3.1 Speed, velocity and motion graphs — Learn

Prerequisites: Section 2.1, especially forces and vectors; the kinetic energy store from Section 1.1; reading a straight-line graph; dividing and multiplying numbers.

By the end of this section you can: tell distance from displacement, and speed from velocity; use typical speeds; calculate speed, distance and time; calculate acceleration; find speed and acceleration from a motion graph, and, at Higher tier, read a tangent and an area; use $v^2 - u^2 = 2as$ ; and describe terminal velocity.

1. Distance and displacement

Distance is how far an object moves. It has a size but no direction, so it is a scalar. A runner who completes a 400 m lap has run a distance of 400 m.

Displacement is the distance in a straight line from the start point to the finish point, together with the direction of that line. Displacement is a vector. The same runner finishes a 400 m lap back at the start, so the displacement is zero.

2. Speed

Speed is how fast an object moves. It has a size but no direction, so it is a scalar. For an object that moves at a steady speed,

\text{speed} = \frac{\text{distance}}{\text{time}} \qquad s = v t

where $s$ is the distance in metres, $v$ is the speed in metres per second, and $t$ is the time in seconds.

The speed of a moving object is rarely steady. A car speeds up, slows down and stops. For a journey with changing speed, the average speed describes the whole journey:

\text{average speed} = \frac{\text{total distance}}{\text{total time}}

Worked example 1. A cyclist travels 90 m in 15 s. Calculate her average speed.

\begin{aligned} v &= \frac{s}{t} \\ &= \frac{90}{15} \\ &= 6.0\,\text{m/s} \end{aligned}

Worked example 2. A car travels at a steady 24 m/s for 5.0 s. Calculate the distance it travels.

\begin{aligned} s &= v t \\ &= 24 \times 5.0 \\ &= 120\,\text{m} \end{aligned}

3. Typical speeds

Some speeds are worth remembering, because they help you judge whether an answer is sensible.

Object	Typical speed
A person walking	1.5 m/s
A person running	3 m/s
A person cycling	6 m/s
Sound in air	330 m/s

A car in a town travels at about 12 m/s. A car on a motorway travels at about 30 m/s. A train travels at about 60 m/s. The speed of the wind changes from day to day, so there is no single typical value for it.

The speed of a person walking, running or cycling depends on age, the ground, and how far they travel. Each value above is a guide, not a rule.

4. Velocity

Velocity is speed in a given direction. It is a vector. Two cars can have the same speed and different velocities, because they travel in different directions.

A car on a roundabout can move at a steady speed and still change velocity. The speed stays the same, but the direction of travel keeps changing, so the velocity keeps changing. The same is true for a satellite in a circular orbit. It moves at a constant speed, and its velocity changes all the time, because its direction changes all the time.

5. Distance-time graphs

A distance-time graph shows how far an object has travelled. Time goes along the horizontal axis. Distance goes along the vertical axis.

A horizontal line means the object is stationary.
A straight sloping line means a steady speed.
A steeper line means a greater speed.

The speed is the gradient of the line:

\text{speed} = \frac{\text{change in distance}}{\text{change in time}}

Worked example 3. A distance-time graph for a walker is a straight line. The line passes through $(0\,\text{s}, 0\,\text{m})$ , $(4.0\,\text{s}, 32\,\text{m})$ and $(8.0\,\text{s}, 64\,\text{m})$ . Calculate the speed.

\begin{aligned} v &= \frac{\text{change in distance}}{\text{change in time}} \\ &= \frac{64 - 0}{8.0 - 0} \\ &= 8.0\,\text{m/s} \end{aligned}

If the object is accelerating, the distance-time graph is curved. To find the speed at one moment, draw a tangent to the curve at that time. The gradient of the tangent is the speed at that moment.

6. Acceleration

Acceleration is the rate at which velocity changes. For motion in a straight line,

\text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}} \qquad a = \frac{\Delta v}{t}

where $a$ is the acceleration in metres per second squared, $\Delta v$ is the change in velocity in metres per second, and $t$ is the time in seconds.

An object that slows down has a negative acceleration. We often say that it decelerates.

Worked example 4. A bus speeds up from rest to 9.0 m/s in 6.0 s. Calculate its acceleration.

\begin{aligned} a &= \frac{\Delta v}{t} \\ &= \frac{9.0 - 0}{6.0} \\ &= 1.5\,\text{m/s}^2 \end{aligned}

Near the Earth's surface, an object that falls freely under gravity accelerates at about $9.8\,\text{m/s}^2$ . A car pulling away from traffic lights accelerates at about $2\,\text{m/s}^2$ . A train braking at a station decelerates at about $1\,\text{m/s}^2$ .

7. Velocity-time graphs

A velocity-time graph shows how the velocity changes. The gradient of the line is the acceleration:

a = \frac{\text{change in velocity}}{\text{change in time}}

Worked example 5. A velocity-time graph is a straight line through $(0\,\text{s}, 0\,\text{m/s})$ , $(5.0\,\text{s}, 15\,\text{m/s})$ and $(10\,\text{s}, 30\,\text{m/s})$ . Calculate the acceleration.

\begin{aligned} a &= \frac{30 - 0}{10 - 0} \\ &= 3.0\,\text{m/s}^2 \end{aligned}

The area under a velocity-time graph is the distance travelled. When the graph is a straight line from rest, the area is the area of a triangle:

\text{distance} = \tfrac{1}{2} \times \text{base} \times \text{height}

Worked example 6. A velocity-time graph rises in a straight line from $0\,\text{m/s}$ at $0\,\text{s}$ to $12\,\text{m/s}$ at $6.0\,\text{s}$ . Calculate the distance travelled.

\begin{aligned} \text{distance} &= \tfrac{1}{2} \times \text{base} \times \text{height} \\ &= 0.5 \times 6.0 \times 12 \\ &= 36\,\text{m} \end{aligned}

You can also find the area by counting the squares under the line.

Try it: acceleration on a velocity-time graph

The figure draws a velocity-time graph for an object that starts from rest and speeds up at a steady rate. You can change the acceleration and the time.

Predict first. Do not move the sliders yet. Write down your answers.

If you double the acceleration, what happens to the final velocity?
If you double the time, what happens to the distance travelled?

Test. Set the acceleration to $2.0\,\text{m/s}^2$ and the time to 4.0 s. Read the final velocity and the distance. Now double the time and read them again.

Explain. The gradient of the line is the acceleration and the area under the line is the distance. Why does doubling the time make the distance four times as large, when the acceleration has not changed? Write your answer in two or three sentences.

8. Motion with uniform acceleration

For motion with a steady acceleration, one equation links the speeds, the acceleration and the distance:

v^2 - u^2 = 2 a s

where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $s$ is the distance. This equation is given on the equations sheet, so you choose it and use it; you do not need to remember it.

Worked example 7. A car travels at 20 m/s. The driver brakes with a steady deceleration of $4.0\,\text{m/s}^2$ until the car stops. Calculate the distance the car travels while braking.

\begin{aligned} v^2 - u^2 &= 2 a s \\ 0^2 - 20^2 &= 2 \times (-4.0) \times s \\ -400 &= -8.0 s \\ s &= \frac{400}{8.0} \\ &= 50\,\text{m} \end{aligned}

9. Terminal velocity

When an object falls through a fluid, two forces act on it. Its weight pulls it down, and the drag force from the fluid pushes it up. At the start the weight is larger than the drag, so the object speeds up.

As the object gets faster, the drag force increases. Eventually the drag force equals the weight. The resultant force is then zero, so the object stops speeding up. It falls at a steady velocity called the terminal velocity.

A velocity-time graph for a falling object shows a steep rise at first. The gradient then gets smaller, and the line becomes horizontal at the terminal velocity.

Watch out. A horizontal line on a velocity-time graph does not mean the object is stationary. It means the velocity is steady. The object is moving at a constant velocity. A horizontal line on a distance-time graph is the one that means the object is stationary. Check which quantity is on the vertical axis before you read the line.

Progress check

Return to the chapter problem. The bus travels at 12 m/s. The driver takes 0.60 s to react before she brakes. Work out the distance the bus travels in that time. This distance is the thinking distance.

You should now be able to

tell distance from displacement, and speed from velocity;
recall typical speeds, including the speed of sound in air;
calculate speed, distance and time, and work out an average speed;
find the speed from the gradient of a distance-time graph;
calculate acceleration from a change in velocity, and tell acceleration from deceleration;
find the acceleration from the gradient of a velocity-time graph;
find the distance from the area under a velocity-time graph, and read a speed from the tangent to a curved distance-time graph (Higher tier);
explain why velocity changes in circular motion (Higher tier);
use $v^2 - u^2 = 2as$ for motion with uniform acceleration;
describe terminal velocity and the forces on a falling object.