5.6.11Forces

Velocity-Time Graphs

5 min read 5 parts Lesson 47 of 67 Unit quiz

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Section 1 of 5Reading the Shape of the Graph

The shape of a velocity-time graph tells you whether an object is moving at constant velocity or accelerating.

Horizontal line

Horizontal line Constant velocity, so the acceleration is zero and the object moves at a steady speed in one direction.

Upward sloping line

Upward sloping line Velocity is increasing, so the object is accelerating in the positive direction.

Downward sloping line

Downward sloping line Velocity is decreasing, so the object is decelerating, or accelerating in the negative direction.

Steeper line

Steeper line A greater gradient means a larger acceleration, so velocity changes more quickly.

Two features of the graph carry the information needed for calculations.

Gradient and Area

Gradient

Gradient Calculated by dividing the change in velocity by the time taken, giving the acceleration in m/s².

Sign of gradient

Sign of gradient A negative gradient means acceleration in the negative direction, which may speed up or slow down the object.

Area under the line

Area under the line The area between the line and the time axis gives the displacement during that time.

Feature	Meaning	Unit
Gradient	Acceleration	m/s²
Area under the line	Displacement during that time	m
Height of the line at a point	Velocity at that moment	m/s

Each section of the graph describes one stage of the journey, and each stage can be analysed separately.

From 0 to 10 s the line rises from 0 to 20 m/s, so the object accelerates uniformly.

From 10 s to 20 s the line is horizontal, so the object moves at a constant velocity of 20 m/s.

From 20 s to 25 s the line falls to zero, so the object decelerates uniformly until it stops.

The gradient of each section gives the acceleration for that stage of the journey.

Question

The velocity-time graph of a car rises uniformly from 0 to 20 m/s in 10 s, stays at 20 m/s until 20 s, then falls to 0 in a further 5 s.
Calculate the acceleration in the first 10 s, the acceleration in the final stage, and the average speed for the whole 25 s.

Acceleration in the first 10 s = (20 − 0) ÷ (10 − 0) = 2.0 m/s².
Acceleration in the final stage = (0 − 20) ÷ (25 − 20) = −4.0 m/s².
Total displacement = ½ × 10 × 20 + 20 × 10 + ½ × 5 × 20 = 100 + 200 + 50 = 350 m.
Average speed = 350 ÷ 25 = 14 m/s.

Answer Acceleration 2.0 m/s² in the first stage, −4.0 m/s² in the final stage, and an average speed of 14 m/s.

Velocity-time graph A graph showing how velocity changes with time, with velocity on the vertical axis.
Gradient The steepness of a line, found by dividing the change in the vertical value by the change in the horizontal value.
Acceleration The rate of change of velocity, measured in m/s².
Deceleration A decrease in velocity over time, meaning acceleration in the opposite direction to motion.
Displacement The distance moved in a stated direction.

Reading a velocity-time graph as a distance-time graph,
- so a horizontal line is taken to mean the object is stationary rather than moving at constant velocity.
Calculating the gradient using the wrong axes, so distance is divided by velocity instead of velocity by time.
Forgetting the sign of the acceleration when the velocity falls to zero from a positive value.
Writing the units of acceleration as m/s instead of m/s².

Physics 0625 & 9702 SUVAT Mastery Graph Interpretation

v = u + at | s = ut + ½at² | v² = u² + 2as | s = ½(u + v)t

Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
Free Fall in vacuum: a = g ≈ 9.81 m/s² downwards (air resistance negligible).