SPH3U Kinematics Practice: Motion Graphs and Equations With Answers

In kinematics, choose a direction convention, identify the known quantities, and decide whether acceleration is constant before selecting an equation. On a position–time graph, slope gives velocity. On a velocity–time graph, slope gives acceleration and signed area gives displacement.

This original SPH3U practice set addresses both calculations and interpretation. Assume straight-line motion and constant acceleration only where stated. Use g = 9.8 m/s² for the vertical-motion question and neglect air resistance there.

Why knowing the formulas may not be enough

A public Ontario student discussion specifically identified motion graphs and projectile motion as sticking points. That is a useful reminder to practise interpreting a situation before substituting numbers; it is not evidence that the course has one universal difficulty level. Student discussion

Write what each symbol means. Displacement is change in position, while distance is total path length. A negative velocity describes direction relative to your chosen axis; it does not automatically mean an object is slowing down.

Which graph relationship should I use?

Position–time

• Slope tells you: Velocity
• Area under the graph tells you: Not the standard displacement relationship

Velocity–time

• Slope tells you: Acceleration
• Area under the graph tells you: Displacement, with signs

Acceleration–time

• Slope tells you: Rate of change of acceleration
• Area under the graph tells you: Change in velocity

Check the axis labels every time. A line that rises on a position–time graph means something different from a line that rises on a velocity–time graph.

Eight original practice questions

• 1. A student walks 120 m east in 30 s at constant velocity. Find the average velocity, taking east as positive.
• 2. A student walks 40 m east and then 30 m west in a total of 20 s. Find displacement, average velocity and average speed.
• 3. An object’s velocity changes uniformly from 4.0 m/s to 16.0 m/s in 6.0 s. Find acceleration.
• 4. A cart starts from rest and accelerates at 2.0 m/s² for 5.0 s. Find final velocity and displacement.
• 5. A car moving at 20.0 m/s slows uniformly at −4.0 m/s² until it stops. Find stopping time and displacement during braking.
• 6. On a velocity–time graph, velocity is +3.0 m/s from t = 0 to 4.0 s, then −2.0 m/s from 4.0 to 7.0 s. Find net displacement and distance travelled in this idealized piecewise model.
• 7. A position–time line passes through (2.0 s, 5.0 m) and (6.0 s, 17.0 m). Find its velocity.
• 8. A ball is thrown vertically upward at 19.6 m/s. Take upward as positive. Find its time to maximum height and height gained before that point.

Worked solutions

1. Divide displacement by elapsed time

v = 120/30 = +4.0 m/s, or 4.0 m/s east. Include direction because velocity is a vector quantity.

2. Separate path length from position change

Displacement = +40 − 30 = +10 m. Average velocity = 10/20 = +0.50 m/s. Distance = 40 + 30 = 70 m, so average speed = 70/20 = 3.5 m/s. Average speed is not always the magnitude of average velocity.

3. Calculate change in velocity

a = (v − u)/t = (16.0 − 4.0)/6.0 = +2.0 m/s². Dividing the final velocity alone by time would ignore the initial velocity.

4. Find two quantities from one motion model

v = u + at = 0 + 2.0(5.0) = 10 m/s. Displacement is s = ut + ½at² = 0 + ½(2.0)(5.0²) = 25 m. The velocity increases from zero, so using final velocity × time would overestimate displacement.

5. Respect the braking interval

At the stop, v = 0. From 0 = 20.0 − 4.0t, t = 5.0 s. Average velocity during uniform braking is (20.0 + 0)/2 = 10.0 m/s. Displacement is 10.0 × 5.0 = 50 m. Do not extend the constant negative acceleration beyond the stated stopping event.

6. Add signed areas for displacement

The first rectangle contributes 3.0 × 4.0 = +12 m. The second contributes −2.0 × 3.0 = −6.0 m. Net displacement is +6.0 m. Distance adds the magnitudes of these segments: 18 m. The instantaneous jump is an idealized graph feature, not a claim of physically instantaneous acceleration.

7. Use the slope, not a single coordinate ratio

v = Δposition/Δtime = (17.0 − 5.0)/(6.0 − 2.0) = +3.0 m/s. Dividing 17 by 6 would silently assume the line began at the origin.

8. Set velocity to zero at the top

With a = −9.8 m/s², 0 = 19.6 − 9.8t gives t = 2.0 s. Height gained is 19.6(2.0) − ½(9.8)(2.0²) = 19.6 m, approximately 20 m to two significant figures. The acceleration remains downward at the highest point; it does not become zero when velocity is momentarily zero.

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Curriculum reference: Ontario Grades 11–12 Science. This is original practice, not an official test.