrelevant distances specifically to see whether you know which quantities matter.
- ✓Time your practice from the start. A student who solves Kinematics correctly in four minutes per question is not yet ready, because the paper does not allow that.
- ✓Work through past-year questions from this chapter specifically. The pattern of what gets asked is far more informative than any predicted weightage table.
Turn this into active practice
Reading solutions creates the feeling of understanding without the retrieval that actually builds it. The reliable way to find out what you know is to be tested on it, get it wrong, and fix it while the reasoning is still fresh.
The JEE Kinematics quiz on QUFF generates fresh questions on this chapter each time, marks them instantly, and explains every answer - so a wrong answer becomes a correction rather than a mystery. Because the questions are generated rather than drawn from a fixed bank, replaying the chapter gives you a new set instead of the ones you have already memorised. Retry the ones you missed a day later, which is when spaced retrieval does the most work.