A Level Physics 9702 · AS Level · Topic 2 · ~9 min read
Kinematics
Displacement, velocity and acceleration, motion graphs, deriving the equations of uniformly accelerated motion, free fall and measuring g, and projectiles with uniform velocity in one direction and uniform acceleration in the other.
Distance & displacementSpeed & velocityAccelerationMotion graphsDeriving suvatFree fallMeasuring gProjectiles
RSby Riyaz Sir
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Paper tag = where it is examinedDATA & FORMULAE given in the examC1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end
In 30 seconds
- Displacement, velocity and acceleration are vectors; distance and speed are scalars. Choose a positive direction and keep the signs.
- Gradient of s–t = velocity. Gradient of v–t = acceleration. Area under v–t = displacement.
- The suvat equations follow from the definitions of velocity and acceleration, but only for constant acceleration.
- Free fall near Earth: a = g = 9.81 m s⁻² downwards, if air resistance is negligible.
- Projectiles: treat horizontal and vertical motion separately. Horizontal velocity is constant; vertical acceleration is g. Time links the two.
By the end you can
- Define and use displacement, speed, velocity and acceleration.
- Interpret s–t and v–t graphs using gradients and areas.
- Derive and use the equations of uniformly accelerated motion.
- Describe an experiment to measure g.
- Solve projectile problems by resolving into horizontal and vertical motion.
1Definitions and graphs
Key definitions
Displacement — distance in a stated direction from a fixed point (vector).
Speed — distance travelled per unit time (scalar).
Velocity — rate of change of displacement, v = Δs/Δt (vector).
Acceleration — rate of change of velocity, a = Δv/Δt (vector).
- "Rate of change" always means "per unit time". Definitions in words must be dimensionally correct: "velocity is displacement per unit time", not "displacement over time taken".
Motion graphs
| Graph | Gradient gives | Area gives |
|---|---|---|
| displacement–time | velocity | (nothing useful) |
| velocity–time | acceleration | displacement |
| acceleration–time | (nothing useful) | change in velocity |
- Instantaneous value from a curve: draw a tangent at the point and find its gradient with a large triangle. In 2024 many candidates calculated an average acceleration instead of the gradient at the given time.
- Area below the time axis on a v–t graph is a displacement in the negative direction.
Worked example
A train accelerates uniformly from rest to 8.0 m s⁻¹ in 4.0 s, travels at constant speed for 6.0 s, then decelerates uniformly to rest in 2.0 s. Find the total distance.
- area = ½ × 4.0 × 8.0 + 6.0 × 8.0 + ½ × 2.0 × 8.0 C1
- = 16 + 48 + 8 = 72 m A1
2The equations of uniformly accelerated motion
Deriving them
- From the definition of acceleration: a = (v − u)/t ⇒ v = u + at.
- Displacement = average velocity × time (area of the trapezium under the v–t graph): s = ½(u + v)t.
- Substitute v = u + at: s = ½(u + u + at)t ⇒ s = ut + ½at².
- Eliminate t using t = (v − u)/a: s = ½(u + v)(v − u)/a ⇒ v² = u² + 2as.
v = u + at ; s = ut + ½at² ; v² = u² + 2as DATA & FORMULAE
s = ½(u + v)t must be recalled; all four need constant accelerationCommon mistakeUsing suvat when acceleration changes, for example with air resistance or a spring. The equations are only valid for constant acceleration.
Using them
- List s, u, v, a, t, fill in the three you know, choose the equation without the one you don't need.
- Sign convention: decide which direction is positive. For something thrown upwards, if up is positive then a = −9.81 m s⁻².
Worked example
A car accelerates uniformly from 12 m s⁻¹ to 30 m s⁻¹ in 6.0 s. Find its acceleration and the distance travelled.
- a = (30 − 12)/6.0 = 3.0 m s⁻² A1
- s = ½(12 + 30) × 6.0 = 126 m C1 A1
Worked example
A ball is thrown vertically upwards at 15 m s⁻¹ from a point 1.2 m above the ground. Find its maximum height above the ground and the time until it hits the ground.
- at the top v = 0: 0 = 15² − 2 × 9.81 × s ⇒ s = 11.5 m, so height = 11.5 + 1.2 = 12.7 m C1 A1
- ground is s = −1.2 m (up positive): −1.2 = 15t − 4.905t² C1
- 4.905t² − 15t − 1.2 = 0 ⇒ t = 3.1 s (positive root) A1
Common mistakeForgetting the release height, or giving g the wrong sign. Both were common in 2024 Paper 2.
3Free fall and measuring g
The acceleration of free fall
- Without air resistance, all objects near Earth's surface fall with the same acceleration g ≈ 9.81 m s⁻², whatever their mass.
- Method: a steel ball is held by an electromagnet and released from rest; a timer starts when the current is switched off and stops when the ball passes a light gate (or opens a trapdoor). Measure the height h with a metre rule.
- Analysis: with u = 0, h = ½gt². Repeat for several heights. Plot h against t²: a straight line through the origin with gradient g/2.
- Improving accuracy: repeat each timing and average; use a dense, small ball to reduce air resistance; use light gates instead of a hand-held stopwatch to remove reaction time; measure h to the bottom of the ball.
4Projectiles
Horizontal and vertical are independent
| Horizontal | Vertical | |
|---|---|---|
| Initial velocity | u cos θ | u sin θ |
| Acceleration | 0 | −g (up positive) |
| Equation | x = (u cos θ)t | y = (u sin θ)t − ½gt² |
- Time is the link. Find the time from the vertical motion, then use it horizontally.
- Never put horizontal and vertical values into one suvat equation, and never use suvat with a horizontal acceleration that isn't there. These were 2024 Paper 2 errors.
Worked example
A ball is kicked horizontally at 12 m s⁻¹ from the top of a cliff 20 m high. Find the time of flight, the horizontal distance and the speed on landing.
- vertical: 20 = ½ × 9.81 × t² ⇒ t = 2.0 s C1 A1
- horizontal: x = 12 × 2.02 = 24 m A1
- vertical speed = 9.81 × 2.02 = 19.8 m s⁻¹; v = √(12² + 19.8²) = 23 m s⁻¹ C1 A1 (at 59° below the horizontal)
Worked example
A ball is projected at 25 m s⁻¹ at 40° above the horizontal from level ground. Find the maximum height and the range.
- u_x = 25 cos 40° = 19.2 m s⁻¹; u_y = 25 sin 40° = 16.1 m s⁻¹ C1
- max height: 0 = 16.1² − 2 × 9.81 × h ⇒ h = 13.2 m A1
- time of flight = 2 × 16.1/9.81 = 3.28 s; range = 19.2 × 3.28 = 63 m C1 A1
- With air resistance (qualitatively): the range and maximum height are smaller, the path is no longer symmetrical, and the descent is steeper than the ascent.
Linking questionForces cause the accelerations in Dynamics; air resistance and terminal velocity are in the same note; and the energy approach to falling objects is in Work, energy and power.
Test yourself — cover the answers first
Quick check
- Define acceleration.
- What does the area under a velocity–time graph represent?
- State the condition for using the suvat equations.
- A stone is dropped from a bridge and hits the water 2.4 s later. How high is the bridge?
- What is the horizontal acceleration of a projectile (no air resistance)?
Exam-style
- Derive v² = u² + 2as from v = u + at and s = ½(u + v)t.
- A student drops a ball from heights h and records the fall time t. A graph of h against t² is a straight line through the origin with gradient 4.85 m s⁻². Find g and suggest why it is less than 9.81 m s⁻².
- A ball is thrown at 20 m s⁻¹ at 30° above the horizontal from a cliff 15 m high. Find the time for it to reach the sea below.
Show answers
- rate of change of velocity
- displacement
- constant (uniform) acceleration
- ½ × 9.81 × 2.4² = 28 m
- zero
- t = (v − u)/a; s = ½(u + v)(v − u)/a = (v² − u²)/(2a) ⇒ v² = u² + 2as
- g = 2 × 4.85 = 9.7 m s⁻²; air resistance, or a delay in the release mechanism (the electromagnet takes time to release the ball)
- −15 = 10t − 4.905t² ⇒ 4.905t² − 10t − 15 = 0 ⇒ t = 3.0 s
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