Mini-Lesson

Kinematics

OCR section 3.02 describes motion. You need the suvat equations for constant acceleration, velocity–time graphs, calculus for when acceleration varies, and projectiles in two dimensions.

Throughout, take g = 9.8 m/s² downwards. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.

Kinematics · suvat

The suvat equations

Five quantities: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time). Each equation leaves one of them out — so list what you know, spot what is missing, and pick the equation accordingly.

v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)tTHESE ARE ONLY VALID FOR CONSTANT ACCELERATION

Worked example — u = 4 m/s, a = 2 m/s², t = 5 s

v = u + at = 4 + 2(5) = 14 m/s.

s = ut + ½at² = 4(5) + ½(2)(25) = 20 + 25 = 45 m.

Check with a fourth equation: s = ½(u + v)t = ½(4 + 14)(5) = ½(18)(5) = 45 m. ✓

Set a positive direction and stick to it. If ‘up’ is positive then a = −9.8, and a downward displacement comes out negative. Sign errors are the single biggest source of lost marks in this topic.

Calculate

Your turn — find v

1A particle starts at u = 4 m/s with constant acceleration a = 2 m/s². Find its velocity v after t = 5 s.

m/s

Hint: v = u + at = 4 + 2 × 5.

Kinematics · gravity

Motion under gravity

Ignoring air resistance, an object in free flight has a constant acceleration of g = 9.8 m/s² vertically downwards — so suvat applies.

Worked example — a ball thrown vertically upwards at 21 m/s

Take up as positive: u = +21, a = −9.8.

At the highest point, v = 0. That is the key insight.

v² = u² + 2as ⇒ 0 = 21² + 2(−9.8)s ⇒ 0 = 441 − 19.6s.

s = 441 / 19.6 = 22.5 m.

Time to the top: v = u + at ⇒ 0 = 21 − 9.8t ⇒ t = 2.14 s.

At the top, the velocity is zero but the acceleration is not. Gravity is still pulling at 9.8 m/s² downwards — that is exactly why the ball does not stay up there.

Calculate

Your turn — maximum height

2A ball is thrown vertically upwards at 21 m/s. Taking g = 9.8 m/s² and ignoring air resistance, find the maximum height reached.

m

Hint: at the top v = 0. Use v² = u² + 2as: 0 = 441 − 19.6s, so s = 441 ÷ 19.6.

Kinematics · graphs

Motion graphs

Graphs turn kinematics into geometry.

On a displacement–time graph, the gradient is the velocity.
On a velocity–time graph, the gradient is the acceleration, and the area under the graph is the displacement.

Worked example — reading a velocity-time graph

A car accelerates uniformly from 0 to 20 m/s in 8 s, holds 20 m/s for 10 s, then decelerates uniformly to rest in 4 s.

The graph is a trapezium. Total time = 22 s; the parallel sides are the top (10 s) and the base (22 s), and the height is 20 m/s.

Displacement = area = ½(10 + 22) × 20 = ½(32)(20) = 320 m.

Area below the time axis is negative displacement — the object is going backwards. To find the total distance travelled, add the magnitudes of the areas instead.

Quick check

Reading the graph

?On a velocity–time graph, what does the area under the graph represent?

Kinematics · calculus

Variable acceleration — use calculus

When the acceleration changes, suvat is useless. Use calculus instead.

v = ds/dt · a = dv/dt = d²s/dt² · s = ∫v dt · v = ∫a dtdifferentiate to go DOWN the chain (s → v → a); integrate to go back UP

Worked example — v = 3t² − 8t + 5 (m/s)

Acceleration at t = 2: a = dv/dt = 6t − 8. At t = 2: a = 12 − 8 = 4 m/s².

Displacement from t = 0 to t = 3:

s = ∫ from 0 to 3 of (3t² − 8t + 5) dt = [t³ − 4t² + 5t] from 0 to 3

= (27 − 36 + 15) − 0 = 6 m.

Displacement is not always distance. Here 3t² − 8t + 5 = (3t − 5)(t − 1), so v is negative between t = 1 and t = 5/3 — the particle briefly reverses. The integral gives the net displacement of 6 m. For the total distance you must split the integral at t = 1 and t = 5/3 and add the magnitudes.

Calculate

Your turn — acceleration by differentiating

3A particle has velocity v = 3t² − 8t + 5 (m/s). Find its acceleration at t = 2 s.

m/s²

Hint: a = dv/dt = 6t − 8. Substitute t = 2.

Calculate

Your turn — displacement by integrating

4The same particle has v = 3t² − 8t + 5 (m/s). Find its displacement from t = 0 to t = 3.

m

Hint: s = [t³ − 4t² + 5t] from 0 to 3 = 27 − 36 + 15.

Quick check

Which method?

?A particle moves with velocity v = t³ − 4t. Why can you not use the suvat equations here?

Kinematics · projectiles

Projectile motion

A projectile is in free flight under gravity alone. The trick is to treat the horizontal and vertical motions completely separately — they share only the time.

HORIZONTAL: a = 0, so velocity is CONSTANTVERTICAL: a = −g = −9.8 m/s² — use suvat on this component

Worked example — launched at 20 m/s at 30° above the horizontal, from ground level

Resolve the initial velocity:

Horizontal: u_x = 20 cos 30° = 20 × 0.8660 = 17.32 m/s (constant throughout).

Vertical: u_y = 20 sin 30° = 20 × 0.5 = 10 m/s.

Time of flight (vertical displacement returns to 0): using s = ut + ½at² with s = 0:

0 = 10t − 4.9t² = t(10 − 4.9t) ⇒ t = 0 (the launch) or t = 10/4.9 = 2.041 s.

Range = horizontal velocity × time = 17.32 × 2.041 = 35.3 m (3 s.f.).

At the highest point the vertical velocity is zero — but the horizontal velocity is not. The projectile is still travelling at 17.32 m/s sideways at the top of its arc. Saying ‘the velocity is zero at the top’ is wrong, and costs marks.

Calculate

Your turn — the range of a projectile

5A projectile is launched from ground level at 20 m/s at 30° above the horizontal, with g = 9.8 m/s² and no air resistance. Find its range, to 3 significant figures.

m

Hint: u_y = 10, so the time of flight is t = 2 × 10 ÷ 9.8 = 2.041 s. u_x = 20 cos 30° = 17.32. Range = 17.32 × 2.041.

Quick check

At the top of the arc

?At the highest point of a projectile's path (no air resistance), which statement is true?

Quick check

Modelling a projectile

?A projectile question says to ignore air resistance. What effect does this assumption have on the calculated range?

Sort it

Constant, variable or zero acceleration?

Tap a situation, then tap the type of acceleration it has.

➡️ Constant (use suvat)

📈 Variable (use calculus)

0️⃣ Zero

Match it

Which suvat equation?

Each suvat equation leaves out one variable. Tap the equation, then the variable it is missing.

Equation

Missing variable

Recap

The big ideas to know

suvat: v = u + at; s = ut + ½at²; v² = u² + 2as; s = ½(u + v)t — constant a only

Signs: choose a positive direction and declare it; up positive ⇒ a = −9.8

Graphs: v–t gradient = acceleration; v–t area = displacement

Calculus: v = ds/dt, a = dv/dt; s = ∫v dt — use whenever a varies

Displacement vs distance: if v changes sign, split the integral to get distance

Projectiles: horizontal a = 0 (constant velocity); vertical a = −9.8; time links them

At the top: vertical velocity = 0, horizontal velocity unchanged, acceleration still 9.8 down

That is OCR 3.02 — and forces are what cause all of it. Press Finish to see your score.

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