Learning Objectives
3 objectivesBy the end of this note, you should be able to:
- Use the four equations of uniformly accelerated motion in one dimension.
- Draw and interpret displacement-time, velocity-time and acceleration-time graphs.
- Determine physical quantities from the slopes and areas of motion graphs, including non-uniform acceleration.
Equations of Uniformly Accelerated Motion
Key Equations
Equations of uniformly accelerated motion (SUVAT):
$$v=u+at$$
$$s=ut+\frac{1}{2}a{t}^{2}$$
$${v}^{2}={u}^{2}+2as$$
$$s=\frac{1}{2}(u+v)t$$
Variables:
- s = displacement (m)
- u = initial velocity (m s⁻¹)
- v = final velocity (m s⁻¹)
- a = acceleration (m s⁻²)
- t = time interval (s)
SI unit: depends on the quantity calculated.
Rearrangements:
$$a=\frac{v-u}{t}$$
$$t=\frac{v-u}{a}$$
$$s=\frac{{v}^{2}-{u}^{2}}{2a}$$
$$a=\frac{{v}^{2}-{u}^{2}}{2s}$$
$$t=\frac{2s}{u+v}$$
ProportionalityFor constant acceleration from rest, displacement is proportional to time squared, so doubling the time quadruples the displacement. Final velocity squared is proportional to displacement, so doubling the displacement doubles v².
The four equations of uniformly accelerated motion describe motion in one dimension when acceleration stays constant throughout the time interval. Each equation links four of the five quantities s, u, v, a and t, and omits one. Choose the equation whose missing quantity is the one not given and not required.
Displacement, velocity and acceleration are all vectors, so direction must be tracked using a sign convention. Before substituting values, choose one direction as positive. A negative result means the quantity points opposite to the chosen positive direction. For an object thrown vertically upwards, taking upwards as positive gives a positive u but a = –9.81 m s⁻², because gravity acts downwards.
The equations only apply when acceleration is uniform. They cannot be used directly across a stage where a changes, such as a vehicle that accelerates and then brakes. In those cases, split the motion into stages and apply the equations to each stage separately.
Examiner InsightAlways list the five SUVAT quantities and circle which are known. The unused quantity selects the correct equation immediately. Forgetting the sign of a vector is the single most common cause of lost marks.
Exam TipState the positive direction in writing before any substitution.
Worked Example: Motorcycle on a Straight Track
A motorcycle accelerates uniformly from 8.0 m s⁻¹ to 22 m s⁻¹ over a distance of 150 m on a straight road. Calculate the acceleration and the time taken.
Step 1: Finding the acceleration
Equation used
$${v}^{2}={u}^{2}+2as$$
Rearranging for a:
$$a=\frac{{v}^{2}-{u}^{2}}{2s}$$
Given
$$u=8.0{\text{ m s}}^{-1}$$
$$v=22{\text{ m s}}^{-1}$$
$$s=150\text{ m}$$
Substitution:
$$a=\frac{{22}^{2}-{8.0}^{2}}{2\times 150}$$
$$a=\frac{484-64}{300}$$
$$a=1.40{\text{ m s}}^{-2}$$
Step 2: Finding the time taken
Equation used
$$v=u+at$$
Rearranging for t:
$$t=\frac{v-u}{a}$$
Substitution:
$$t=\frac{22-8.0}{1.40}$$
$$t=10.0\text{ s}\approx 10\text{ s (2 s.f.)}$$
The motorcycle takes 10 s to gain 14 m s⁻¹ at a steady acceleration of 1.4 m s⁻². The positive sign confirms acceleration is in the same direction as the motion.
Drawing and Interpreting Motion Graphs
Motion in one dimension is represented graphically using displacement-time, velocity-time and acceleration-time graphs. Time is plotted on the horizontal axis and the motion quantity on the vertical axis. The shape of each line reveals how the motion changes over the recorded interval.
On a displacement-time graph, a horizontal line shows the object is stationary, a straight sloping line shows uniform velocity, and a curve shows changing velocity. A line bending upwards more steeply represents speeding up. A line bending towards the horizontal represents slowing down. A negative gradient indicates motion in the negative direction.
On a velocity-time graph, a horizontal line shows constant velocity, a straight sloping line shows uniform acceleration, and a curve shows changing acceleration. A line crossing the time axis shows the object reversing direction. The y-intercept gives the initial velocity at t = 0.
On an acceleration-time graph, a horizontal line shows constant acceleration. A horizontal line at zero shows constant velocity. A spike or step shows a sudden change in acceleration, such as a collision or a sudden brake.
MisconceptionA horizontal line on a displacement-time graph means zero velocity, because the displacement is not changing. A horizontal line on a velocity-time graph means constant velocity, not zero. The two graphs look similar but mean opposite things.
Exam TipRead the y-axis label before describing motion.
Gradients and Areas of Motion Graphs
The gradient and area of motion graphs give physical quantities directly. The gradient of a displacement-time graph at any point equals the velocity at that instant. The gradient of a velocity-time graph equals the acceleration. The area between a velocity-time graph and the time axis equals the displacement.
For uniform motion, the gradient is found from any two points on the straight line. For non-uniform motion, the instantaneous gradient is found from a tangent drawn at the chosen point. The slope of that tangent gives the velocity or acceleration at that instant.
The area under a velocity-time graph equals displacement, since (velocity) × (time) has units of metres. For a straight-line v-t graph, the area is found using the area of the rectangle, triangle or trapezium beneath it. For a curved v-t graph, the area is estimated by counting squares on the gridded graph or by splitting the area into thin strips and using the trapezium rule.
The area under an acceleration-time graph equals the change in velocity, since (acceleration) × (time) has units of m s⁻¹. Areas below the time axis represent negative changes and must be subtracted. The total signed area gives the net change.
| Graph | Gradient gives | Area under gives |
|---|---|---|
| Displacement-time | velocity (m s⁻¹) | not used |
| Velocity-time | acceleration (m s⁻²) | displacement (m) |
| Acceleration-time | not used | change in velocity (m s⁻¹) |
Examiner InsightFor non-uniform acceleration, the instantaneous gradient is found from a tangent at the chosen point. Using a large triangle when measuring the gradient reduces percentage uncertainty.
Exam TipExtend the tangent across at least half the graph width.
Worked Example: Reading a Velocity-Time Graph
A trolley moves along a straight track. Its velocity-time graph shows a straight line from (0 s, 2.0 m s⁻¹) to (8.0 s, 14 m s⁻¹), then a horizontal line at 14 m s⁻¹ from 8.0 s to 12 s. Calculate the acceleration during the first stage and the total displacement over the 12 s.
Step 1: Acceleration during the first stage
Equation used
$$a=\frac{\Delta v}{\Delta t}$$
Given
$$\Delta v=14-2.0=12{\text{ m s}}^{-1}$$
$$\Delta t=8.0\text{ s}$$
Substitution:
$$a=\frac{12}{8.0}$$
$$a=1.5{\text{ m s}}^{-2}$$
Step 2: Displacement during the first stage (trapezium area)
$${s}_{1}=\frac{1}{2}(u+v)\times t$$
$${s}_{1}=\frac{1}{2}\times (2.0+14)\times 8.0$$
$${s}_{1}=64\text{ m}$$
Step 3: Displacement during the second stage (rectangle area)
$${s}_{2}=v\times t$$
$${s}_{2}=14\times 4.0$$
$${s}_{2}=56\text{ m}$$
Step 4: Total displacement
$${s}_{\text{total}}={s}_{1}+{s}_{2}=64+56=120\text{ m}$$
The trolley accelerates uniformly at 1.5 m s⁻² for 8 s, then continues at constant velocity. The total area under the v-t graph confirms it travels 120 m in 12 s.
QUICK RECAP
Key Points
- Four SUVAT equations apply only when acceleration is constant.
- Each SUVAT equation omits one of the five quantities s, u, v, a, t.
- Always state the positive direction before substituting vectors.
- A negative answer means the vector points the opposite way.
- For changing acceleration, split motion into stages or use graphs.
- d-t graph gradient = velocity at that instant.
- v-t graph gradient = acceleration; v-t graph area = displacement.
- a-t graph area = change in velocity over the interval.
- Tangent gives instantaneous gradient on a curved graph.
- Curved areas are found by counting squares or using trapezium strips.
- Doubling time from rest quadruples displacement (s ∝ t²).
- Doubling initial speed quadruples stopping distance for fixed deceleration.
CAN I…? PROGRESS CHECK
Self-Assessment
- Select the correct SUVAT equation given any three of s, u, v, a, t?
- Apply a sign convention correctly to a vertical projectile problem?
- Use SUVAT to calculate stopping distance, time to fall, or final speed?
- Sketch displacement-time, velocity-time and acceleration-time graphs for a stated motion?
- Describe the motion represented by a given motion graph?
- Determine velocity from the gradient of a displacement-time graph?
- Determine acceleration from the gradient of a velocity-time graph?
- Determine displacement from the area under a velocity-time graph?
- Use a tangent to find an instantaneous gradient on a curved graph?
- Estimate the area under a curved v-t graph using trapezium strips?
- State the proportional effect of doubling time or speed in a SUVAT context?