Acceleration: change in velocity, ∆v, in metres per second, m/s

The idea

Acceleration measures how quickly velocity changes, so the change in velocity ∆v is the difference between the final velocity and the initial velocity. Because velocity is a vector, ∆v can be positive, negative or zero. For example, a car speeds up from 4 m/s to 16 m/s, so ∆v = 16 m/s − 4 m/s = 12 m/s. If it slows from 20 m/s to 8 m/s, ∆v = 8 m/s − 20 m/s = −12 m/s, a negative change showing deceleration. Always subtract the initial velocity from the final velocity, keep the direction consistent, and give the unit as metres per second, m/s. ∆v is then divided by the time taken to find acceleration in m/s².

What earns the marks

• States that ∆v is the change in velocity, found by subtracting the initial velocity from the final velocity.
• Uses the correct order: ∆v = final velocity − initial velocity, not the reverse.
• Recognises that a negative ∆v indicates a decrease in velocity, often called deceleration.
• Includes the unit metres per second, m/s, with the numerical value.
• Applies ∆v correctly within a = ∆v ÷ t when calculating acceleration.

Common mistakes and how to put them right

• Subtracting final velocity from initial velocity.

  Put it right by doing final minus initial.

• Ignoring direction and treating velocity as a scalar.

  Put it right by keeping a consistent positive direction and using negative values for opposite motion.

• Writing the unit as m/s² for ∆v.

  Put it right: Correct because ∆v is a velocity change measured in m/s, while m/s² belongs to acceleration.

Exam tips

• Underline the initial and final velocities in the question before subtracting.
• Show the substitution line, for example ∆v = 16 m/s − 4 m/s = 12 m/s, so the examiner can credit the method.
• Check the sign of your answer against whether the object is speeding up or slowing down.