When the Acceleration Dies Away
A particle moves with acceleration $a = 4 - 2t\ \text{m s}^{-2}$, starting at the origin with velocity $4\ \text{m s}^{-1}$. The acceleration is different at every instant, so no single value of $a$ can go into the constant-acceleration formulae. Instead the definitions are read backwards: $v = \int a\,dt$ and $x = \int v\,dt$. Scrub the time slider and watch the three curves — the velocity and the position are built by integration.
acceleration $a = 4 - 2t$ = 2.00 m s⁻² · velocity $v = 4 + 4t - t^{2}$ = 7.00 m s⁻¹ · position $x = 4t + 2t^{2} - \dfrac{t^{3}}{3}$ = 5.67 m
The amber line is the acceleration: it falls from $4$ to $0\ \text{m s}^{-2}$ at $t = 2\ \text{s}$. The emerald velocity keeps climbing while the acceleration is positive, and the indigo position is the area under the velocity curve.
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. Velocity is the rate of change of position, so:
R2. To find the velocity from the acceleration you must:
R3. $\int 4t\,dt$ equals:
Forwards: differentiate the position to get the velocity, differentiate the velocity to get the acceleration. Backwards: integrate.
| Relation | Read backwards |
|---|---|
| $v = \dfrac{dx}{dt}$ | $x = \int v\,dt$ |
| $a = \dfrac{dv}{dt}$ | $v = \int a\,dt$ |
Every integration brings a constant of integration, and every constant is fixed by an initial condition: the value of $v$ or $x$ at $t = 0$ (or at any other stated instant).
| Function | Integral |
|---|---|
| $\int k\,dt$ | $kt + c$ |
| $\int t^{n}\,dt$ | $\dfrac{t^{n+1}}{n+1} + c$ ($n \neq -1$) |
Example: $\int (6 - 2t)\,dt = 6t - t^{2} + c$. If the particle starts from rest, then $v(0) = 0$ gives $c = 0$, so $v = 6t - t^{2}$.
This is exactly why the SUVAT formulae fail here: they assume one fixed value of $a$ for the whole motion.
Take $a = 6 - 2t\ \text{m s}^{-2}$ with the particle starting from rest at the origin. Then $v = \int (6 - 2t)\,dt = 6t - t^{2}$, and $s = \int (6t - t^{2})\,dt = 3t^{2} - \dfrac{t^{3}}{3}$, both constants being zero.
At $t = 3\ \text{s}$: $v = 18 - 9 = 9\ \text{m s}^{-1}$ and $s = 27 - 9 = 18\ \text{m}$. The velocity is zero again when $t(6 - t) = 0$, that is at $t = 6\ \text{s}$.
Groups of 4: the stages below unlock one at a time — only the current stage can be answered. A wrong choice turns red, locks and shows a hint, so keep discussing within the group until the stage falls. A stage cleared on the first attempt scores; a stage cleared after hints still unlocks the next one. Compute each integral on paper before you tap.
Work alone. Two position functions are loaded — the default is the cubic x = t³ − 3t² − t + 3, and the second tab is x = 15t − 5t². Scrub the time slider and predict the readouts by differentiating mentally before you move it: the velocity and the acceleration shown are the derivatives of the position function. Then answer the three challenges.
An object moves along a straight line with acceleration $a = 10 + 3t - t^{2}\ \text{m s}^{-2}$. It is initially at the origin and at rest. Find expressions for the velocity $v$ (in $\text{m s}^{-1}$) and the position $x$ (in $\text{m}$) of the object at time $t$ seconds.
Integrate once for the velocity and again for the position. Each integration produces a constant; both constants here are zero, because the motion starts at the origin and at rest.
$v = \int (10 + 3t - t^{2})\,dt = 10t + \dfrac{3}{2}t^{2} - \dfrac{t^{3}}{3} + c$.
"At rest" means $v = 0$ at $t = 0$, so $c = 0$ and $v = 10t + \dfrac{3}{2}t^{2} - \dfrac{t^{3}}{3}$.
$x = \int \left(10t + \dfrac{3}{2}t^{2} - \dfrac{t^{3}}{3}\right) dt = 5t^{2} + \dfrac{1}{2}t^{3} - \dfrac{t^{4}}{12} + c_{2}$.
"At the origin" means $x = 0$ at $t = 0$, so $c_{2} = 0$ and $x = 5t^{2} + \dfrac{1}{2}t^{3} - \dfrac{t^{4}}{12}$.
$\dfrac{dx}{dt} = 10t + \dfrac{3}{2}t^{2} - \dfrac{t^{3}}{3}$ — exactly the velocity of Step 1. The chain is consistent.
An object moves along a straight line so that its acceleration (in $\text{m s}^{-2}$) is given by $a = 4 - 2t$. It starts its motion at the origin with speed $4\ \text{m s}^{-1}$ in the direction of increasing $x$. (i) Find as functions of $t$ the velocity and the position of the object. (ii) Describe the motion of the object, including its speed and the value of the acceleration at $t = 2\ \text{s}$.
This time the initial velocity is $4\ \text{m s}^{-1}$, not zero, so the first constant is $4$. The position constant is still zero. To describe the motion, watch when the acceleration changes sign — that is when the velocity stops rising.
$v = \int (4 - 2t)\,dt = 4t - t^{2} + c$, with $v(0) = 4$, so $c = 4$ and $v = 4 + 4t - t^{2}$.
$x = \int (4 + 4t - t^{2})\,dt = 4t + 2t^{2} - \dfrac{t^{3}}{3} + c_{2}$, with $x(0) = 0$, so $c_{2} = 0$ and $x = 4t + 2t^{2} - \dfrac{t^{3}}{3}$.
While $0 \leq t < 2$ the acceleration $4 - 2t$ is positive, so the object keeps speeding up. At $t = 2\ \text{s}$ the acceleration is momentarily zero and the velocity reaches its greatest value: $v = 8\ \text{m s}^{-1}$, with the object at $x = \dfrac{40}{3} = 13.3\ \text{m}$.
After $t = 2$ the acceleration is negative, so the object slows down; it comes to rest when $4 + 4t - t^{2} = 0$, that is at $t = 2 + 2\sqrt{2} \approx 4.8\ \text{s}$, and then moves back in the negative direction.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. For a particle moving along a line, the acceleration is:
Q2. The constant-acceleration formulae may NOT be used when:
Q3. Given $a = 6t\ \text{m s}^{-2}$, the velocity is:
Q4. A particle has $v = 4t + 3$ and starts at the origin. Its position at time $t$ is:
Work each on paper, then reveal the solution to check your working.
A particle has position $s = 10 + 2t - t^{2}$ metres at time $t$ seconds. (a) Find an expression for the velocity. (b) Write down the initial position and the initial velocity. (c) Find the time and the position when the velocity is zero.
Reveal Solution
(a) $v = \dfrac{ds}{dt} = 2 - 2t$. (b) $s(0) = 10\ \text{m}$ and $v(0) = 2\ \text{m s}^{-1}$. (c) $v = 0$ when $t = 1\ \text{s}$, and then $s = 10 + 2 - 1 = 11\ \text{m}$.
A particle has velocity $v = 4t + 3$ metres per second at time $t$ seconds. (a) Find an expression for the acceleration. (b) Write down the initial velocity and the initial acceleration.
Reveal Solution
(a) $a = \dfrac{dv}{dt} = 4\ \text{m s}^{-2}$ — constant. (b) $v(0) = 3\ \text{m s}^{-1}$ and $a(0) = 4\ \text{m s}^{-2}$. Because the acceleration is constant here, the SUVAT formulae would also work.
A particle has velocity $v = 4t + 3$ metres per second, and its initial position is zero. Find an expression for its position at time $t$.
Reveal Solution
$s = \int (4t + 3)\,dt = 2t^{2} + 3t + c$, and $s(0) = 0$ gives $c = 0$, so $s = 2t^{2} + 3t$.
The speed of a ball rolling down a hill is modelled by $v = 1.7t$ metres per second. (i) Describe the speed–time graph of the ball. (ii) How far does the ball travel in $10\ \text{s}$?
Reveal Solution
(i) The speed is proportional to the time, so the speed–time graph is a straight line through the origin, rising $1.7\ \text{m s}^{-1}$ every second. (ii) $s = \int_{0}^{10} 1.7t\,dt = \left[0.85t^{2}\right]_{0}^{10} = 85\ \text{m}$.
During braking the speed of a car is modelled by $v = 40 - 2t^{2}$ metres per second, until it stops moving. (i) How long does the car take to stop? (ii) How far does it move before it stops?
Reveal Solution
(i) The car stops when $v = 0$: $40 - 2t^{2} = 0$, so $t^{2} = 20$ and $t = \sqrt{20} = 4.47\ \text{s}$ (3 s.f.). (ii) $s = \int_{0}^{\sqrt{20}} (40 - 2t^{2})\,dt = \left[40t - \dfrac{2}{3}t^{3}\right]_{0}^{\sqrt{20}} = \dfrac{80}{3}\sqrt{20} = 119\ \text{m}$ (3 s.f.).
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