Motion
Velocity, acceleration and v–t graphs
Velocity
Velocity is speed in a given direction. It is a vector quantity.
\[ \text{velocity} = \frac{\text{displacement}}{\text{time}} \quad\Rightarrow\quad v = \frac{s}{t} \]
SI unit: m/s (always state the direction, e.g. "12 m/s northwards").
Speed vs velocity:
An object moving in a circle at constant speed has changing velocity because the direction constantly changes.
| Speed | Velocity |
|---|---|
| Scalar — magnitude only | Vector — magnitude + direction |
| Always positive (≥ 0) | Can be positive or negative |
| e.g. 12 m/s | e.g. 12 m/s north, or −12 m/s |
Acceleration
Acceleration is the rate of change of velocity. It is a vector quantity.
\[ a = \frac{v - u}{t} \quad \text{or} \quad a = \frac{\Delta v}{\Delta t} \]
Where: \( u \) = initial velocity, \( v \) = final velocity, \( t \) = time taken.
SI unit: m/s²
SI unit: m/s²
- Positive acceleration — velocity is increasing.
- Deceleration (negative acceleration) — velocity is decreasing; \( a \) is negative when the object is slowing down.
- Zero acceleration — object moves at constant velocity (or is stationary).
Worked example — calculating acceleration
A car accelerates from rest to 28 m/s in 7.0 s. Calculate the acceleration.
\( u = 0\,\text{m/s},\quad v = 28\,\text{m/s},\quad t = 7.0\,\text{s} \)
\( a = \dfrac{v - u}{t} = \dfrac{28 - 0}{7.0} = 4.0\,\text{m/s}^2 \)
Speed-time and velocity-time graphs
A speed-time (v-t) graph shows how speed changes over time. The two key rules:
- Gradient = acceleration. A steeper positive gradient means greater acceleration. A negative gradient means deceleration.
- Area under the graph = distance travelled. Split the area into rectangles and triangles and add them up.
Reading a v-t graph
To find acceleration from section A: gradient = \( \dfrac{\Delta v}{\Delta t} = \dfrac{v_2 - v_1}{t_2 - t_1} \)
To find total distance: add the areas of each section.
Area A (triangle) = \( \tfrac{1}{2} \times \text{base} \times \text{height} \)
Area B (rectangle) = \( \text{base} \times \text{height} \)
Area C (triangle) = \( \tfrac{1}{2} \times \text{base} \times \text{height} \)
To find total distance: add the areas of each section.
Area A (triangle) = \( \tfrac{1}{2} \times \text{base} \times \text{height} \)
Area B (rectangle) = \( \text{base} \times \text{height} \)
Area C (triangle) = \( \tfrac{1}{2} \times \text{base} \times \text{height} \)
Worked example — acceleration and distance from a v-t graph
A v-t graph shows: speed rises from 0 to 20 m/s over 5 s (section A), stays at 20 m/s for 10 s (section B), then falls to 0 in 4 s (section C). Find: (a) the acceleration in A; (b) total distance.
(a) \( a = \dfrac{20 - 0}{5} = 4.0\,\text{m/s}^2 \)
(b) Area A = \( \tfrac{1}{2}(5)(20) = 50\,\text{m} \)
Area B = \( (10)(20) = 200\,\text{m} \)
Area C = \( \tfrac{1}{2}(4)(20) = 40\,\text{m} \)
Total = \( 50 + 200 + 40 = 290\,\text{m} \)
Comparing d-t and v-t graphs
| Feature | Distance-time graph | Speed-time graph |
|---|---|---|
| y-axis | Distance (m) | Speed (m/s) |
| Gradient gives | Speed | Acceleration |
| Area under gives | — (not useful) | Distance |
| Horizontal line | Object stationary | Constant speed |
| Positive slope | Moving (constant speed) | Accelerating |
| Negative slope | Moving back towards start | Decelerating |
The most common exam question type: a v-t graph is given with two or three sections. You must (a) calculate acceleration in one section by finding the gradient, and (b) calculate total distance by finding the total area. Always show your gradient working as a fraction with correct units.