M2 1.3, M2 1.4 - Motion using calculus | Edexcel A-Level International Mathematics

Find velocity and acceleration from changing position, recover motion from acceleration and initial conditions, and interpret speed and direction in a plane. Calculus lets you describe motion when acceleration is not constant.

Differentiating position and velocity

Can a particle finish just four metres from its starting point after travelling twelve metres? Position records where it is; distance also records the journey, including any changes of direction.

Let $s$ be signed displacement from a fixed origin along a straight line, positive in a chosen direction. A displacement of $-2\,\mathrm{m}$ places the particle on the negative side of the origin. Velocity is the rate at which displacement changes. Acceleration is the rate at which velocity changes.

Motion derivatives

v=\frac{\mathrm{d}s}{\mathrm{d}t},\qquad a=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}^2s}{\mathrm{d}t^2}.

Speed is $|v|$ . Positive velocity means motion in the positive direction, even when the displacement is negative. These motion relationships are not separately printed in the formula booklet: learn them.

Differentiate the complete expression before substituting the time. If $s$ is in metres and $t$ in seconds, the derivatives have units $\mathrm{m}\,\mathrm{s}^{-1}$ and $\mathrm{m}\,\mathrm{s}^{-2}$ .

Worked example 1: velocity at an instant

A particle has displacement $s=t^3-6t^2+9t$ metres for $0\le t\le4$ , where $t$ is in seconds. Find its velocity and acceleration at $t=2$ .

Plan. Differentiate once for velocity and twice for acceleration.

\begin{aligned} v&=3t^2-12t+9 &&\text{differentiate displacement}\\ a&=6t-12 &&\text{differentiate velocity}\\ v(2)&=12-24+9=-3\\ a(2)&=12-12=0 \end{aligned}

Answer. Velocity $-3\,\mathrm{m}\,\mathrm{s}^{-1}$ , acceleration $0\,\mathrm{m}\,\mathrm{s}^{-2}$ . The particle is moving in the negative direction; it is not at rest.

Check. Writing $t=2+h$ gives $s=2-3h+h^3$ . The coefficient of the small time change is $-3$ , confirming the instantaneous velocity.

Keep the same motion, but change the requested measure from velocity to distance. Turning points now matter.

Worked example 2: distance after reversals

For the particle in Worked example 1, find the total distance travelled from $t=0$ to $t=4$ .

Plan. Find when velocity changes sign, then add the lengths of the separate journeys.

\begin{aligned} v&=3(t-1)(t-3)\\ v=0&\quad\Longrightarrow\quad t=1,\ 3\\ s(0)&=0,\quad s(1)=4,\quad s(3)=0,\quad s(4)=4\\ \text{distance}&=|4-0|+|0-4|+|4-0|=12 \end{aligned}

The velocity is positive before $1$ , negative between $1$ and $3$ , and positive afterwards, so both stationary instants are reversals.

Answer. $12\,\mathrm{m}$ , although the final displacement is $4\,\mathrm{m}$ .

Check. The particle travels from $0$ to $4$ , back to $0$ , then to $4$ : three journeys of four metres.

A zero of velocity need not be a reversal: for $v=(t-1)^2$ , velocity stays nonnegative on both sides of $1$ . Check the sign or the movement before splitting a journey.

Watch out

The constant-acceleration equations do not apply when $a$ depends on time. For this motion, $a=6t-12$ ; use calculus, not one constant value of acceleration over the journey.

Use the same turning-point method with fresh motion.

Integrating acceleration with initial conditions

Acceleration determines how velocity changes, but it does not tell you the velocity at the start. Each integration therefore introduces an unknown constant, fixed by a separate condition.

If $a=6t-8$ , then $v=3t^2-8t+C$ : changing $C$ shifts every velocity by the same amount without changing acceleration. Integrating again introduces a different constant for position.

Recovering motion

v=\int a\,\mathrm{d}t,\qquad s=\int v\,\mathrm{d}t.

Determine one constant from a known velocity and a second from a known position. These motion relationships are not separately printed in the formula booklet: learn them.

A condition at $t=1$ is just as useful as one at $t=0$ ; substitute the time actually given.

Worked example 3: conditions at zero

A particle has acceleration $a=6t-8\,\mathrm{m}\,\mathrm{s}^{-2}$ . At $t=0$ , its velocity is $3\,\mathrm{m}\,\mathrm{s}^{-1}$ and its displacement is $2\,\mathrm{m}$ . Find $v$ and $s$ in terms of $t$ .

Plan. Integrate twice, applying the relevant condition immediately after each integration.

\begin{aligned} v&=3t^2-8t+C\\ 3&=C &&\text{use }v(0)=3\\ s&=t^3-4t^2+3t+D\\ 2&=D &&\text{use }s(0)=2 \end{aligned}

Answer. $v=3t^2-8t+3$ and $s=t^3-4t^2+3t+2$ , in SI units.

Check. Differentiating $s$ gives the stated $v$ , then differentiating $v$ gives $6t-8$ ; both values at zero satisfy the data.

Change only the time at which the same velocity and displacement are specified, from zero to one second.

Worked example 4: conditions at a later time

The acceleration is again $a=6t-8\,\mathrm{m}\,\mathrm{s}^{-2}$ , but now $v(1)=3\,\mathrm{m}\,\mathrm{s}^{-1}$ and $s(1)=2\,\mathrm{m}$ . Find $v$ and $s$ .

Plan. Use $t=1$ in each integrated expression, retaining the constants until then.

\begin{aligned} v&=3t^2-8t+C\\ 3&=3-8+C\quad\Longrightarrow\quad C=8\\ s&=t^3-4t^2+8t+D\\ 2&=1-4+8+D\quad\Longrightarrow\quad D=-3 \end{aligned}

Answer. $v=3t^2-8t+8$ and $s=t^3-4t^2+8t-3$ , in SI units.

Check. Substitution gives $v(1)=3$ , $s(1)=2$ ; differentiating twice recovers the acceleration. The different conditions produce a different motion with the same acceleration.

The power rule also works with fractional and negative powers: for example, integrating $t^{-1/2}$ gives $2t^{1/2}+C$ , on $t>0$ . Keep any stated time domain; a formula involving $1/t$ cannot supply data at $t=0$ .

Watch out

“Initially at rest” fixes velocity, not position. “Initially at the origin” fixes position, not velocity. Neither condition fixes both integration constants by itself.

Apply the given conditions at one second.

Vector differentiation and integration

In a plane, the horizontal and vertical coordinates can change independently, so differentiate or integrate each component separately.

With fixed perpendicular unit vectors $\mathbf{i}$ and $\mathbf{j}$ , a position vector $\mathbf{r}=x\mathbf{i}+y\mathbf{j}$ locates a particle relative to a fixed origin. The vectors themselves do not vary with time.

Vector motion

\mathbf{v}=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}=\dot{\mathbf{r}},\qquad \mathbf{a}=\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=\ddot{\mathbf{r}}.

For integration, each component has its own constant. The dots mean differentiation with respect to time. These motion relationships are not separately printed in the formula booklet: learn them.

A condition such as $\mathbf{v}(0)=2\mathbf{i}-3\mathbf{j}$ is two scalar conditions, one for each component.

Worked example 5: differentiating a vector

A particle has position vector $\mathbf{r}=(t^2+2t)\mathbf{i}+(t^3-3t)\mathbf{j}$ metres. Find its velocity and acceleration at $t=2$ seconds.

Plan. Differentiate the two component functions before substituting the time.

\begin{aligned} \mathbf{v}&=(2t+2)\mathbf{i}+(3t^2-3)\mathbf{j}\\ \mathbf{a}&=2\mathbf{i}+6t\mathbf{j}\\ \mathbf{v}(2)&=6\mathbf{i}+9\mathbf{j}\\ \mathbf{a}(2)&=2\mathbf{i}+12\mathbf{j} \end{aligned}

Answer. Velocity $(6\mathbf{i}+9\mathbf{j})\,\mathrm{m}\,\mathrm{s}^{-1}$ ; acceleration $(2\mathbf{i}+12\mathbf{j})\,\mathrm{m}\,\mathrm{s}^{-2}$ .

Check. At $t=2+h$ , the coordinate changes start with $6h$ and $9h$ . These are the two instantaneous velocity components.

Keep the same motion but reverse the information: start with acceleration and recover position.

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M2 1.3, M2 1.4 - Motion using calculus

Differentiating position and velocity

Integrating acceleration with initial conditions

Vector differentiation and integration

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