Acceleration

1. Definition

A. Acceleration

Acceleration, a, is the rate of change of velocity with time.

a = (Δ v)/(Δ t) = (v-u)/t

• a = acceleration (m s⁻²)
• u = initial velocity (m s⁻¹)
• v = final velocity (m s⁻¹)
• t (or Δ t) = time taken (s)

B. Uniform acceleration

Uniform acceleration means acceleration is constant (velocity changes by equal amounts in equal time intervals).

2. Key Ideas

• O Level kinematics mainly uses straight-line (one-dimensional) motion.
• Average acceleration over a time interval:
  • a = (Δ v)/(Δ t) = (v-u)/t
• SI unit: m s⁻² (also written as m/s²).
• a = 0 means constant velocity (the object can still be moving).
• Uniform acceleration: constant a → equal change in v in equal times.
• Non-uniform acceleration: a changes with time, but you can still calculate average acceleration using Δ v/Δ t.

Scroll diagram horizontally to read all labels.

3. Detailed Explanations

A. How to calculate average acceleration (exam method)

1. Write down u, v and t (include direction/sign).
2. Use a = (v-u)/t.
3. Give the final answer with unit m s⁻² (and direction if asked).
4. Do a sign check: if the object is slowing down, acceleration should be opposite to the velocity direction.

Mini-example (take right as positive): u = +4, v = +10, t = 3.0
a = (10-4)/3.0 = 2.0 m s⁻²

B. Positive/negative acceleration and “deceleration” (1D)

In 1D motion, the sign of velocity/acceleration depends on the positive direction you choose (e.g. “right is +”).

Speed decreases (“deceleration”) when velocity and acceleration have opposite signs.

C. Uniform vs non-uniform acceleration

Uniform acceleration: equal change in velocity in equal time intervals.

Example (uniform acceleration):

Non-uniform acceleration: the change in velocity per second is not constant.

Example (non-uniform acceleration):

Here, the velocity changes by + 10, + 20, -10, + 30, + 20 m s⁻¹ each second (not constant).

D. Link to velocity–time graphs

On a velocity–time graph, acceleration is the gradient:

a = (Δ v)/(Δ t)

See: Reading Kinematics Graphs (Displacement–Time & Velocity–Time).

4. Common Mistakes

A. Using speed instead of velocity

• Acceleration depends on velocity change, so direction (or sign) matters.

B. Wrong sign for Δ v

• Using u-v instead of v-u.
• Forgetting that “slowing down” can be either positive or negative acceleration depending on the direction chosen.

C. Unit and time conversion errors

• Mixing km h⁻¹ with m s⁻¹ without converting.
• Using minutes/hours without converting to seconds when using SI units.

D. “Zero acceleration means zero velocity”

• a = 0 means velocity is constant, not necessarily zero.

5. Exam Tips

A. Definition marks

• Write: “rate of change of velocity per unit time”.
• State the equation: a = Δ v/Δ t.
• Give the SI unit: m s⁻².

B. Calculation method (full marks)

• Choose a sign convention (or state directions clearly).
• Show substitution with units.
• Final answer with unit, and direction if asked.

C. Description questions

• If an object is slowing down: say “acceleration is opposite to velocity”.
• If an object is speeding up: say “acceleration is in the same direction as velocity”.

6. Worked Examples

Modelled example 1

From rest to 20 m s⁻¹ in 10 s

Core

Problem

A bus starts from rest and reaches 20 m s⁻¹ in 10 s, moving right. Find its average acceleration.

Study the worked solution

1. Identify the velocities

   Method

   Use u = 0 and v = +20 m s⁻¹ with right positive.

   Reason

   “Starts from rest” fixes the initial velocity, and the stated direction fixes the sign.

   Working

   a = (v-u)/t

2. Calculate the average acceleration

   Reason

   Average acceleration is velocity change per unit time.

   Working

   a = (20-0)/10 = 2.0 m s⁻² to the right

Guided practice 2

Westwards, then stops (sign convention)

About 5 min

Problem

A car travels west at 30 m s⁻¹ and stops in 5.0 s. Take east as positive and find its average acceleration.

Assign signs before substituting

Signed initial velocity

m s^-1Unit: m s^-1

Average acceleration

m s^-2Unit: m s^-2

Check my attempt

Hints

Hint 1: assign velocity signs

With east positive, u = -30 m s⁻¹ and v = 0.

Hint 2: keep the double negative

Substitute a = [0-(-30)]/5.0.

View solution step by step

1. Apply the sign convention

   Method

   Write westward velocity as negative.

   Reason

   Velocity includes direction, so the chosen positive direction controls its sign.

   Working

   u = -30 m s⁻¹, v = 0

2. Calculate and interpret

   Reason

   The positive result points east, opposite to the westward motion.

   Working

   a = (0-(-30))/5.0 = +6.0 m s⁻² eastwards

Common misconception 3

Can an object move when a = 0?

Find and correct the mistake

Learner response

A student says: If acceleration is zero, the object must be stationary. Locate the error and give a counterexample.

Distinguish velocity from velocity change

What does a = 0 mean?

Velocity is constantVelocity must be zeroDirection must change

Check my attempt

View solution step by step

1. Locate the confused quantities

   Method

   Separate velocity from acceleration.

   Reason

   Acceleration measures how velocity changes, not velocity itself.

   Working

   a = 0 ⇒ Δ v = 0 over the interval

2. Give a counterexample

   Method

   Use steady straight-line motion.

   Reason

   A non-zero velocity can remain constant.

   Working

   An object moving steadily at 5.0 m s⁻¹ has a = 0 while still moving.

Examiner practice 4

Acceleration from a velocity–time line

3 marks

Examination question

A velocity–time graph is a straight line from (0 s,4.0 m s⁻¹) to (5.0 s,14.0 m s⁻¹). Find the acceleration. [3 marks]

Show the gradient calculation

Write your three-mark solution

View solution step by step

1. Identify acceleration as gradient

   1 mark

   Method

   Use rise in velocity divided by run in time.

   Reason

   The gradient of a velocity–time graph is acceleration.

   Working

   a = (Δ v)/(Δ t)

2. Substitute coordinate differences

   1 mark

   Method

   Subtract corresponding coordinates in the same order.

   Reason

   Gradient uses changes, not a single coordinate value.

   Working

   a = (14.0-4.0)/(5.0-0)

3. State the acceleration

   1 mark

   Reason

   The gradient unit is (m s⁻¹)/s.

   Working

   a = 2.0 m s⁻²

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark the gradient principle, substitution and final value separately.

Mark points

1. 1 mark: Identifies acceleration as the velocity–time gradient.
2. 1 mark: Uses (14.0 - 4.0)/(5.0 - 0).
3. 1 mark: Obtains 2.0 m s⁻².My response includes the correct unit.

Calculate my mark

Challenge 5

Moving right but slowing down

Minimal support

Qualitative transfer

A car moves right at 25 m s⁻¹ while braking and slowing down. State the direction of its acceleration and explain without calculating a magnitude.

Infer direction from the speed change

Acceleration direction

LeftRightNo acceleration

Check my attempt

Hints

Hint 1: compare velocity and acceleration

Speed decreases when acceleration and velocity point in opposite directions.

Hint 2: reverse the motion direction

The velocity is rightward, so the acceleration is leftward.

View solution step by step

1. Use the slowing-down condition

   Method

   Place acceleration opposite to velocity.

   Reason

   Opposite signs reduce the magnitude of velocity.

   Working

   The car’s velocity is rightward, so its acceleration is leftward.

Next step

Try Mind Stretcher 1 independently.

7. Mind Stretchers

Mind stretcher 1: Direction reversal (average acceleration)Extension

Take right as positive. A trolley has velocity + 6.0 m s⁻¹ at one instant. After 4.0 s, its velocity is -2.0 m s⁻¹. Find the average acceleration.

Show Answer

a = (v-u)/t; = (-2.0-6.0)/4.0; = -2.0 m s⁻²

The negative sign means acceleration is to the left.

Mind stretcher 2: “Deceleration” does not mean negativeExtension

Take right as positive. An object moves left and slows down: u = -12 m s⁻¹, v = -4.0 m s⁻¹ in 4.0 s. Find a and state whether it is accelerating or decelerating.

Show Answer

a = (v-u)/t; = (-4.0-(-12))/4.0; = +2.0 m s⁻²

Acceleration is to the right (positive). Because velocity is to the left (negative), acceleration is opposite to velocity, so the object is decelerating (speed decreasing).

Mind stretcher 3: Is the acceleration uniform?Extension

An object’s velocities are:

Time / s	Velocity / m s⁻¹
0	0
1	3
2	7
3	12

Is the acceleration uniform? Explain.

Show Answer

No. The change in velocity each second is not constant:

• from 0 to 1 s: Δ v = 3
• from 1 to 2 s: Δ v = 4
• from 2 to 3 s: Δ v = 5

So acceleration increases with time (non-uniform).