Year 11 · Science · Physical sciences
Describing motion with equations
Students use vectors, displacement–time and velocity–time graphs and the equations of uniformly accelerated motion to analyse straight-line motion, including free fall.
Awaiting educator review. This material was drafted by our content team with AI assistance and is waiting for review by a qualified educator. Please check it suits your child before using it. Created 10 October 2026.
Learning goals
Learners will
- Distinguish between scalar and vector quantities.
- Use the equations of uniformly accelerated motion to solve problems.
- Interpret the gradient and area of a velocity–time graph.
Success looks like
- I can identify u, v, a, s and t in a problem and choose the right equation.
- I can solve free-fall problems using g = 9.8 m/s².
- I can find acceleration and displacement from a velocity–time graph.
The big idea
Scalars have magnitude only; vectors have magnitude and direction. Distance and speed are scalars. Displacement and velocity are vectors, so we choose a positive direction (for example, up = positive) and use signs consistently.
For uniform acceleration in a straight line, we use five variables: u (initial velocity), v (final velocity), a (acceleration), s (displacement) and t (time).
- v = u + at
- s = ut + ½at²
- v² = u² + 2as
To solve a problem: list the known variables, identify the unknown, choose the equation that links them, substitute with units, then calculate and check that the answer is sensible.
On a velocity–time graph, the gradient gives acceleration and the area under the graph gives displacement.
In free fall near Earth's surface, all objects accelerate downwards at about 9.8 m/s² if air resistance is negligible. A dropped tennis ball is a reasonable approximation over short distances.
Worked examples
Dropping a ball
A ball is dropped from rest from a height of 1.5 m. How long does it take to reach the floor? (Ignore air resistance, g = 9.8 m/s², down positive.)
- Known: u = 0, a = 9.8 m/s², s = 1.5 m. Unknown: t.
- Use s = ut + ½at². With u = 0: 1.5 = ½ × 9.8 × t².
- t² = 1.5 ÷ 4.9 ≈ 0.306.
- t ≈ 0.55 s.
Answer: About 0.55 s.
Braking distance
A car travelling at 20 m/s brakes with a constant deceleration of 5 m/s². How far does it travel before stopping?
- Known: u = 20 m/s, v = 0, a = −5 m/s². Unknown: s.
- Use v² = u² + 2as: 0 = 20² + 2 × (−5) × s.
- 0 = 400 − 10s, so s = 40 m.
Answer: 40 m.
Reading a velocity–time graph
A cyclist's velocity rises steadily from 0 to 6 m/s over 4 s. Find the acceleration and displacement.
- Acceleration = gradient = (6 − 0) ÷ 4 = 1.5 m/s².
- Displacement = area of triangle = ½ × 4 × 6 = 12 m.
Answer: a = 1.5 m/s², s = 12 m.
Practice check
Have a go, then check your answers. Each answer comes with an explanation.
Answer guide for parents
Which is a vector quantity?
Velocity — Velocity has both magnitude and direction.
What does the gradient of a velocity–time graph represent?
Acceleration — Gradient = change in velocity ÷ time = acceleration.
Which equation would you use to find v when you know u, a and s but not t?
v² = u² + 2as — v² = u² + 2as does not include t.
A ball is dropped from rest. Ignoring air resistance, what is its speed after 2.0 s? (g = 9.8 m/s²)
19.6 m/s — v = u + at = 0 + 9.8 × 2.0 = 19.6 m/s.
A car accelerates uniformly from 10 m/s to 30 m/s in 5 s. What is its acceleration?
4 m/s² — a = (30 − 10) ÷ 5 = 4 m/s².
A ball is dropped from rest and falls for 2.0 s. How far does it fall in metres? (g = 9.8 m/s², ignore air resistance)
19.6 — s = ½ × 9.8 × 2.0² = 4.9 × 4 = 19.6 m.
A car at 30 m/s brakes at a constant 6 m/s². How many seconds does it take to stop? (number)
5 — v = u + at: 0 = 30 − 6t, so t = 5 s.
On a velocity–time graph, a train moves at a constant 15 m/s for 20 s. What displacement in metres does the area represent? (number)
300 — Area of rectangle = 15 × 20 = 300 m.
Watch out for
- Students often treat deceleration as a positive number in equations. Choose a positive direction and give acceleration a negative sign when it opposes motion.
- Some think an object at the top of its flight has zero acceleration. Its velocity is momentarily zero, but its acceleration is still 9.8 m/s² downwards.
- Some think heavier objects fall faster in free fall. Without air resistance, all fall with the same acceleration.
Tips for parents
- Encourage a consistent layout: knowns, unknown, equation, substitution, answer with units.
- Record a ball drop in slow motion and compare the measured time with the calculated time; discuss reasons for any difference.
- Safety: drop only soft balls, away from windows and people, and never lean out from heights.
Go further
- Derive s = ut + ½at² from the area under a velocity–time graph.
- Investigate how reaction time adds to stopping distance at 50 km/h and 100 km/h.
- Research how air resistance leads to terminal velocity for skydivers.