From the Lights: v = at and s = ½at²
A car waits at the lights and pulls away with a constant acceleration $a$. After $5\ \text{s}$ its speed is $v = at$ and it has covered $s = \dfrac{1}{2}at^{2}$. Drag the acceleration slider: the line on the left is the car's velocity–time graph, and the little car on the right sits at exactly the shaded area of that graph — whatever $a$ is.
speed after $5\ \text{s}$: $v = at$ = 7.50 m s⁻¹ · distance covered: $s = \dfrac{1}{2}at^{2}$ = 18.75 m
Speed grows linearly with time, distance grows with the square of time. Every constant-acceleration problem today is a matter of writing down the five letters and choosing the one formula that fits the numbers you are given.
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. The gradient of a velocity–time graph gives:
R2. The area between a velocity–time graph and the time axis gives:
R3. In a calculation, a deceleration of $2\ \text{m s}^{-2}$ is entered as:
Every constant-acceleration problem is built from five letters: $u$ (initial velocity), $v$ (final velocity), $a$ (acceleration), $s$ (displacement) and $t$ (time). Each formula uses four of them and leaves exactly one out — that missing letter is the key to choosing the formula.
| Formula | Uses | Leaves out |
|---|---|---|
| $v = u + at$ | $u, a, t$ | $s$ |
| $s = ut + \dfrac{1}{2}at^{2}$ | $u, a, t$ | $v$ |
| $v^{2} = u^{2} + 2as$ | $u, a, s$ | $t$ |
| $s = \dfrac{1}{2}(u + v)t$ | $u, v, t$ | $a$ |
Each formula is the straight-line velocity–time picture in disguise: $v = u + at$ is the height of the line, and the other three are the trapezium area $\dfrac{1}{2}(u + v)t$ rearranged.
Do not reach for the formula you remember best. Write down the five letters and mark each one known or unknown.
Write $u, v, a, s, t$ and put a value next to every letter the question gives you. Deceleration goes in as a negative $a$.
The right formula contains every known letter and the letter you want — so the one letter it leaves out is the one the question never mentions.
A van travelling at $20\ \text{m s}^{-1}$ brakes to rest in $40\ \text{m}$. Known: $u = 20$, $v = 0$, $s = 40$; wanted: $a$; no $t$ is mentioned, so use $v^{2} = u^{2} + 2as$: $0 = 400 + 80a$, giving $a = -5\ \text{m s}^{-2}$ — a deceleration of $5\ \text{m s}^{-2}$.
Two phrases appear in almost every question, and each one removes a letter before you start.
| Phrase | Means | Formulae in their simplest dress |
|---|---|---|
| starting from rest | $u = 0$ | $v = at$, $s = \dfrac{1}{2}at^{2}$, $v^{2} = 2as$ |
| coming to rest | $v = 0$ | $0 = u + at$, $u^{2} = 2as$ |
Mini-example: a trolley from rest with $a = 3\ \text{m s}^{-2}$ for $6\ \text{s}$ reaches $v = 3 \times 6 = 18\ \text{m s}^{-1}$ and covers $s = \dfrac{1}{2} \times 3 \times 36 = 54\ \text{m}$. The third formula checks it: $v^{2} = 2 \times 3 \times 54 = 324$, so $v = 18$ again.
Groups of 4: each card lists what is given and what is wanted. Tap a card, then tap the formula zone that solves the problem — two cards belong to each formula. When all eight cards are filed, press Check cards. If a card goes red, talk over why that formula cannot be the one, then move it and check again.
Work alone. Four pieces of student working are shown. Decide for each one whether it is correct or contains an error, then read the explanation. Two are sound and two hide a classic slip — look hard at every ½ and every 2.
A car is moving at $20\ \text{m s}^{-1}$ when the driver begins to increase speed. Every $10\ \text{s}$ the car gains $5\ \text{m s}^{-1}$, until it reaches its maximum speed of $50\ \text{m s}^{-1}$, which it then keeps. (i) Find the acceleration while the car is speeding up. (ii) How long after it begins to speed up does the car reach $50\ \text{m s}^{-1}$? (iii) Find the distance travelled by the car in the first $150\ \text{s}$. (iv) Write down an expression for the speed of the car $t$ seconds after it begins to speed up.
An extra $5\ \text{m s}^{-1}$ every $10\ \text{s}$ is a constant acceleration of $0.5\ \text{m s}^{-2}$. Find the time to reach the maximum speed first, then split the $150\ \text{s}$ into the speeding-up stage and the steady stage.
Gaining $5\ \text{m s}^{-1}$ every $10\ \text{s}$ means $a = \dfrac{5}{10} = 0.5\ \text{m s}^{-2}$.
Use $v = u + at$ with $u = 20$, $v = 50$ and $a = 0.5$: $50 = 20 + 0.5t$, so $t = 60\ \text{s}$.
First $60\ \text{s}$: $s = ut + \dfrac{1}{2}at^{2} = 20(60) + \dfrac{1}{2}(0.5)(3600) = 1200 + 900 = 2100\ \text{m}$.
Remaining $150 - 60 = 90\ \text{s}$ at $50\ \text{m s}^{-1}$: $90 \times 50 = 4500\ \text{m}$.
Total $= 2100 + 4500 = 6600\ \text{m}$.
For $0 \leq t \leq 60$ the speed obeys $v = u + at$, so $v = 20 + 0.5t$ for $0 \leq t \leq 60$; after that the speed is a constant $50\ \text{m s}^{-1}$.
A car is travelling at $36\ \text{km h}^{-1}$ when the driver has to make an emergency stop. During the time the driver takes to appreciate the situation and apply the brakes the car travels $7\ \text{m}$ (the 'thinking distance'). It then pulls up with constant deceleration in a further $8\ \text{m}$ (the 'braking distance'), a total stopping distance of $15\ \text{m}$. (i) Find the initial speed of the car in metres per second and the driver's reaction time. (ii) Sketch the velocity–time graph for the car. (iii) Calculate the deceleration once the car starts braking.
Convert $36\ \text{km h}^{-1}$ into $\text{m s}^{-1}$ first. During the reaction time the speed does not change, so the thinking distance is steady speed × reaction time. For the braking stage, $t$ is not known — that is the signal to reach for $v^{2} = u^{2} + 2as$.
$36\ \text{km h}^{-1} = \dfrac{36}{3.6} = 10\ \text{m s}^{-1}$.
The car travels $7\ \text{m}$ at a steady $10\ \text{m s}^{-1}$, so $t = \dfrac{7}{10} = 0.7\ \text{s}$.
The velocity–time graph is a horizontal line at $10\ \text{m s}^{-1}$ for $0.7\ \text{s}$, then a straight line falling to zero. The first strip is the rectangle $10 \times 0.7 = 7\ \text{m}$; the triangle under the falling line has area $8\ \text{m}$. The braking takes $\dfrac{2 \times 8}{10} = 1.6\ \text{s}$.
Use $v^{2} = u^{2} + 2as$ with $v = 0$, $u = 10$ and $s = 8$: $0 = 100 + 16a$, so $a = -6.25\ \text{m s}^{-2}$ — a deceleration of $6.25\ \text{m s}^{-2}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. The formula $v^{2} = u^{2} + 2as$ does not contain:
Q2. A body that starts from rest has:
Q3. A car moving at $12\ \text{m s}^{-1}$ accelerates at $2\ \text{m s}^{-2}$ for $5\ \text{s}$. Its speed is now:
Q4. For motion with constant acceleration, the velocity–time graph is always:
Work each on paper, then reveal the solution to check your working.
A parachutist jumps from a helicopter hovering above an airfield. Her speed increases at a constant rate to $28\ \text{m s}^{-1}$ in the first $3\ \text{s}$ of her fall. (i) Find her acceleration during these $3\ \text{s}$. (ii) Find the distance she falls in this time.
Reveal Solution
(i) She starts from rest, so $u = 0$. Using $v = u + at$: $a = \dfrac{28 - 0}{3} = 9.33\ \text{m s}^{-2}$ (2 d.p.). (ii) Using $s = ut + \dfrac{1}{2}at^{2}$: $s = \dfrac{1}{2} \times 9.33 \times 9 = 42\ \text{m}$ — the same answer as the trapezium area $\dfrac{1}{2}(0 + 28)(3)$.
Continuing the parachutist's fall: after the first $3\ \text{s}$ her speed decreases uniformly from $28\ \text{m s}^{-1}$ to $8\ \text{m s}^{-1}$ over the next $6\ \text{s}$. (i) Find her deceleration during these $6\ \text{s}$. (ii) Find the distance she falls while slowing down.
Reveal Solution
(i) $a = \dfrac{v - u}{t} = \dfrac{8 - 28}{6} = -3.33\ \text{m s}^{-2}$ (2 d.p.), a deceleration of $3.33\ \text{m s}^{-2}$. (ii) The average speed is $\dfrac{28 + 8}{2} = 18\ \text{m s}^{-1}$, so $s = 18 \times 6 = 108\ \text{m}$.
The complete parachute jump takes $1\ \text{minute}$: the first $3\ \text{s}$ speeding up to $28\ \text{m s}^{-1}$, the next $6\ \text{s}$ slowing to $8\ \text{m s}^{-1}$, and after that a steady $8\ \text{m s}^{-1}$ until she reaches the ground. Find the height of the helicopter.
Reveal Solution
The three stages cover $42\ \text{m}$ and $108\ \text{m}$, and the steady stage lasts $60 - 3 - 6 = 51\ \text{s}$, giving $8 \times 51 = 408\ \text{m}$. Total height $= 42 + 108 + 408 = 558\ \text{m}$.
A train takes $10\ \text{minutes}$ to travel between two stations. It accelerates at $0.5\ \text{m s}^{-2}$ for $30\ \text{s}$, then travels at a constant speed, and is finally brought to rest in $15\ \text{s}$ with a constant deceleration. Find (i) the steady speed, (ii) the rate of deceleration and (iii) the distance between the two stations.
Reveal Solution
(i) $v = u + at = 0 + 0.5 \times 30 = 15\ \text{m s}^{-1}$. (ii) The speed falls from $15\ \text{m s}^{-1}$ to zero in $15\ \text{s}$, so the deceleration is $\dfrac{15}{15} = 1\ \text{m s}^{-2}$. (iii) The steady stage lasts $600 - 30 - 15 = 555\ \text{s}$. Areas under the graph: $\dfrac{1}{2} \times 30 \times 15 + 555 \times 15 + \dfrac{1}{2} \times 15 \times 15 = 225 + 8325 + 112.5 = 8662.5\ \text{m}$.
A second car travels at $60\ \text{km h}^{-1}$. Its driver has the same reaction time of $0.7\ \text{s}$ and the brakes give the same deceleration of $6.25\ \text{m s}^{-2}$ as in the worked example. Find the total stopping distance, and comment on why it grows faster than the speed itself — the speed rises by a factor of $\dfrac{5}{3}$.
Reveal Solution
$60\ \text{km h}^{-1} = \dfrac{60}{3.6} = 16.67\ \text{m s}^{-1}$ (2 d.p.). Thinking distance $= 16.67 \times 0.7 = 11.67\ \text{m}$. Braking distance from $u^{2} = 2as$: $s = \dfrac{16.67^{2}}{2 \times 6.25} = \dfrac{277.8}{12.5} = 22.22\ \text{m}$. Total $= 11.67 + 22.22 = 33.9\ \text{m}$ (3 s.f.). The speed is larger by a factor $\dfrac{5}{3}$, so the braking distance is larger by $\left(\dfrac{5}{3}\right)^{2} = \dfrac{25}{9}$ — almost three times — while the thinking distance grows only by the factor $\dfrac{5}{3}$.
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