P9.2 Velocity and Acceleration

A free GCSE Combined Science: Trilogy lesson for AQA (8464) from OBA School: speed in a given direction, changing velocity, acceleration and deceleration, and the equation v² − u² = 2as. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

• Explain the difference between speed and velocity, and between distance and displacement
• (Higher tier) explain why an object moving in a circle at constant speed is accelerating
• Calculate acceleration using a = Δv ÷ t, and rearrange it to find a velocity or a time
• Describe deceleration as negative acceleration, and estimate everyday accelerations
• Use v² − u² = 2as for uniform acceleration, including free fall with g = 9.8 m/s²

Key words

• Velocity: Speed in a given direction. Velocity is a vector quantity.
• Displacement: Distance travelled in a given direction (a vector), measured in metres.
• Vector: A quantity with both magnitude (size) and direction, e.g. velocity, displacement, acceleration, force.
• Scalar: A quantity with magnitude only, e.g. speed, distance, time, mass, energy.
• Acceleration: The rate of change of velocity: change in velocity ÷ time taken. Unit m/s².
• Deceleration: Slowing down: a negative acceleration.
• Uniform acceleration: Acceleration that stays the same, so the velocity changes by the same amount every second.

Speed and velocity

Velocity is speed in a given direction. Speed only tells you how fast; velocity tells you how fast and which way. In the same way, displacement is distance in a given direction.

Two objects can have the same speed but different velocities. It is common to choose one direction as positive, so a velocity in the opposite direction is given a minus sign.

An object has a constant velocity only if it keeps the same speed and keeps moving in a straight line in the same direction.

Worked example: distance or displacement?

A delivery drone flies 600 m due north to drop a parcel, then flies straight back to its base.
distance travelled = 600 + 600 = 1200 m (a scalar: no direction)
displacement after the round trip = 0 m, because it ends where it started

Answer: distance 1200 m; displacement 0 m (halfway: 600 m north)

Key fact: Vectors: velocity, displacement, acceleration, force, weight, momentum. Scalars: speed, distance, time, mass, energy, temperature.

Moving in a circle (Higher tier)

When an object moves round a circle, its direction of motion changes all the time. So even if its speed is constant, its velocity is changing. A changing velocity means the object is accelerating.

Examples include a satellite in a circular orbit, a car driving round a roundabout at a steady speed, and a stone whirled on a string.

Exam tip: Constant speed does not mean constant velocity. In a circle, the speed can stay the same while the velocity changes.

Acceleration

Acceleration tells you how quickly the velocity changes. It is the change in velocity per second:

acceleration, a (m/s²) = change in velocity, Δv (m/s) ÷ time taken, t (s)

For a velocity that changes from an initial value u to a final value v: Δv = v − u, so a = (v − u) ÷ t

Rearranged: v = u + a t and t = (v − u) ÷ a

The unit m/s² means “metres per second, every second”.

Worked example: an electric scooter

A scooter speeds up from 1.0 m/s to 7.0 m/s in 3.0 s. Calculate its acceleration.
a = (v − u) ÷ t = (7.0 − 1.0) ÷ 3.0 = 6.0 ÷ 3.0

Answer: 2.0 m/s²

Worked example: finding a final velocity

A train moving at 5.0 m/s accelerates at 0.80 m/s² for 20 s. Calculate its final velocity.
v = u + a t = 5.0 + (0.80 × 20) = 5.0 + 16

Answer: 21 m/s

Deceleration and everyday accelerations

When an object slows down, v is smaller than u, so the acceleration comes out negative. A negative acceleration is called a deceleration.

Worked example: a cyclist braking

A cyclist moving at 10 m/s brakes and stops in 4.0 s.
a = (v − u) ÷ t = (0 − 10) ÷ 4.0 = −2.5 m/s²

Answer: deceleration = 2.5 m/s² (no minus sign needed when you call it a deceleration)

You should be able to estimate the size of everyday accelerations:

• bus pulling away from a stop; Estimated acceleration about 1 m/s²
• family car speeding up from 0 to 27 m/s in about 10 s; Estimated acceleration about 3 m/s²
• sprinter leaving the blocks; Estimated acceleration about 5 m/s²
• car making an emergency stop on a dry road; Estimated acceleration about 7 m/s² (deceleration)
• any object falling freely near the Earth’s surface; Estimated acceleration 9.8 m/s²

The equation v² − u² = 2as

For uniform (constant) acceleration you can link the velocities to the distance travelled, without needing the time:

(final velocity)² − (initial velocity)² = 2 × acceleration × distance

v² − u² = 2 a s

v and u in m/s, a in m/s², s in m.

To find v: v = √(u² + 2as). To find a: a = (v² − u²) ÷ 2s. To find s: s = (v² − u²) ÷ 2a.

Worked example: a stone dropped into a well

A stone is dropped from rest and falls 25 m. Ignore air resistance. Calculate its speed when it hits the water.
u = 0, a = g = 9.8 m/s², s = 25 m
v² = u² + 2as = 0 + 2 × 9.8 × 25 = 490
v = √490

Answer: 22.1 m/s (≈ 22 m/s)

Worked example: braking to a stop

A car moving at 20 m/s brakes steadily and stops in a distance of 25 m. Calculate its deceleration.
a = (v² − u²) ÷ 2s = (0 − 20²) ÷ (2 × 25) = −400 ÷ 50 = −8.0 m/s²

Answer: deceleration = 8.0 m/s²

Exam tip: Square the velocities before you subtract. When you finish with v², remember to take the square root to get v.

Common mistakes

• Saying an object moving in a circle at constant speed is not accelerating. Its direction, and so its velocity, is changing.
• Writing v − u the wrong way round. Final minus initial: a slowing object has a negative acceleration.
• Giving the unit of acceleration as m/s. It is m/s².
• In v² − u² = 2as, subtracting the velocities first and then squaring, or forgetting the final square root.
• Using minutes or km/h in the equations without converting to seconds and m/s.

Revision questions

Try these, then check your answers below each one.

Question 1. Which of these quantities is a vector? (Choose from: mass, displacement, energy, distance.)

Answer: displacement

Working: Displacement is distance in a given direction, so it is a vector. Mass, energy and distance have size only (scalars).

Question 2. A car speeds up from 3.0 m/s to 21 m/s in 4.5 s. Calculate its acceleration.

Answer: 4 m/s²

Working: a = (v − u) ÷ t = (21 − 3.0) ÷ 4.5 = 18 ÷ 4.5 = 4.0 m/s².

Question 3. A runner moving at 9.0 m/s stops in 1.5 s. Calculate the size of her deceleration.

Answer: 6 m/s²

Working: a = (0 − 9.0) ÷ 1.5 = −6.0 m/s², so the deceleration is 6.0 m/s².

Related lessons

• Previous lesson: P9.1 Speed and Distance–Time Graphs
• Next lesson: P9.3 More about Velocity–Time Graphs
• In this chapter: P9.4 Analysing Motion Graphs

Continue in the full course

This lesson is section P9.2 (AGCSC3802) of our free GCSE Combined Science course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Combined Science course.