Mechanics

Motion Graphs: Displacement, Velocity, and Acceleration

Master the interpretation of kinematic graphs. Learn to convert between displacement-time, velocity-time, and acceleration-time graphs using gradients and areas under the curve.

V
Vectora Team
STEM Education
11 min read
2025-10-10

Why Study Motion Graphs?

Motion graphs translate the movement of an object into a visual, mathematical format. They allow us to instantly understand an object's complete history — where it was, how fast it was moving, and whether it was speeding up or slowing down — at a single glance.

There are three primary types of motion graphs:

  1. Displacement-Time (sts-t)
  2. Velocity-Time (vtv-t)
  3. Acceleration-Time (ata-t)

Learning Goals: By the end of this guide, you should be able to:

  1. Interpret the physical meaning of the gradient (slope) for each graph.
  2. Interpret the physical meaning of the area under the curve for each graph.
  3. Transform one type of graph into another for the same motion.
  4. Describe the motion of an object from standard graph shapes.

1. Displacement-Time (sts-t) Graphs

Plotting position (or displacement, ss) on the y-axis against time (tt) on the x-axis.

Key Rules:

  • y-value: The position of the object relative to the starting point.
  • Gradient (slope): The velocity of the object (ΔsΔt=v\frac{\Delta s}{\Delta t} = v).
  • Area under the curve: Meaningless.

Interpreting Shapes:

  • Horizontal flat line: The object is stationary (velocity is zero).
  • Straight diagonal line: The object is moving with constant velocity.
  • Curved line: The object is accelerating (changing slope means changing velocity).

2. Velocity-Time (vtv-t) Graphs

Plotting velocity (vv) on the y-axis against time (tt) on the x-axis. This is the most information-dense motion graph.

Key Rules:

  • y-value: The instantaneous velocity. If the line crosses the x-axis, the object has reversed direction.
  • Gradient (slope): The acceleration of the object (ΔvΔt=a\frac{\Delta v}{\Delta t} = a).
  • Area under the curve: The change in displacement (Δs=v×t\Delta s = v \times t).

Interpreting Shapes:

  • Horizontal flat line: The object is moving at constant velocity (zero acceleration).
  • Straight diagonal line: The object is experiencing constant uniform acceleration.
  • Curved line: The object has changing (non-uniform) acceleration.

3. Acceleration-Time (ata-t) Graphs

Plotting acceleration (aa) on the y-axis against time (tt) on the x-axis.

Key Rules:

  • y-value: The instantaneous acceleration.
  • Gradient (slope): The rate of change of acceleration (jerk), rarely assessed in high school physics.
  • Area under the curve: The change in velocity (Δv=a×t\Delta v = a \times t).

Interpreting Shapes:

  • Horizontal flat line on x-axis: Zero acceleration (constant velocity).
  • Horizontal flat line above/below x-axis: Constant uniform acceleration.

Transforming Graphs (The Calculus Connection)

To move "down" the chain from displacement \rightarrow velocity \rightarrow acceleration, you find the gradient (derivative).

To move "up" the chain from acceleration \rightarrow velocity \rightarrow displacement, you find the area under the curve (integral).

Graph TypeGradient represents...Area under curve represents...
Displacement-Time (sts-t)Velocity(Nothing useful)
Velocity-Time (vtv-t)AccelerationChange in Displacement
Acceleration-Time (ata-t)Rate of change of acceleration (Jerk)Change in Velocity

Worked Examples

Example 1: Calculating Distance from a Velocity Graph

Question: A car accelerates from rest to 20 m/s20\ \text{m/s} in 5 s5\ \text{s}, travels at constant velocity for 10 s10\ \text{s}, and then brakes to a stop over 4 s4\ \text{s}. Draw the vtv-t graph and calculate total displacement.

Step 1: The graph forms a trapezium.

  • Base = 5+10+4=19 s5 + 10 + 4 = 19\ \text{s}.
  • Top parallel side = 10 s10\ \text{s}.
  • Height (velocity) = 20 m/s20\ \text{m/s}.

Step 2: Area of trapezium = 12(a+b)h\frac{1}{2}(a + b)h

Area=12(19+10)×20=12(29)(20)=290 m\text{Area} = \frac{1}{2}(19 + 10) \times 20 = \frac{1}{2}(29)(20) = 290\ \text{m}

The total displacement is 290 m290\ \text{m}.

Example 2: Graphing Free Fall (with Bouncing)

Question: Describe the vtv-t graph of a ball dropped from a height, which bounces perfectly elastically.

Answer:

  1. The ball falls downward, accelerating at 9.8 m/s2-9.8\ \text{m/s}^2. The graph is a straight straight line starting at 00, sloping downward into negative velocity values.
  2. Upon impact, the velocity instantly flips from a large negative value to a large positive value (the bounce). The line jumps vertically upwards.
  3. The ball travels upward, slowing down under gravity. The line slopes downward from the positive maximum back to 00 at the exact same gradient of 9.8 m/s2-9.8\ \text{m/s}^2.

Common Mistakes

  1. Confusing sts-t and vtv-t graph shapes — A flat horizontal line on an sts-t graph means "stationary". A flat horizontal line on a vtv-t graph means "moving at a constant speed". Always check the y-axis label!
  2. Ignoring negative areas — If the line on a vtv-t graph drops below the x-axis, the object is moving backward. The area under that section must be treated as negative displacement when calculating the final position. Distance (scalar) would treat it as positive.
  3. Misinterpreting zero acceleration — Acceleration is zero when velocity is constant. It does NOT mean the velocity is necessarily zero. A car cruising at 100 km/h100\ \text{km/h} has zero acceleration.

Exam Tips (A-Level / AP / IB)

  • Count the squares: If the curve on a vtv-t graph isn't a triangle or rectangle, you must estimate the area under the curve by counting grid squares. State your method clearly.
  • Tangents: To find the instantaneous velocity at a specific point on a curved sts-t graph, draw a straight tangent line at that point and calculate its gradient (m=ΔyΔxm = \frac{\Delta y}{\Delta x}).
  • Exam questions love objects that change direction. Remember: an object changes direction exactly when the vtv-t graph crosses the horizontal x-axis (v=0v=0).

Frequently Asked Questions

What happens if the ata-t graph isn't horizontal?

If acceleration is changing over time (a diagonal or curved ata-t graph), the object is experiencing "jerk". Standard high-school kinematic equations (SUVAT) only apply when acceleration is constant (horizontal ata-t lines). If it changes, you must use calculus to find velocity and displacement.

Why doesn't the area under an sts-t graph mean anything?

The area would have units of "metre-seconds" (ms\text{m}\cdot\text{s}). There is no physical quantity or useful concept in classical mechanics that corresponds to distance multiplied by time.


  • Projectile Motion — Apply your knowledge of motion graphs to 2D trajectory splitting.
  • Kinetic Theory of Gases — Examine the statistical distribution of velocities of billions of particles simultaneously.