The Trapezium That Carries Its Own Average
A car's speed rises steadily from $5\ \text{m s}^{-1}$ to $v\ \text{m s}^{-1}$ over $8\ \text{s}$. The region under the velocity–time graph is a trapezium, and its area is the distance travelled. Because the line is straight, that area equals the average of the two speeds multiplied by the time: $s = \dfrac{1}{2}(u + v)t$. Drag the final-speed slider and watch the dashed average-velocity line sit halfway up the trapezium.
distance $s = \dfrac{1}{2}(u + v)t$ = 56.0 m · average velocity $\dfrac{1}{2}(u + v)$ = 7.0 m s⁻¹ · initial speed fixed at $u = 5\ \text{m s}^{-1}$
The dashed line is the average velocity — exactly halfway between the two speeds. The rectangle under it and the trapezium have the same area, so distance = average velocity × time.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. Which constant acceleration formula does not contain the time $t$?
R2. A car travelling at $12\ \text{m s}^{-1}$ decelerates at $2\ \text{m s}^{-2}$. Its speed after $4\ \text{s}$ is:
R3. A body starts from rest, so $u = 0$. The formula $s = ut + \dfrac{1}{2}at^{2}$ then becomes:
When a speed changes uniformly from $u$ to $v$, the average velocity is simply the mean of the two: $\dfrac{u + v}{2}$. So a car going from $5\ \text{m s}^{-1}$ to $9\ \text{m s}^{-1}$ has the same average velocity as a car sitting at a steady $7\ \text{m s}^{-1}$.
Multiplying the average velocity by the time gives the distance travelled:
$s = \dfrac{1}{2}(u + v)t$
This is the area of the trapezium under the velocity–time graph — the formula is a picture, not a new fact.
The formula contains $u, v, t$ and $s$ — but no $a$. That makes it the first choice whenever two speeds and a time are known, and the acceleration is neither given nor wanted.
| You know | You want | Use |
|---|---|---|
| $u, v, t$ | $s$ | $s = \dfrac{1}{2}(u + v)t$ |
| $s, u, t$ | $v$ | $v = \dfrac{2s}{t} - u$ |
| $s, u, v$ | $t$ | $t = \dfrac{2s}{u + v}$ |
The formula only works while the speed changes uniformly — for constant acceleration. A journey with stages has to be split into stages.
A cyclist speeds up uniformly from $3\ \text{m s}^{-1}$ to $9\ \text{m s}^{-1}$ over $12\ \text{s}$: $s = \dfrac{1}{2}(3 + 9)(12) = 72\ \text{m}$. No acceleration needed.
A train covers $800\ \text{m}$ in $50\ \text{s}$ and finishes at $25\ \text{m s}^{-1}$. Then $800 = \dfrac{1}{2}(u + 25)(50)$, so $u + 25 = 32$ and $u = 7\ \text{m s}^{-1}$.
Groups of 4: tap a journey on the left, then the distance it covers on the right to draw a colour-coded line. Every journey gives two speeds and a time, so the distance is always ½(u + v)t — work it out in your group before you link. When all six lines are drawn, press Check to score. Discuss any red links, then press Reset to try the shuffle again.
The straight line is a velocity–time graph over a 12 s window. Set the initial velocity u and the acceleration a, and the shaded trapezium shows the displacement. Your task: adjust the two sliders until the shaded area equals the target 180 m — the badge fires when you are close enough. Watch how the two sliders control the two parts of the trapezium, then answer the three challenges.
A train travels from A to B, a distance of $20000\ \text{m}$, taking $1000\ \text{s}$. The journey has three stages. In the first stage the train starts from rest at A and accelerates uniformly until its speed is $V\ \text{m s}^{-1}$. In the second stage the train travels at the constant speed $V\ \text{m s}^{-1}$ for $600\ \text{s}$. During the third stage the train decelerates uniformly, coming to rest at B. (i) Sketch the velocity–time graph for the journey. (ii) Find the value of $V$.
The graph is a triangle, a rectangle and a triangle. The two triangles have the same height $V$, and their two times add up to $1000 - 600 = 400\ \text{s}$. Write the total distance as a single expression in $V$.
A straight rise from $0$ to $V$ over $t_1$ seconds, a horizontal line at $V$ for $600\ \text{s}$, then a straight fall from $V$ to $0$ over $t_3$ seconds, where $t_1 + 600 + t_3 = 1000$.
First stage: $s_1 = \dfrac{1}{2}(0 + V)t_1 = \dfrac{1}{2}Vt_1$. Second stage: $s_2 = 600V$. Third stage: $s_3 = \dfrac{1}{2}(V + 0)t_3 = \dfrac{1}{2}Vt_3$.
Total distance $= \dfrac{1}{2}V(t_1 + t_3) + 600V = \dfrac{1}{2}V(400) + 600V = 800V$.
So $800V = 20000$ and $V = 25\ \text{m s}^{-1}$.
A train leaves a station where it has been at rest and picks up speed at a constant rate for $60\ \text{s}$ until it reaches a speed of $17\ \text{m s}^{-1}$. It then remains at this constant speed for $60\ \text{s}$ before it begins to slow down uniformly as it approaches a set of signals. After $45\ \text{s}$ of slowing it is travelling at $10\ \text{m s}^{-1}$ and the signal changes. The train again increases speed uniformly for $75\ \text{s}$ until it reaches $20\ \text{m s}^{-1}$. A second set of signals then orders the train to stop, which it does after slowing down uniformly for $30\ \text{s}$. (i) Draw a speed–time graph for the train. (ii) Use your graph to find the distance the train has travelled from the station, and its average speed for the whole journey.
Split the journey into five straight strips. Each strip is a trapezium, and each area is $\dfrac{1}{2}(u + v)t$ with that strip's two speeds and its time. Add the five areas, then divide by the total time.
The speeds at the joins are $0 \to 17 \to 17 \to 10 \to 20 \to 0\ \text{m s}^{-1}$, over the five times $60, 60, 45, 75, 30\ \text{s}$.
$s_1 = \dfrac{1}{2}(0 + 17)(60) = 510\ \text{m}$ · $s_2 = 17 \times 60 = 1020\ \text{m}$ · $s_3 = \dfrac{1}{2}(17 + 10)(45) = 607.5\ \text{m}$
$s_4 = \dfrac{1}{2}(10 + 20)(75) = 1125\ \text{m}$ · $s_5 = \dfrac{1}{2}(20 + 0)(30) = 300\ \text{m}$
$510 + 1020 + 607.5 + 1125 + 300 = 3562.5\ \text{m}$.
Total time $= 60 + 60 + 45 + 75 + 30 = 270\ \text{s}$, so the average speed is $\dfrac{3562.5}{270} = 13.2\ \text{m s}^{-1}$ (3 s.f.).
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. For a speed changing uniformly from $u$ to $v$, the average velocity over the interval is:
Q2. A car speeds up uniformly from $6\ \text{m s}^{-1}$ to $14\ \text{m s}^{-1}$ in $10\ \text{s}$. The distance covered is:
Q3. $s = \dfrac{1}{2}(u + v)t$ is the best formula to use when:
Q4. The average velocity and the average speed of a journey are equal when:
Work each on paper, then reveal the solution to check your working.
A lift travels between the ground floor G and the roof garden R of a hotel. It starts from rest, takes $5\ \text{s}$ to increase its speed uniformly to $2\ \text{m s}^{-1}$, maintains this speed for $5\ \text{s}$, then slows down uniformly to rest in another $5\ \text{s}$. It waits $5\ \text{s}$ at R, then repeats the same pattern (speeding up, steady, slowing down, each for $5\ \text{s}$) on the way back down to G. Taking upwards as positive, calculate the acceleration in each of the seven five-second intervals of the complete round trip. Take care with the signs.
Reveal Solution
Upwards: speeding up $a = \dfrac{2 - 0}{5} = 0.4\ \text{m s}^{-2}$; steady $a = 0$; slowing $a = \dfrac{0 - 2}{5} = -0.4\ \text{m s}^{-2}$. The wait is another $a = 0$.
Coming down the velocities are negative: accelerating downwards $a = \dfrac{-2 - 0}{5} = -0.4\ \text{m s}^{-2}$; steady $a = 0$; slowing to rest $a = \dfrac{0 - (-2)}{5} = +0.4\ \text{m s}^{-2}$.
So the seven values are $0.4,\ 0,\ -0.4,\ 0,\ -0.4,\ 0,\ 0.4\ \text{m s}^{-2}$.
For the same lift: it is at rest, speeds up uniformly to $2\ \text{m s}^{-1}$ in $5\ \text{s}$, travels at $2\ \text{m s}^{-1}$ for $5\ \text{s}$, then comes to rest in a further $5\ \text{s}$. Find the distance from G to R and the average speed for the outward $15\ \text{s}$ journey.
Reveal Solution
Three stages using $\dfrac{1}{2}(u + v)t$: $\dfrac{1}{2}(0 + 2)(5) + 2 \times 5 + \dfrac{1}{2}(2 + 0)(5) = 5 + 10 + 5 = 20\ \text{m}$. Average speed $= \dfrac{20}{15} = 1.33\ \text{m s}^{-1}$ (3 s.f.).
In Exercise 1D Q9 the car's velocity–time graph was found to be: velocity rising uniformly from $0$ to $10\ \text{m s}^{-1}$ during the first $80\ \text{s}$, constant at $10\ \text{m s}^{-1}$ from $t = 80$ to $t = 560$, then falling uniformly to zero at $t = 600\ \text{s}$. Use $s = \dfrac{1}{2}(u + v)t$ for the two sloping stages to find (i) the total distance of the journey and (ii) the average velocity for the whole journey.
Reveal Solution
(i) $\dfrac{1}{2}(0 + 10)(80) + 10 \times 480 + \dfrac{1}{2}(10 + 0)(40) = 400 + 4800 + 200 = 5400\ \text{m}$. (ii) Average velocity $= \dfrac{5400}{600} = 9\ \text{m s}^{-1}$.
A lorry enters a busy road travelling at $20\ \text{m s}^{-1}$. Its speed–time graph is steep at first and flattens out as the traffic thins: over the first $10\ \text{s}$ the speed falls steadily from $20\ \text{m s}^{-1}$ to $8\ \text{m s}^{-1}$, and over the next $10\ \text{s}$ it falls from $8\ \text{m s}^{-1}$ to $6\ \text{m s}^{-1}$. Treating each $10\ \text{s}$ section as straight, (i) describe the journey, (ii) estimate the acceleration at the beginning and towards the end, and (iii) estimate the distance travelled and the average speed over the $20\ \text{s}$.
Reveal Solution
(i) The lorry is slowing down, gently at first and then more gently still as the road clears. (ii) Beginning: $a = \dfrac{8 - 20}{10} = -1.2\ \text{m s}^{-2}$; towards the end: $a = \dfrac{6 - 8}{10} = -0.2\ \text{m s}^{-2}$. (iii) $\dfrac{1}{2}(20 + 8)(10) + \dfrac{1}{2}(8 + 6)(10) = 140 + 70 = 210\ \text{m}$, so the average speed is $\dfrac{210}{20} = 10.5\ \text{m s}^{-1}$.
Return to the train of the worked example: it travels $20000\ \text{m}$ from A to B in $1000\ \text{s}$, with a $600\ \text{s}$ steady stage at $V = 25\ \text{m s}^{-1}$, and uniform acceleration from rest in the first stage. Given that this first-stage acceleration is $0.15\ \text{m s}^{-2}$, find the distance travelled during the third (braking) stage.
Reveal Solution
First stage: $t_1 = \dfrac{25}{0.15} = 166.7\ \text{s}$ (1 d.p.), so $s_1 = \dfrac{1}{2}(0 + 25)(166.7) = 2083\ \text{m}$ (nearest metre). The steady stage covers $600 \times 25 = 15000\ \text{m}$, leaving $20000 - 2083 - 15000 = 2917\ \text{m}$ for the third stage.
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