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SUVAT practice

Choose equations and solve constant-acceleration problems with signs, braking and vertical motion.

Practise

Find acceleration

Velocity changes from +11 to +17 m s⁻¹ in 2 s. Calculate the constant acceleration, including its sign. Give your answer to 3 significant figures.

One-dimensional motion with constant acceleration. Positive is the stated direction; displacement and velocity are signed. Air resistance is neglected in vertical motion.

Positive direction →Initiallyu = 11 m/sFinallyv = 17 m/sAcceleration unknownElapsed time 2 s
One-dimensional motion, positive to the right. Initial velocity 11 m/s; final velocity 17 m/s. Acceleration unknown. Arrows show direction only; positions are schematic.
Larger diagram
Positive direction →Initiallyu = 11 m/sFinallyv = 17 m/sAcceleration unknownElapsed time 2 s
Work through the calculation

Check intermediate steps, then use your unrounded working for the final answer. Each numerical check uses 3 significant figures. These steps provide support; the main answer is checked separately.

Step 1

Find the signed change v − u. Give your answer to 3 significant figures.

Show this step

Subtract initial velocity from final velocity.

Answer: 6.00 m s⁻¹ (3 significant figures).

Step 2

Now calculate the final answer to the main question.

Show this step

Use a = (v − u)/t.

Answer: 3.00 m s⁻² (3 significant figures).

Treat the given values as exact unless an uncertainty is stated.

Use decimals or e notation, e.g. 4.2e-3. Enter the unit separately.

Hint

Use a = (v − u)/t.

Worked solution
  1. a = (17 − 11) / 2 = 3 m s⁻².
  2. Answer: 3.00 m s⁻² (3 significant figures).
Another question
Question type: Mixed practice
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Choose a SUVAT equation from the known quantities

A vehicle slows uniformly from 20 m s⁻¹ to rest with acceleration −4.0 m s⁻², taking forwards as positive. Time is not given, so use v² = u² + 2as. Distance = (0² − 20²)/(2 × −4.0) = 50 m.

Use a signed acceleration in the equation. At the highest point of a vertical throw, velocity is zero but acceleration remains downwards. SUVAT assumes constant acceleration; it does not describe every motion. This stopping-distance example excludes the driver’s thinking distance.

Apply constant-acceleration equations to free fall

Choose an equation from the known quantities

For constant acceleration, v = u + at; s = ut + ½at²; s = (u + v)t/2; v² = u² + 2as. Choose a relation that contains the unknown and the known values without an additional unknown.

Keep the signs

Choose a positive direction and use it throughout. Negative acceleration does not always mean slowing down. A negative final velocity means motion in the opposite direction; displacement differs from total distance if the object reverses.

Stopping and vertical motion

For braking, set v = 0 and use a negative acceleration when initial motion is positive. Braking distance excludes reaction distance. For a vertical throw, velocity is zero at the highest point while acceleration remains downward. These tasks neglect air resistance and use g = 9.81 m s⁻².

A-level relevance

AQA 7408: 3.4.1.3 (selected); OCR A H556: 3.1.2 (selected). Tasks cover the selected skills listed above.

AQA mechanics and materials specification · OCR A specification

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