Kinematics for JEE Main

Kinematics describes motion without asking what causes it. It is the first real chapter of JEE physics, and every later mechanics chapter leans on it. Projectiles, circular motion, SHM and even charged particles in fields all end with a kinematics step.

JEE Main questions here tend to be of four kinds: equations of motion with a twist (such as a velocity reversal), graphs, position or velocity given as a function of time, and relative motion. This guide covers each and shows where candidates usually lose the mark.

The basic quantities

• Displacement is the change in position, a vector. Distance is the total path length, a scalar. They differ whenever the body turns back.
• Average velocity = total displacement ÷ total time. Average speed = total distance ÷ total time.
• Instantaneous velocity v = dx/dt; acceleration a = dv/dt.
• Using the chain rule, a = v (dv/dx). This form is useful when acceleration is given in terms of position.

Average speed is never less than the magnitude of average velocity. They are equal only if the body moves in one direction along a straight line.

Equations for constant acceleration

For motion along a straight line with constant acceleration (constant in magnitude and direction):

Where v² = u² + 2as comes from: write a = v (dv/dx), separate as v dv = a dx and integrate from u to v and 0 to s. With a constant, ½(v² − u²) = as.

Three habits prevent most errors:

1. Choose a positive direction and give every vector quantity a sign. For a ball thrown up with up as positive, u is positive and a = −g.
2. Check that acceleration is constant for the whole stretch. If it changes, split the motion into parts or use calculus.
3. Remember s is displacement. If the body turns back, find the turning point separately to get distance.

Motion graphs

Shape tells a story. A straight x–t line means constant velocity; a curve bending upward means increasing velocity. On a v–t graph, total distance is the sum of the magnitudes of the areas above and below the time axis.

Variable acceleration

When a is not constant, the equations of motion do not apply. Use calculus instead:

• Given x(t): differentiate to get v, then a.
• Given a(t): integrate, v = u + ∫a dt, then x = ∫v dt.
• Given a(v): write a = dv/dt and separate variables.
• Given a(x) or v(x): use a = v (dv/dx).

Relative motion

The velocity of A relative to B is v_AB = v_A − v_B (vector subtraction). Relative motion makes problems about two moving bodies into problems about one.

River crossing. A swimmer's speed in still water is v; the river of width d flows at u.

• Shortest time: swim perpendicular to the bank. Time = d/v, and the drift downstream = ud/v.
• Shortest path (straight across): possible only if v > u. Head upstream at angle θ to the perpendicular, with sin θ = u/v. Time = d / √(v² − u²).

Rain and umbrella. Hold the umbrella along the velocity of rain relative to you, v_rain − v_you.

Worked problems

Problem 1: displacement versus distance. A particle starts with u = 20 m/s and a = −4 m/s². Find its displacement and the distance travelled in the first 8 s.

It stops when 0 = 20 − 4t, at t = 5 s. By then, x = 20(5) − 2(25) = 50 m.
At t = 8 s: x = 20(8) − 2(64) = 160 − 128 = 32 m.
Displacement = 32 m. Distance = 50 + (50 − 32) = 68 m.

Problem 2 (numerical answer): position as a function of time. x = t³ − 6t² + 9t (x in m, t in s). Find the distance covered in the first 4 s.

v = 3t² − 12t + 9 = 3(t − 1)(t − 3), so the particle turns at t = 1 s and t = 3 s.
x(0) = 0, x(1) = 1 − 6 + 9 = 4, x(3) = 27 − 54 + 27 = 0, x(4) = 64 − 96 + 36 = 4.
Distance = 4 + 4 + 4 = 12 m. (Displacement is only 4 m.)

Problem 3: stone from a tower. A stone is thrown upward at 20 m/s from the top of a 25 m tower. When does it hit the ground, and how fast? (g = 10 m/s²)

Take up as positive; the ground is at s = −25 m.
−25 = 20t − 5t², so t² − 4t − 5 = 0, giving (t − 5)(t + 1) = 0 and t = 5 s.
v² = 20² + 2(−10)(−25) = 400 + 500 = 900, so the speed is 30 m/s.

Problem 4: river crossing. A river is 100 m wide and flows at 3 m/s. A swimmer can swim at 5 m/s in still water.

Shortest time: 100/5 = 20 s, with drift 3 × 20 = 60 m downstream.
Shortest path: sin θ = 3/5, so θ = 37° upstream of the perpendicular. Speed across = √(25 − 9) = 4 m/s, so the time is 100/4 = 25 s.

Practice set

1. A ball is thrown up at 30 m/s (g = 10 m/s²). Find the time to return and the maximum height.
2. A v–t graph is a triangle with a base of 10 s and a height of 20 m/s. What is the displacement?
3. A car moving at 72 km/h stops in 40 m with uniform deceleration. Find the deceleration.
4. A body starts from rest with uniform acceleration. What is the ratio of the distances covered in the 1st, 2nd and 3rd seconds?
5. The stopping distance of a car at 36 km/h is 10 m. With the same braking, what is it at 72 km/h?
6. Two trains, each 150 m long, move in opposite directions at 20 m/s and 30 m/s. How long do they take to cross each other?
7. A particle starts from rest at x = 0 with a = 2t (SI units). Find its velocity and position at t = 3 s.
8. Rain falls vertically at 6 m/s. A cyclist rides at 8 m/s. At what angle should the cyclist tilt the umbrella?

Answers

1. Time = 2u/g = 6 s; height = u²/2g = 900/20 = 45 m.
2. Area = ½ × 10 × 20 = 100 m.
3. 72 km/h = 20 m/s; a = v²/2s = 400/80 = 5 m/s².
4. 1 : 3 : 5, from sₙ = a(2n − 1)/2.
5. Stopping distance is proportional to u². Doubling speed gives 4 × 10 = 40 m.
6. Relative speed = 50 m/s; distance = 300 m; time = 6 s.
7. v = t² = 9 m/s; x = t³/3 = 9 m.
8. Rain relative to the cyclist has components 6 m/s down and 8 m/s backwards: magnitude 10 m/s. Tilt the umbrella forward, at tan⁻¹(8/6) ≈ 53° to the vertical.

What to do next

• Solve 30 problems on the equations of motion, flagging every velocity reversal.
• Draw the x–t, v–t and a–t graphs for a ball thrown up and caught again.
• Do 15 problems on relative motion, including river and rain cases.
• Move on to projectile motion, which is two kinematics problems at once, and then Newton's laws.