VCE Physics 2026: Motion Graphs, Signed Area and Impulse

A velocity–time graph can tell you where an object finishes, how far it travels and when it turns around. Those answers come from different features of the same graph. Keeping gradient, signed area and direction separate makes the working much easier to follow.

This Loav Tuitions guide supports 2026 VCE Physics Units 3 & 4 revision, connecting foundational graph skills to motion and impulse in Unit 3 Area of Study 1 of the current VCAA study design. Every example and practice question below is original.

Read the axes before choosing a method

First choose a positive direction. In one-dimensional motion, a negative velocity means movement in the opposite direction; it does not automatically mean slowing down.

• Position–time: gradient gives velocity. A horizontal section means the object is stationary.
• Velocity–time: gradient gives acceleration; signed area gives displacement.
• Acceleration–time: signed area gives change in velocity. Add the initial velocity to find the final velocity.
• Net force–time: signed area gives net impulse, equal to change in momentum.

For straight-line sections, use gradient = change in vertical value ÷ change in time. Rectangles, triangles and trapeziums handle the areas without calculus.

On a velocity–time graph, count areas above zero as positive and areas below zero as negative. Add their magnitudes for distance; add their signed values for displacement. A direction change requires velocity to change sign. Merely reaching zero is not enough.

Worked example 1: one graph, two different totals

A trolley moves along a straight track. Take right as positive. Its velocity is +4 m/s from 0 to 2 s, falls uniformly to −4 m/s at 6 s, then stays at −4 m/s until 7 s. Find its acceleration from 2 to 6 s, turning time, displacement and distance.

1. Calculate the gradient. Between 2 and 6 s, acceleration = [−4 − (+4)] ÷ (6 − 2) = −2 m/s². The velocity decreases throughout that interval.

2. Find the turning point. Starting from +4 m/s at 2 s, it takes 4 ÷ 2 = 2 s to reach zero. The trolley therefore changes direction at t = 4 s. It slows before this time and speeds up afterwards, because its negative velocity then grows in magnitude.

3. Split the areas at zero velocity. From 0 to 2 s, the rectangle gives 4 × 2 = +8 m. From 2 to 4 s, the triangle gives ½ × 2 × 4 = +4 m. The 4–6 s triangle gives −4 m, and the 6–7 s rectangle gives −4 m.

Displacement = 8 + 4 − 4 − 4 = +4 m, or 4 m right of the starting point. Distance = 8 + 4 + 4 + 4 = 20 m. The trolley travels 12 m right, then 8 m left.

Over the whole 7 s, average velocity is 4 ÷ 7 ≈ +0.57 m/s; average speed is 20 ÷ 7 ≈ 2.9 m/s. These averages use different numerators.

Worked example 2: impulse during a collision

A 0.20 kg trolley initially moves right at 4.0 m/s and collides with a second trolley on a level, low-friction track. Neglect external horizontal impulse on the two-trolley system. The horizontal contact force on the first trolley forms a triangular force–time pulse: zero at 0 s, −60 N at 0.020 s and zero at 0.040 s. Find its final velocity.

Take right as positive. For the first trolley, contact force is the net horizontal force. The signed triangular area is:

Impulse = ½ × 0.040 × (−60) = −1.2 N s.

Using impulse = m(v − u), we obtain −1.2 = 0.20[v − 4.0], so v = −2.0 m/s. It rebounds left at 2.0 m/s.

Check with momentum: initially p = 0.20 × 4.0 = +0.80 kg m/s. Adding −1.2 N s gives −0.40 kg m/s, which agrees with the final velocity.

The average force is −1.2 ÷ 0.040 = −30 N. Using the peak force of −60 N for the entire contact time would double the impulse. Momentum is conserved for the isolated two-trolley system; each individual trolley can still change momentum.

Worked example 3: a positive force while moving left

A 0.50 kg glider initially has velocity −3.0 m/s. A constant net force of +1.0 N acts for 2.0 s. Find when it turns and how far it travels during this interval.

Acceleration = F_net/m = 1.0/0.50 = +2.0 m/s². With v = u + at, zero velocity occurs when 0 = −3.0 + 2.0t, giving t = 1.5 s. At 2.0 s, its velocity is +1.0 m/s.

The velocity–time line therefore crosses zero inside the interval. Its first triangular area is −½ × 1.5 × 3.0 = −2.25 m. Its second is +½ × 0.50 × 1.0 = +0.25 m.

Displacement is −2.00 m, while distance is 2.50 m: to two significant figures, −2.0 m and 2.5 m. The positive force initially slows the leftward motion, then speeds up the rightward motion. Force direction alone does not tell you the direction of travel.

Four practice questions

Try these before reading the solutions. Show your sign convention, substitution and units.

1. A position–time graph joins (0 s, 2 m), (3 s, 11 m) and (5 s, 7 m) with straight lines. Find each section’s velocity, total displacement and distance.
2. Velocity decreases uniformly from +6 m/s to −2 m/s over 4 s. Find acceleration, turning time, displacement and distance.
3. A 0.40 kg trolley initially moves at −2.0 m/s. Its net force–time graph is a positive triangular pulse of duration 0.10 s and peak force 12 N. Does it reverse direction?
4. A 0.15 kg ball changes velocity from −8.0 m/s to +6.0 m/s during 0.035 s of contact. Find its net impulse and average net force.

Fully explained solutions

1. Use gradients for velocity. The first is (11 − 2)/(3 − 0) = +3 m/s; the second is (7 − 11)/(5 − 3) = −2 m/s. The idealised graph changes direction at 3 s. Displacement is final minus initial position: 7 − 2 = +5 m. Distance adds both legs: 9 + 4 = 13 m.

2. Split the graph where velocity reaches zero. Acceleration = (−2 − 6)/4 = −2 m/s², so the object turns after 6/2 = 3 s. The positive triangle gives +9 m; the final one-second triangle gives −1 m. Displacement is +8 m and distance is 10 m. Treating both areas as positive would answer the distance question only.

3. Compare impulse with initial momentum. Impulse = ½ × 0.10 × 12 = +0.60 N s. Initial momentum is 0.40 × (−2.0) = −0.80 kg m/s, so final momentum is −0.20 kg m/s and final velocity is −0.50 m/s. It still moves in the negative direction. The pulse supplies less than the +0.80 N s needed just to stop it.

4. Subtract signed velocities. Change in velocity = 6.0 − (−8.0) = +14 m/s. Net impulse = 0.15 × 14 = +2.1 N s. Average net force = 2.1/0.035 = +60 N. The reversal makes the velocity-change magnitude 14 m/s; subtracting the two speeds would give the wrong result.

Your final graph check

Before moving on, ask: Have I read the axes? Chosen a positive direction? Split any area at zero? Distinguished distance from displacement? Checked the answer’s units and direction?

For further revision, use VCAA’s official Physics exam resources and assessment reports, starting with materials listed for the current study design. Keep the 2026 Physics formula sheet beside you while practising. The examples here are independent teaching material, not VCAA questions or predictions of the exam.