Kinematics

Motion in a straight line & in a plane · Class 11 Physics

1The big idea

Kinematics describes how things move — position, velocity and acceleration — without yet asking why (that is dynamics, the next chapter). One idea sits underneath everything: once you know an object's acceleration together with its starting position and velocity, its entire future motion is fixed.

In the physics you will be tested on, the acceleration is almost always one of just two things: zero (steady motion in a straight line) or g ≈ 9.8 m s⁻² (free fall and projectiles, directed downward). Get fluent with those two cases and most problems open up.

2Getting the words right

Distance vs displacement. Distance is the total path length and is always positive. Displacement is the straight-line change in position with direction, so it can even be zero after a long round trip. This is exactly why velocity carries a sign and speed does not.

Speed vs velocity. Speed tells you how fast; velocity tells you how fast and in which direction. That sign is not decoration — a ball thrown up has positive velocity going up and negative velocity coming down, and it is that automatic sign flip that makes the equations give the correct answer without extra thought.

Acceleration is the rate of change of velocity, not of speed. A car rounding a bend at constant speed is still accelerating, because its direction — and therefore its velocity — is changing. Hold on to that; it is the seed of circular motion later in this very chapter.

3Equations of motion (constant acceleration)

When the acceleration a is constant, three equations tie together the five quantities: u (initial velocity), v (final velocity), a, t (time) and s (displacement).

The three equations of motion v = u + a t · s = u t + ½ a t² · v² = u² + 2 a s

Do not pick one at random — choose by what is missing. Each equation leaves out exactly one quantity:

No displacement s ? → use v = u + a t
No final velocity v ? → use s = u t + ½ a t²
No time t ? → use v² = u² + 2 a s — the one students forget, and the one to reach for whenever a question never mentions time.

For free fall just set a = g. A stone dropped from rest: v = g t and the distance fallen is ½ g t².

Reading the graphs

On a position–time graph the slope is the velocity. On a velocity–time graph the slope is the acceleration and the area under the line is the displacement.

A straight, sloping v–t line means constant acceleration. Its slope gives a; the shaded area gives s.

4Motion in a plane — the key trick

Here is the single most useful idea in the chapter, and it is the exact idea behind the Castle Siege game:

Remember this

Horizontal and vertical motion are completely independent. They share only one thing — the clock.

A cannonball has no horizontal force on it (ignoring air), so its horizontal velocity never changes. Its vertical motion is pure free fall — slowing on the way up, stopping for an instant at the top, speeding up on the way down — precisely as if you had thrown it straight up. The two motions run side by side, unaware of one another, ticking along on the same time t.

5Projectile motion, built up from that idea

Split the launch velocity u at angle θ into its two independent parts:

The launch velocity splits into a constant horizontal part u cos θ and a free-falling vertical part u sin θ. H = maximum height, R = range.

Write the position at any time t:

x = (u cos θ) t · y = (u sin θ) t − ½ g t²

Now the point — you need not memorise the range, height and time formulas. You can build them from those two lines in three short steps:

Time of flight T — the ball is back at launch height when y = 0, giving T = 2 u sin θg.
Maximum height H — at the top the vertical velocity is zero, which happens at t = T/2. Substitute: H = u² sin²θ2g.
Range R — horizontal distance covered in the whole flight: R = (u cos θ) T = u² sin 2θg.

Seeing where the formulas come from is the difference between looking them up and actually knowing them. Reconstruct them once or twice and they stay with you.

Two facts for the exam

Range is largest at θ = 45°, because sin 2θ peaks when 2θ = 90°. And two angles that add to 90° (say 30° and 60°) give the same range. Because the two motions share only time, a ball fired horizontally and one simply dropped from the same height land at the same instant.

◆ Activity — range vs angle

Speed is fixed at u = 25 m s⁻¹. Drag the angle and watch the range. Notice how the peak sits at 45°, and how the complementary angle (shown faint) lands in the same place.

Worked example

A ball is thrown at 20 m s⁻¹ at 30° above the horizontal. Take g = 10 m s⁻². Find the time of flight, the maximum height and the range.

Split the velocity: u_y = 20 sin 30° = 10 m s⁻¹, u_x = 20 cos 30° = 17.3 m s⁻¹.

Time of flight: T = 2u_y/g = 2×10/10 = 2 s
Maximum height: H = u_y²/2g = 100/20 = 5 m
Range: R = u_x × T = 17.3 × 2 = 34.6 m

Notice the range came straight out of “horizontal speed × total time” — no memorised formula needed. (Check: R = u² sin 2θ / g = 400 × sin 60° / 10 = 34.6 m ✓.)

6Uniform circular motion (the bridge ahead)

An object going round a circle at constant speed is still accelerating, because its direction keeps changing. That acceleration points toward the centre and has magnitude

a_c = v²r = ω² r

It is called centripetal (“centre-seeking”) acceleration. In the next chapter you will multiply it by the mass to get the centripetal force that holds the object in its circle.

7Relative velocity

Velocity is always measured relative to something. The velocity of A as seen from B is v_A − v_B (a vector subtraction). Two everyday cases the exam likes:

Boat crossing a river: to travel straight across, the boat must aim upstream so its upstream component cancels the current. The time to cross depends only on the across-river speed, not on the current.
Rain and a moving person: you tilt the umbrella into the relative velocity of the rain, which combines the rain's fall with your own motion.

8Common confusions to clear up

At the highest point of a projectile the ball is not at rest. Its vertical velocity is zero, but it is still moving horizontally at u cos θ. The speed is minimum there, not zero.
Zero velocity does not mean zero acceleration. At the top of a thrown ball v = 0 for an instant, yet a = g throughout — which is exactly why it does not stay up there.
Displacement can be negative or zero; distance cannot. Watch the signs, especially in up-and-down free-fall problems.

9Check yourself

Class 11 Physics · Kinematics · aligned to NCERT Chapters 3–4 · SmartStudy.School