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Velocity and acceleration practice: keep direction in the answer
Velocity tells you how position changes over time, including direction. Acceleration tells you how velocity changes. A negative acceleration does not always mean an object is slowing down.
Updated October 15, 2026 · 11 free practice questions
Choose your positive direction first
For straight-line motion, label one direction positive and keep it consistent. If east is positive, an eastward velocity is positive and a westward velocity is negative. Average acceleration is (final velocity − initial velocity) ÷ elapsed time. Its SI unit is m/s².
A change in speed, direction or both can produce acceleration. Read OpenStax's explanation of acceleration for background. The numbers and exercises below are original.
Worked example: speeding up eastward
A cart changes from +3 m/s to +15 m/s in 4 s. Its average acceleration is (15 − 3) ÷ 4 = +3 m/s². The positive result means its velocity changes toward the chosen positive direction.
If that acceleration is constant throughout the interval, average velocity is (3 + 15) ÷ 2 = 9 m/s, giving displacement 9 × 4 = 36 m east. Without the constant-acceleration assumption, the endpoint average is not generally enough to calculate displacement.
Worked example: negative does not mean slower
A second cart changes from −2 m/s to −8 m/s in 3 s. Average acceleration is [−8 − (−2)] ÷ 3 = −2 m/s². Its speed increases from 2 to 8 m/s while it travels in the negative direction.
Compare a cart changing from +8 to +2 m/s over 3 s: it also has −2 m/s² average acceleration, but it slows down. For one-dimensional motion, acceleration opposite to velocity reduces speed; acceleration in the same direction increases it.
Three motion questions to try
1. A bicycle changes from +12 to +4 m/s in 2 s. Find average acceleration.
Reveal the answer
(4 − 12) ÷ 2 = −4 m/s². The bike is slowing while moving in the positive direction.
2. An object travels 20 m east and then 20 m west, returning to its starting point in 10 s. Find average velocity and average speed.
Reveal the answer
Displacement is zero, so average velocity is 0 m/s. Total distance is 40 m, so average speed is 4 m/s. The return trip does not cancel distance traveled.
3. A cart changes from −6 to +6 m/s in 4 s. Find average acceleration. Must its acceleration have been constant?
Reveal the answer
[6 − (−6)] ÷ 4 = +3 m/s². These endpoints determine average acceleration, but do not establish the acceleration at every instant. Many motion histories can share the same endpoints.
Eight more velocity and acceleration practice problems
Work from signed quantities, then state the direction. Questions 4–9 progress from unit conversion to reversals; questions 10–11 use the graph below. All answers are free to reveal.
4. Convert before calculating
A car changes from 36 km/h east to 72 km/h east in 5 s. Find average acceleration.
Reveal answer 4 and working
Divide by 3.6: the velocities are +10 and +20 m/s. Then (20 − 10)/5 = +2 m/s². Using km/h directly with seconds would mix units.
5. Find stopping time
A bicycle moves at +12 m/s and has constant acceleration −3 m/s² until it stops. How long does stopping take?
Reveal answer 5 and working
Use v = u + at: 0 = 12 − 3t, so t = 4 s. This model ends at the stop; do not assume the bicycle then reverses.
6. Find stopping distance
For the bicycle in question 5, how far does it travel before stopping?
Reveal answer 6 and working
With constant acceleration, average velocity is (12 + 0)/2 = 6 m/s. Displacement is 6 × 4 = 24 m. Velocity stays nonnegative, so distance is also 24 m.
7. Read a position–time graph
A straight position–time segment runs from (2 s, 5 m) to (6 s, −3 m). Find velocity on that segment.
Reveal answer 7 and working
Slope = (−3 − 5)/(6 − 2) = −2 m/s. The straight segment means constant velocity and zero acceleration within the interval, even though position changes.
8. Separate distance and displacement
A walker travels 30 m east in 10 s, then 10 m west in 5 s. Find average velocity and average speed for the whole trip.
Reveal answer 8 and working
Displacement is +20 m and elapsed time is 15 s: average velocity = +20/15 ≈ +1.33 m/s. Distance is 40 m: average speed = 40/15 ≈ 2.67 m/s. Do not average the two segment speeds without time weighting.
9. Constant acceleration with a reversal
A cart starts at −4 m/s and accelerates constantly at +2 m/s² for 4 s. Find final velocity, displacement and distance.
Reveal answer 9 and working
Final velocity is −4 + 2 × 4 = +4 m/s. Displacement is (−4 + 4)/2 × 4 = 0 m. It stops at 2 s; each half covers 4 m, so total distance is 8 m. Zero displacement does not mean no movement.
10. Read the velocity–time graph
For the plotted motion, find acceleration from 0–2 s, displacement from 0–4 s and displacement from 4–8 s.
Reveal answer 10 and working
The first slope is (4 − 0)/2 = +2 m/s². The first triangle contributes ½ × 2 × 4 = 4 m, then the rectangle contributes 2 × 4 = 8 m: total +12 m through 4 s. From 4–8 s the equal positive and negative triangles cancel, giving 0 m displacement.
11. Use absolute areas for distance
For that same velocity–time graph, find distance and average speed over all 8 s.
Reveal answer 11 and working
Areas above zero are 4 + 8 + 4 = 16 m; the triangle below zero contributes another 4 m of distance. Total distance = 20 m and average speed = 20/8 = 2.5 m/s. Signed displacement is 12 m, so average velocity is +1.5 m/s.
Check the graph method in OpenStax’s graphical analysis lesson. Then connect acceleration to force with Newton’s second law practice, or compare energy before and after a speed change in work and energy practice.
A quick motion-check routine
- Sketch an axis and mark the positive direction.
- Write signed velocities before substituting numbers.
- Subtract the initial value, including its sign.
- Include time and the resulting unit.
- Describe the result in words: which way does velocity change?
If you mix distance with displacement, draw the journey. If a question asks for a full motion description, an average alone may hide a stop or reversal. Make a practice card asking what information is missing, rather than assuming every problem permits a constant-acceleration formula.
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