Velocity-Time Graphs and Terminal Velocity

What a velocity–time graph represents

A velocity–time graph records an object's velocity at different times. Time, in seconds (s), goes on the horizontal axis. Velocity, in metres per second (m/s), goes on the vertical axis. Because velocity includes direction, a negative value means motion in the direction chosen as negative; it does not mean that the object has a “negative speed”.

To draw a graph from measurements, choose scales that use most of the available axes, label both axes with quantities and units, and plot each measured pair of time and velocity values. Join points with straight segments when the data represent distinct stages, or use a suitable smooth curve when velocity changes continuously. Do not automatically draw a line from the origin: the object may already have an initial velocity at $t=0$.

The height of the graph at any instant gives the velocity. A point at $(5\text{ s},12\text{ m/s})$, for example, means that the velocity at 5 s is 12 m/s.

Reading motion from the line

The direction and steepness of the line reveal how velocity is changing:

• a horizontal line shows constant velocity and zero acceleration;
• a straight line sloping upwards shows constant positive acceleration;
• a straight line sloping downwards shows constant negative acceleration;
• a curve shows that the acceleration is changing;
• a line on the time axis shows zero velocity, so the object is stationary;
• a line below the time axis shows motion in the negative direction.

A downward slope is sometimes described as deceleration, but take care. Negative acceleration reduces speed only when the object has positive velocity. If both velocity and acceleration are negative, the object's speed is increasing in the negative direction.

Gradient gives acceleration

Acceleration is the change in velocity per unit time, so it is the gradient of a velocity–time graph:

$a=\frac{\Delta v}{\Delta t}$

The gradient's units are

$\frac{\text{m/s}}{\text{s}}=\text{m/s}^2.$

For a straight line, choose two points on the line and calculate “change in vertical coordinate divided by change in horizontal coordinate”. If velocity rises from 0 m/s to 12 m/s in 4 s, the acceleration is

$a=\frac{12-0}{4-0}=3\text{ m/s}^2.$

A straight line has one constant gradient, so this is uniform acceleration. A steeper line represents an acceleration of greater magnitude.

On a curved section, the average acceleration over a time interval comes from the gradient of the chord joining its endpoints. To estimate the acceleration at one instant, draw a tangent to the curve at that point and find the tangent's gradient. As a curve becomes less steep, its acceleration decreases.

Area under the graph: displacement and distance — Higher Tier

For constant velocity, displacement is velocity multiplied by time. On a velocity–time graph, these two quantities are the height and width of the region beneath the line. Their product is its area:

$\text{area}=\text{velocity}\times\text{time}$

The units confirm the meaning:

$\text{m/s}\times\text{s}=\text{m}.$

If an object accelerates uniformly from 0 m/s to 12 m/s in 4 s, the region under the line is a triangle. Its area is

$\frac{1}{2}\times4\times12=24\text{ m}.$

Therefore, the object moves 24 m during those four seconds, provided its velocity remains in one direction.

For a graph made from straight sections, split the area into rectangles, triangles or trapezia and add their areas. For an irregular curve, estimate the enclosed area by counting squares. First determine what one small square represents: multiply its time width by its velocity height. Count complete squares, combine partial squares sensibly, and multiply by the value of one square.

Strictly, signed area gives displacement. Area above the time axis is positive and area below it is negative. If the graph crosses the axis, subtract the opposing signed areas to find displacement. To find total distance travelled, add the magnitudes of all the separate areas instead. When the velocity never changes sign, the magnitude of the displacement equals the distance travelled.

Falling through a fluid

An object falling through air or another fluid has weight acting downwards and drag acting upwards, opposite to its motion. Just after release, its speed and drag are small, so weight is greater than drag. There is a downward resultant force, and the object accelerates.

As the object speeds up, drag increases. Weight is approximately constant, so the difference between weight and drag becomes smaller. The resultant force and acceleration therefore decrease. On a velocity–time graph, the curve continues to rise but its gradient gradually becomes shallower.

Eventually drag equals weight. The resultant force is then zero, so acceleration is zero. The object continues at a constant velocity called its terminal velocity. This appears as a horizontal section on the velocity–time graph.

Data for Reaching terminal velocity

The graph's changing gradient connects the motion to the forces: a steep initial gradient means a large acceleration when weight greatly exceeds drag; the flattening curve means the forces are becoming more balanced; the horizontal section means they are balanced. Terminal velocity does not mean that no forces act—it means that the forces balance.

Opening a parachute

Before a skydiver opens a parachute, they may be falling at a high terminal velocity. Opening the parachute greatly increases drag. For a time, drag is greater than weight, giving an upward resultant force. The skydiver is still moving downwards but slows rapidly, so a graph using downward as positive has a steep negative gradient.

As the skydiver slows, drag decreases. Eventually it again equals weight. Acceleration returns to zero and the skydiver reaches a new, lower terminal velocity. If upward had instead been chosen as the positive direction, the plotted velocities and gradient signs would reverse, but the physical sequence of forces and changes in speed would be the same.

Read the graph

• Horizontal axis: time in s.
• Vertical axis: velocity in m/s.
• Graph height: velocity at that instant.
• Horizontal line: constant velocity and zero acceleration.
• Line on the time axis: stationary.
• Straight slope: uniform acceleration.
• Curve: changing acceleration.
• Below the time axis: motion in the chosen negative direction.

Find acceleration

$a=\text{gradient}=\frac{\Delta v}{\Delta t}$

Acceleration is measured in m/s². Use two widely separated points on a straight line. For instantaneous acceleration on a curve, find the gradient of a tangent.

Find distance or displacement — Higher Tier

$\text{displacement}=\text{signed area under the velocity–time graph}$

Use rectangles, triangles and trapezia. For an irregular region, count squares and account for the axis scales.

If velocity stays on one side of the time axis, the magnitude of the area gives distance travelled. If it crosses the axis:

• displacement = positive area minus negative-area magnitude;
• distance = sum of the magnitudes of all areas.

Recognise terminal velocity

1. Initially: weight exceeds drag, so there is a downward resultant force and the object accelerates.
2. As speed rises: drag increases, so resultant force and gradient decrease.
3. At terminal velocity: drag equals weight, resultant force is zero, acceleration is zero and the graph is horizontal.
4. Parachute opens: drag can exceed weight, producing rapid slowing followed by a lower terminal velocity.