Motion graphs: displacement, velocity and acceleration

Physics · Senior school physics · about 10 min

Motion graphs show the same journey in different ways. A position-time graph tracks where an object is, a velocity-time graph shows how quickly position changes and an acceleration-time graph describes how velocity changes. Reading their gradients and areas is an efficient way to solve many kinematics questions.

The key idea

Gradients give rates of change, while areas give accumulated changes:

v=\frac{dx}{dt},\qquad a=\frac{dv}{dt},\qquad \Delta x=\int v\,dt.

The gradient of a position-time graph is velocity. The gradient of a velocity-time graph is acceleration. Signed area under a velocity-time graph is displacement.

Three aligned coordinate graphs stacked vertically with a common horizontal t axis running from zero at the left to a later time at the right. The top graph has vertical axis x and a smooth increasing upward-curving parabola, the middle has vertical axis v and a straight increasing line beginning above zero, and the bottom has vertical axis a and a flat horizontal line above zero. Between graphs draw a downward linking arrow from position to velocity labelled gradient and another downward arrow from velocity to acceleration labelled gradient. Show no numbered axes or equations. All three share the same time domain and describe constant positive acceleration. — Gradients connect position to velocity and velocity to acceleration.

Position-time graphs

A position-time graph has position $x$ vertically and time $t$ horizontally. Its gradient gives velocity in $\mathrm{m/s}$ . A straight line with positive gradient shows constant positive velocity; a horizontal line shows rest. A curved line indicates that velocity is changing, and its instantaneous gradient is the tangent slope at that time. A negative slope means movement in the negative direction, not an impossible speed. Do not interpret the height of the position graph as speed. To find displacement between two times, subtract the initial position from the final position.

Velocity-time graphs

The gradient of a velocity-time graph gives acceleration in $\mathrm{m/s^2}$ . A horizontal line has zero acceleration, even when velocity is not zero. The signed area between the velocity curve and the time axis gives displacement: regions above the axis contribute positively, and regions below contribute negatively. For constant velocity, area is a rectangle; for uniform acceleration, use a rectangle and triangle or a trapezium. Total distance travelled is found by adding the absolute areas, whereas displacement retains their signs. A downward sloping line is negative acceleration, but an object only slows down when acceleration opposes its velocity.

Acceleration-time graphs

On an acceleration-time graph, signed area over an interval is the change in velocity, $\Delta v=\int a\,dt$ . A constant positive acceleration produces an upward-sloping straight line on a velocity-time graph and a curved position-time graph. If acceleration is zero, velocity is constant and position changes linearly. Always distinguish initial values from changes: an area under an acceleration graph gives $v_f-v_i$ , not $v_f$ by itself. For a piecewise journey, analyse each time interval separately and carry the final velocity and position into the next interval.

Worked example: A vehicle accelerating, cruising and stopping

A vehicle starts at position $x=0\,\mathrm{m}$ with velocity $+2.0\,\mathrm{m/s}$ . It accelerates uniformly at $+1.5\,\mathrm{m/s^2}$ for $4.0\,\mathrm{s}$ , then travels at constant speed for $3.0\,\mathrm{s}$ . It finally brakes uniformly at $-2.0\,\mathrm{m/s^2}$ for $4.0\,\mathrm{s}$ . Find its velocity after the first stage, the displacement in each stage and its final position.

Draw one velocity versus time plot with a horizontal t (s) axis and vertical v (m/s) axis. The polyline begins at (t=0 s, v=2 m/s), rises straight to (4 s, 8 m/s), remains horizontal at v=8 m/s until t=7 s, then falls straight to (11 s, 0 m/s). Put small solid dots at these four endpoints. Project faint dotted vertical guide lines from t=4 s and t=7 s to the axis, and shade the three regions under the curve separately without printing their areas. The axes must be scaled so the t values 0, 4, 7 and 11 are in proportion and the v values 0, 2 and 8 are in proportion. Within the regions write A_1, A_2 and A_3. The picture must not show any calculated displacement or final position. — The signed areas of the three velocity-time segments give their displacements.

Show the solution

Step 1: First stage, accelerate. From the gradient of the velocity-time graph:

v_1=2.0+(1.5)(4.0)=8.0\,\mathrm{m/s}.

The area under the first segment is a trapezium:

\Delta x_1=\frac{2.0+8.0}{2}(4.0)=20\,\mathrm{m}.

Step 2: Second stage, constant speed. The rectangular area is

\Delta x_2=(8.0)(3.0)=24\,\mathrm{m}.

Step 3: Third stage, brake. The final velocity is

v_3=8.0+(-2.0)(4.0)=0\,\mathrm{m/s}.

The triangular area is

\Delta x_3=\frac{8.0+0}{2}(4.0)=16\,\mathrm{m}.

Step 4: Add signed displacements. All velocities are non-negative during the journey:

x_f=0+20+24+16=60\,\mathrm{m}.

The stage boundaries occur at $t=4.0\,\mathrm{s}$ , $t=7.0\,\mathrm{s}$ and $t=11.0\,\mathrm{s}$ .

Answer: $8.0\,\mathrm{m/s}$ after stage one; displacements of $20\,\mathrm{m}$ , $24\,\mathrm{m}$ and $16\,\mathrm{m}$ ; final position $60\,\mathrm{m}$ .

Common mistakes

Finding distance from signed area without checking for negative velocity

Use the signed area for displacement; add the magnitudes of areas for total distance.

Reading acceleration from the height of a velocity-time graph

Acceleration is the gradient of the velocity graph, not its vertical coordinate.

Assuming negative acceleration always means slowing down

Check the signs of both velocity and acceleration; opposing signs reduce speed.

Resetting initial velocity or position at each stage

Carry each stage's final value into the next stage before finding its gradient or area.

Check your understanding

1. A position-time graph is a straight line from $(0\,\mathrm{s},2\,\mathrm{m})$ to $(4\,\mathrm{s},14\,\mathrm{m})$ . What is the velocity?

2. An object maintains a velocity of $-3.0\,\mathrm{m/s}$ for $5.0\,\mathrm{s}$ . What is its displacement?

3. Velocity decreases uniformly from $+10\,\mathrm{m/s}$ to $+2.0\,\mathrm{m/s}$ over $4.0\,\mathrm{s}$ . What is the acceleration?

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