The suvat equations only work when the acceleration is constant. When displacement, velocity or acceleration is given as a function of time, you need calculus (DfE content area Q · Kinematics, using the differentiation and integration of content areas G and H).
$a \;\xrightarrow{\;\int \mathrm{d}t\;}\; v = \displaystyle\int a\,\mathrm{d}t \;\xrightarrow{\;\int \mathrm{d}t\;}\; s = \displaystyle\int v\,\mathrm{d}t$
A particle moves along a straight line so that its displacement from $O$ at time $t$ s is $s = t^3 - 6t^2 + 9t$ m, for $t \geqslant 0$. Find expressions for $v$ and $a$, and find when the particle is at rest.
For the particle in Example 1, find the total distance travelled in the first $4$ seconds.
A particle's displacement is $s = 2 + 12t - t^3$ m for $t \geqslant 0$. Find its greatest displacement from $O$ in the positive direction.
A particle has velocity $v = 8t - 2t^2$ m s$^{-1}$ for $0 \leqslant t \leqslant 4$. Find its maximum velocity.
Integrating introduces a constant of integration. You find it from an initial condition — usually the velocity or position at $t = 0$.
A particle has acceleration $a = (6t - 4)$ m s$^{-2}$. At $t = 0$ it is at $O$ with velocity $3$ m s$^{-1}$. Find $v$ and $s$ in terms of $t$, and its displacement when $t = 2$.
A particle has velocity $v = 12 - 3t^2$ m s$^{-1}$. Find (a) its displacement and (b) the distance it travels between $t = 0$ and $t = 3$.
Show that if $a$ is constant, with $v = u$ and $s = 0$ at $t = 0$, then $v = u + at$ and $s = ut + \tfrac12 at^2$.
Velocity
$v = \dfrac{\mathrm{d}s}{\mathrm{d}t}$
Acceleration
$a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}^2 s}{\mathrm{d}t^2}$
Velocity from $a$
$v = \int a\,\mathrm{d}t$ + use initial velocity.
Displacement from $v$
$s = \int v\,\mathrm{d}t$ + use initial position.
At rest
$v = 0$.
At the origin
$s = 0$.
Max/min displacement
$v = 0$; check with $\dfrac{\mathrm{d}^2s}{\mathrm{d}t^2}$.
Max/min velocity
$a = 0$; check with $\dfrac{\mathrm{d}a}{\mathrm{d}t}$.
Distance
Split the integral where $v = 0$; add the sizes.
In every question the particle moves in a straight line, $t$ is in seconds, $s$ in metres and $v$ in m s$^{-1}$.
$s = 2t^3 - 5t^2 + 4$. Find the velocity and acceleration when $t = 2$.
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$v = 6t^2 - 10t$, so $v(2) = 24 - 20 = 4$ m s$^{-1}$.
$a = 12t - 10$, so $a(2) = 14$ m s$^{-2}$.
$v = t^2 - 6t + 8$. Find the times when the particle is at rest and its acceleration at each of these times.
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$(t - 2)(t - 4) = 0$, so $t = 2$ s and $t = 4$ s.
$a = 2t - 6$: $a(2) = -2$ m s$^{-2}$, $a(4) = 2$ m s$^{-2}$.
$a = 4 - 2t$, and $v = 5$ when $t = 0$. Find $v$ in terms of $t$ and the maximum velocity.
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$v = 4t - t^2 + c$, with $c = 5$: $v = 5 + 4t - t^2$.
Maximum when $a = 0$: $t = 2$, $v = 5 + 8 - 4 = 9$ m s$^{-1}$ (a maximum since $\tfrac{\mathrm{d}a}{\mathrm{d}t} = -2 < 0$).
$v = 6t - 3t^2$, and the particle starts at $O$. Find $s$ in terms of $t$, and the time when it returns to $O$.
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$s = 3t^2 - t^3 + k$, with $k = 0$: $s = 3t^2 - t^3 = t^2(3 - t)$.
$s = 0$ when $t = 0$ or $t = 3$. It returns to $O$ at $t = 3$ s.
$v = 3t^2 + 2t$. Find the displacement between $t = 1$ and $t = 3$.
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$\displaystyle\int_1^3 (3t^2 + 2t)\,\mathrm{d}t = \Big[t^3 + t^2\Big]_1^3 = 36 - 2 = 34$ m.
($v > 0$ throughout, so this is also the distance.)
$v = 4t - t^2$. Find the displacement and the distance travelled in the interval $0 \leqslant t \leqslant 6$.
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$\int v\,\mathrm{d}t = 2t^2 - \tfrac13 t^3$. $v = 0$ at $t = 4$.
$\displaystyle\int_0^4 = 32 - \tfrac{64}{3} = \tfrac{32}{3}$. $\displaystyle\int_4^6 = (72 - 72) - \tfrac{32}{3} = -\tfrac{32}{3}$.
Displacement $= 0$ m (back at the start). Distance $= \tfrac{64}{3} = 21.3$ m (3 s.f.).
$s = t^3 - 9t^2 + 24t$. Find the distance travelled in the first $5$ seconds.
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$v = 3t^2 - 18t + 24 = 3(t - 2)(t - 4)$, zero at $t = 2$ and $t = 4$.
$s(0) = 0$, $s(2) = 8 - 36 + 48 = 20$, $s(4) = 64 - 144 + 96 = 16$, $s(5) = 125 - 225 + 120 = 20$.
Distance $= 20 + 4 + 4 = 28$ m.
$v = t^3 - 6t^2 + 9t + 2$ for $t \geqslant 0$. Find the minimum velocity for $t > 0$, and explain why the particle is never at rest.
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$a = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$, zero at $t = 1$ and $t = 3$.
$\tfrac{\mathrm{d}a}{\mathrm{d}t} = 6t - 12$: negative at $t = 1$ (maximum), positive at $t = 3$ (minimum).
$v(3) = 27 - 54 + 27 + 2 = 2$ m s$^{-1}$.
Also $v(0) = 2$, and $v$ increases for $t > 3$. So $v \geqslant 2 > 0$ for all $t \geqslant 0$: the particle is never at rest.
A particle starts from rest at $O$. For $0 \leqslant t \leqslant 2$ its velocity is $v = 3t^2$; after $t = 2$ it moves with constant velocity. Find its distance from $O$ when $t = 5$.
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First stage: $\displaystyle\int_0^2 3t^2\,\mathrm{d}t = \Big[t^3\Big]_0^2 = 8$ m. At $t = 2$, $v = 12$ m s$^{-1}$.
Second stage: $12 \times 3 = 36$ m.
Distance from $O$ $= 8 + 36 = 44$ m.
A particle $P$ starts from rest at $O$ and moves in a straight line with acceleration $a = (12 - 6t)$ m s$^{-2}$.
(a) Find $v$ in terms of $t$. (b) Find the maximum velocity. (c) Find when $P$ next comes to rest, and its distance from $O$ then. (d) Find when $P$ returns to $O$, and its speed at that moment.
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(a) $v = 12t - 3t^2 + c$; $v = 0$ at $t = 0$, so $v = 12t - 3t^2$.
(b) $a = 0$ at $t = 2$; $v(2) = 24 - 12 = 12$ m s$^{-1}$ (maximum, since $\tfrac{\mathrm{d}a}{\mathrm{d}t} = -6 < 0$).
(c) $3t(4 - t) = 0 \Rightarrow t = 4$ s.
$s = 6t^2 - t^3$ (constant zero since $s = 0$ at $t = 0$), so $s(4) = 96 - 64 = 32$ m.
(d) $s = t^2(6 - t) = 0 \Rightarrow t = 6$ s.
$v(6) = 72 - 108 = -36$, so the speed is $36$ m s$^{-1}$ (moving back towards and through $O$).