Kinematics (DfE content area Q) describes motion without asking what causes it. On this page all motion is in a straight line, so direction is shown by a sign. Choose a positive direction at the start and keep it.
| Quantity | Symbol | Meaning | Can it be negative? |
|---|---|---|---|
| Position / displacement | $s$ or $x$ | Distance from a fixed origin, in the positive or negative direction | Yes |
| Distance | — | Total length of path travelled | No |
| Velocity | $v$ | Rate of change of displacement | Yes — negative means moving in the negative direction |
| Speed | — | Magnitude of velocity, $|v|$ | No |
| Acceleration | $a$ | Rate of change of velocity | Yes |
The graph above shows a cyclist's displacement from home. Describe the motion and find the average speed and average velocity for the whole $30$ s.
Area between a $v$–$t$ graph and the $t$-axis $=$ displacement
A car starts from rest and accelerates uniformly to $12$ m s$^{-1}$ in $4$ s, travels at this speed for $10$ s, then decelerates uniformly to rest in $6$ s (graph above). Find the accelerations and the total distance.
A train accelerates uniformly from rest to $20$ m s$^{-1}$ in $25$ s, travels at constant speed for $T$ s, then decelerates uniformly to rest in $15$ s. The total distance is $2$ km. Find $T$.
A particle's velocity decreases uniformly from $6$ m s$^{-1}$ at $t = 0$ to $-4$ m s$^{-1}$ at $t = 10$. Find (a) the acceleration, (b) when it is instantaneously at rest, (c) its displacement and (d) the distance travelled in the $10$ s.
A ball is thrown vertically upwards at $9.8$ m s$^{-1}$ and caught at the same height. Taking upwards as positive and ignoring air resistance, sketch its $v$–$t$ graph.
$s$–$t$ gradient
Velocity.
$v$–$t$ gradient
Acceleration.
$v$–$t$ area
Displacement (below axis counts negative).
$a$–$t$ area
Change in velocity.
Horizontal $s$–$t$
At rest.
Horizontal $v$–$t$
Constant velocity, $a = 0$.
Average speed
Total distance ÷ total time.
Average velocity
Displacement ÷ total time.
Trapezium
Start and end at rest: area $= \tfrac12(\text{top} + \text{bottom}) \times \text{height}$.
Sketch a graph for every question, even when it is not asked for. Take $g = 9.8$ m s$^{-2}$.
A straight-line displacement–time graph passes through $(0, 0)$ and $(8, 120)$, with $t$ in seconds and $s$ in metres. Find the velocity.
▶ Show solution
Gradient $= \dfrac{120}{8} = 15$ m s$^{-1}$.
A cyclist's velocity increases uniformly from $4$ m s$^{-1}$ to $16$ m s$^{-1}$ in $6$ s. Find the acceleration and the distance travelled.
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$a = \dfrac{16 - 4}{6} = 2$ m s$^{-2}$.
Distance = trapezium area $= \tfrac12(4 + 16)(6) = 60$ m.
A lift starts from rest, accelerates uniformly to $2.5$ m s$^{-1}$ in $2$ s, moves at constant speed for $8$ s and decelerates uniformly to rest in $3$ s. Find the distance travelled.
▶ Show solution
Trapezium with parallel sides $8$ and $2 + 8 + 3 = 13$, height $2.5$.
Distance $= \tfrac12(8 + 13)(2.5) = 26.25$ m.
Check: $2.5 + 20 + 3.75 = 26.25$ ✓
A walker's displacement–time graph has three straight sections: from $(0, 0)$ to $(600, 900)$, then to $(900, 900)$, then to $(1500, 0)$, with $t$ in s and $s$ in m. Describe the motion and find the walker's speed in each section.
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$0$–$600$ s: walks away at $\dfrac{900}{600} = 1.5$ m s$^{-1}$.
$600$–$900$ s: at rest, $900$ m from the start.
$900$–$1500$ s: gradient $\dfrac{-900}{600} = -1.5$, so walks back to the start at $1.5$ m s$^{-1}$.
For the walker in Question 4, find the average speed and the average velocity over the whole $1500$ s.
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Average speed $= \dfrac{1800}{1500} = 1.2$ m s$^{-1}$.
Displacement is $0$, so average velocity $= 0$ m s$^{-1}$.
A bus accelerates uniformly from rest to $12$ m s$^{-1}$ in $8$ s, travels at $12$ m s$^{-1}$ for $T$ s, then decelerates uniformly to rest in $6$ s. It travels $840$ m altogether. Find $T$.
▶ Show solution
$\tfrac12(8)(12) + 12T + \tfrac12(6)(12) = 840$
$48 + 12T + 36 = 840 \Rightarrow 12T = 756 \Rightarrow T = 63$ s.
A particle's velocity decreases uniformly from $8$ m s$^{-1}$ at $t = 0$ to $-4$ m s$^{-1}$ at $t = 6$. Find its displacement and the distance it travels in these $6$ s.
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$a = \dfrac{-4 - 8}{6} = -2$ m s$^{-2}$, so $v = 8 - 2t$, which is zero when $t = 4$.
Area above: $\tfrac12(4)(8) = 16$. Area below: $\tfrac12(2)(4) = 4$.
Displacement $= 16 - 4 = 12$ m. Distance $= 16 + 4 = 20$ m.
A stone is thrown vertically upwards at $14.7$ m s$^{-1}$. Taking upwards as positive and ignoring air resistance, sketch the $v$–$t$ graph until it returns to its starting point, and use it to find the greatest height reached.
▶ Show solution
Straight line of gradient $-9.8$ from $(0,\,14.7)$, crossing the axis at $t = \dfrac{14.7}{9.8} = 1.5$ s and ending at $(3,\,-14.7)$.
Greatest height = area of the first triangle $= \tfrac12(1.5)(14.7) = 11.025 \approx 11.0$ m.
A car passes a stationary motorbike at a constant $15$ m s$^{-1}$. At that instant the motorbike sets off in the same direction, accelerating uniformly at $2.5$ m s$^{-2}$ until it reaches $25$ m s$^{-1}$, then continuing at that speed. Find when and where the motorbike overtakes the car.
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The motorbike reaches $25$ m s$^{-1}$ after $\dfrac{25}{2.5} = 10$ s, having travelled $\tfrac12(10)(25) = 125$ m. The car has travelled $150$ m — still ahead.
For $t > 10$: motorbike at $125 + 25(t - 10)$, car at $15t$.
$125 + 25t - 250 = 15t \Rightarrow 10t = 125 \Rightarrow t = 12.5$ s.
Distance $= 15 \times 12.5 = 187.5$ m from the starting point.
A train travels between two stations $3$ km apart. It starts from rest and accelerates uniformly at $0.5$ m s$^{-2}$ until reaching $20$ m s$^{-1}$, travels at $20$ m s$^{-1}$, then decelerates uniformly at $1$ m s$^{-2}$ to stop at the second station.
(a) Sketch the $v$–$t$ graph. (b) Find the time spent accelerating and decelerating. (c) Find the total journey time. (d) Find the average speed for the journey.
▶ Show solution
(a) A trapezium: rising line from $(0,0)$, horizontal at $v = 20$, falling line to the $t$-axis.
(b) Accelerating: $\dfrac{20}{0.5} = 40$ s. Decelerating: $\dfrac{20}{1} = 20$ s.
(c) Distance in those stages: $\tfrac12(40)(20) + \tfrac12(20)(20) = 400 + 200 = 600$ m.
Remaining $3000 - 600 = 2400$ m at $20$ m s$^{-1}$ takes $120$ s.
Total time $= 40 + 120 + 20 = 180$ s.
(d) Average speed $= \dfrac{3000}{180} = 16.7$ m s$^{-1}$ (3 s.f.).