Acceleration and velocity as functions of time

Highlights only0 highlighted

This first section opens Mechanics 3, the WME03 paper, which is an A Level unit and the last of the three mechanics units in the IAL course.

On the M3 paper, motion with a variable acceleration has appeared as one full question in every recent series, usually as the third question and worth about 9 marks.

Here the acceleration or velocity is given as a function of time $t$ ; the next section deals with an acceleration that depends on the displacement $x$ .

The idea: calculus does the work that suvat cannot

In M1 every acceleration was constant, and the five suvat formulae were enough.

In M2 the displacement, velocity or acceleration was given as a function of time, and you differentiated or integrated it.

M3 keeps those equations, sets them inside longer questions (velocities in pieces, greatest speeds, exact answers with $\ln$ terms), and in the next section adds an acceleration that depends on the displacement.

The velocity of a parachutist approaching a steady speed, of a train whose engine power fades, or of a ball moving through oil is not a polynomial: it involves exponentials, reciprocals, roots and trigonometric functions.

The calculus needed matches P1 to P4, so any function you can differentiate or integrate in the pure units can appear.

The whole topic rests on two definitions:

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

Going down the chain (displacement to velocity to acceleration) you differentiate.

Going up the chain you integrate, and each integration brings a constant that must be fixed from information in the question.

Key terms

Further kinematics (wording to learn): Kinematics of a particle moving in a straight line when the acceleration is a function of the displacement ( $x$ ), or time ( $t$ ). The setting up and solution of equations where $\dfrac{\mathrm{d}v}{\mathrm{d}t} = \mathrm{f}(t)$ , $v\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{f}(x)$ , $\dfrac{\mathrm{d}x}{\mathrm{d}t} = \mathrm{f}(x)$ or $\dfrac{\mathrm{d}x}{\mathrm{d}t} = \mathrm{f}(t)$ will be consistent with the level of calculus required in units P1, P2, P3 and P4.

Displacement $x$ : the position of the particle measured from a fixed origin $O$ on the line, positive in a stated direction.

Velocity $v$ : the rate of change of displacement, $\dfrac{\mathrm{d}x}{\mathrm{d}t}$ ; speed is its magnitude.

Acceleration $a$ : the rate of change of velocity, $\dfrac{\mathrm{d}v}{\mathrm{d}t}$ .

Initial conditions: the values of $x$ or $v$ at a stated time, used to find each constant of integration.

Why integration gives the velocity and the displacement

If $\dfrac{\mathrm{d}v}{\mathrm{d}t} = \mathrm{f}(t)$ , then $v$ is a function whose derivative is $\mathrm{f}(t)$ , which is exactly what an indefinite integral is: $v = \displaystyle\int \mathrm{f}(t)\,\mathrm{d}t$ .

Many functions have the same derivative, differing by a constant, and that is why the velocity is not fixed until you know its value at one instant.

The area idea from P1 and P2 gives the same result another way.

Over a short time $\delta t$ the velocity hardly changes, so the particle moves about $v\,\delta t$ .

Adding these small steps from $t = t_1$ to $t = t_2$ and letting $\delta t \to 0$ gives

x(t_2) - x(t_1) = \int_{t_1}^{t_2} v\,\mathrm{d}t,

the area under the velocity-time graph.

A definite integral needs no constant, because the constant cancels when you subtract.

Use the definite form when the question gives a starting position; use the indefinite form with a constant when you need $x$ as a function of $t$ .

Two kinds of equation in this section

Given

Write

Then

$a = \mathrm{f}(t)$

$\dfrac{\mathrm{d}v}{\mathrm{d}t} = \mathrm{f}(t)$

integrate to get $v$ , fix the constant from a known velocity

$v = \mathrm{f}(t)$

$\dfrac{\mathrm{d}x}{\mathrm{d}t} = \mathrm{f}(t)$

integrate to get $x$ , or use a definite integral for a displacement

Writing the first line of the table, the differential equation, is worth a method mark in itself.

Integrating with no differential equation shown can lose that mark, as the next section's exam tip explains.

The calculus you will need from P1 to P4

These are the integrals that M3 kinematics questions use most.

None of them is new; what is new is meeting them inside a mechanics story.

Powers, including negative and fractional ones: $\displaystyle\int (t + 2)^{-2}\,\mathrm{d}t = -(t + 2)^{-1} + C$ , $\displaystyle\int 4t^{1/2}\,\mathrm{d}t = \tfrac{8}{3}t^{3/2} + C$ .

The reciprocal: $\displaystyle\int \frac{k}{t}\,\mathrm{d}t = k\ln t + C$ for $t > 0$ . This produces the $\ln$ terms that M3 answers often contain.

Exponentials: $\displaystyle\int \mathrm{e}^{-kt}\,\mathrm{d}t = -\frac{1}{k}\mathrm{e}^{-kt} + C$ .

Trigonometric functions: $\displaystyle\int \sin kt\,\mathrm{d}t = -\frac{1}{k}\cos kt + C$ , with $t$ in radians.

Recognition and parts: $\displaystyle\int \frac{2t}{t^2 + 4}\,\mathrm{d}t = \ln(t^2 + 4) + C$ ; a product such as $t\mathrm{e}^{-t/2}$ needs integration by parts (P4).

Velocity in pieces: continuity

Some questions give the velocity by one formula for an early stage and a different formula afterwards, with an unknown constant in the second.

A particle cannot change its velocity in an instant unless it receives an impulse, so the two formulae must agree at the changeover time.

Setting them equal there, continuity, finds the constant.

The graph below shows a velocity of this kind: the curve changes formula at $t = 4$ but the velocity does not jump.

Velocity-time graph in two pieces. From t = 0 to t = 4 the curve v = 4√t rises from the origin to the point (4, 8). From t = 4 the curve v = 12 − 16/t carries on from the same point with no jump, passes through (8, 10) and climbs towards the dashed line v = 12 without reaching it.

You met continuity in M2 with polynomial pieces.

A velocity given in pieces is examined on M3 even though the syllabus does not name it: one recent M3 paper set exactly this structure, with a $\ln$ term in the displacement.

The examiner report on that paper noted that candidates who did not use continuity gave their answers in terms of the unknown constant.

Distance and displacement

Displacement is a signed area; distance is the total area with every piece counted as positive.

If $v$ changes sign during the interval, find where $v = 0$ , integrate over each piece separately, and add the sizes.

A single integral across a change of direction gives the displacement, which can be much less than the distance.

Connecting it

This builds on the M2 equations $\dfrac{\mathrm{d}x}{\mathrm{d}t} = \mathrm{f}(t)$ and $\dfrac{\mathrm{d}v}{\mathrm{d}t} = \mathrm{g}(t)$ , now examined in longer, less structured M3 questions.

The same calculus reappears in further dynamics, where a variable force gives the acceleration through Newton's second law, and in simple harmonic motion, whose velocity is a trigonometric function of time.

In physics and engineering, any motion with drag, a fading engine or a spring involves an acceleration that changes with time, and differential equations of exactly this kind describe it.

What the answers mean

A velocity such as $v = 10 - \dfrac{20}{t + 2}$ tells a story before you integrate it: at $t = 0$ it is $0$ , it always increases, and as $t$ grows it approaches $10\ \text{m s}^{-1}$ without ever reaching it.

A limiting value like this is a terminal velocity in the model.

When an answer contains $\ln 3$ or $\mathrm{e}^{-2}$ , leave it exact if the question asks for an exact value, and also convert it to a decimal for yourself to check it is a sensible size.

The marks on the M3 paper

How the marks work

M1 is for a correct method applied to the question, such as integrating with at least one power going up. A1 is for an accurate result and needs its M1. B1 is an independent mark for a stated value or fact.

DM1 counts only if the earlier M1 it depends on was earned. The June 2024 and January 2025 M3 schemes write it with a capital D and the June 2025 scheme writes dM1; these notes use DM1. A1* marks a printed answer: every step must be shown, with at least one line before the result, and the result written exactly as printed.

A1ft follows through from your own earlier wrong value. isw means a correct simplified answer keeps its marks even if wrong working follows. A misread costs two of the A or B marks in that part.

When a stem names the route ("using the work-energy principle", "using algebraic integration"), only that route scores.

Later in M3, in simple harmonic motion, $a$ stands for the amplitude, and the acceleration is written $\ddot{x}$ .

Exact or decimal?

Most M3 answers are exact: in letters such as $m$ , $g$ , $a$ and $l$ , or as surds, fractions and $\ln$ terms written as a single simplified expression.

Give an exact answer whenever the question says "exact" or the data are all exact.

Where $g = 9.8$ is used numerically, give the answer to 2 or 3 significant figures.

Where no $g$ appears and the question asks for a decimal, 3 significant figures is safe.

Method: acceleration or velocity as a function of $t$

1. Write the differential equation, $\dfrac{\mathrm{d}v}{\mathrm{d}t} = \mathrm{f}(t)$ or $\dfrac{\mathrm{d}x}{\mathrm{d}t} = \mathrm{f}(t)$ .

2. Integrate with respect to $t$ , including $+\,C$ .

3. Find $C$ from the condition the question gives, at the time it gives it. A particle can start at $t = 1$ , not $t = 0$ .

4. For a displacement over an interval, a definite integral avoids a second constant.

5. If the velocity changes sign, split the interval at $v = 0$ before finding a distance.

6. Read the answer back against the story: direction, size, limiting value.

Worked example 1: a velocity that approaches a limit

A particle $P$ moves along the positive $x$ -axis.

At time $t$ seconds, $t \geq 0$ , the acceleration of $P$ is $\dfrac{20}{(t + 2)^2}\ \text{m s}^{-2}$ in the positive $x$ direction.

When $t = 0$ , $P$ is at rest at the origin $O$ .

(a) Show that the velocity of $P$ at time $t$ is $\dfrac{10t}{t + 2}\ \text{m s}^{-1}$ . (4)
(b) Find the exact distance of $P$ from $O$ when $t = 4$ . (4)

1. (a) $\dfrac{\mathrm{d}v}{\mathrm{d}t} = 20(t + 2)^{-2}$ M1

2. $v = -20(t + 2)^{-1} + C$ M1 A1

3. $t = 0$ , $v = 0$ : $0 = -10 + C$ , so $C = 10$ and $v = 10 - \dfrac{20}{t + 2} = \dfrac{10(t + 2) - 20}{t + 2} = \dfrac{10t}{t + 2}$ A1*

4. (b) $x = \displaystyle\int_0^4 \left(10 - \frac{20}{t + 2}\right)\mathrm{d}t$ M1

5. $= \Big[10t - 20\ln(t + 2)\Big]_0^4$ A1

6. $= (40 - 20\ln 6) - (0 - 20\ln 2)$ DM1

7. $= 40 - 20\ln 3$ m A1

Answer: (b) $40 - 20\ln 3$ m, about $18.0$ m.

Line 3 shows the working between the constant and the printed form, which an A1* needs.

In (b) the velocity is never negative, so the displacement is also the distance.

As a check: $P$ cannot have gone further than $10 \times 4 = 40$ m, because its velocity is always below $10\ \text{m s}^{-1}$ , and the $\ln$ term is what it loses by starting from rest.

The velocity-time graph for the next example is the one drawn above.

Worked example 2: velocity in two pieces (exam style)

[In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.]

A particle $P$ starts from rest at the origin $O$ and moves along the positive $x$ -axis.

At time $t$ seconds the velocity of $P$ is $v\ \text{m s}^{-1}$ , where

v = 4\sqrt{t} \quad (0 \leq t < 4), \qquad v = 12 - \frac{k}{t} \quad (t \geq 4),

and $k$ is a constant.

The velocity of $P$ does not jump at $t = 4$ .

(a) Find the value of $k$ . (2)
(b) Find the acceleration of $P$ when $t = 8$ . (3)
(c) Find the exact value of $x$ when $t = 8$ . (5)

1. (a) At $t = 4$ : $4\sqrt{4} = 12 - \dfrac{k}{4}$ M1

2. $8 = 12 - \dfrac{k}{4}$ , so $k = 16$ A1

3. (b) For $t \geq 4$ : $a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{16}{t^2}$ M1 A1ft

4. At $t = 8$ : $a = \dfrac{16}{64} = \dfrac{1}{4}\ \text{m s}^{-2}$ A1

5. (c) $x = \displaystyle\int_0^4 4t^{1/2}\,\mathrm{d}t + \int_4^8 \left(12 - \frac{16}{t}\right)\mathrm{d}t$ M1

6. $\displaystyle\int_0^4 4t^{1/2}\,\mathrm{d}t = \Big[\tfrac{8}{3}t^{3/2}\Big]_0^4 = \frac{64}{3}$ A1

7. $\displaystyle\int_4^8 \left(12 - \frac{16}{t}\right)\mathrm{d}t = \Big[12t - 16\ln t\Big]_4^8 = 48 - 16\ln 2$ M1 A1

8. $x = \dfrac{64}{3} + 48 - 16\ln 2 = \dfrac{208}{3} - 16\ln 2$ m A1

Answer: (a) $k = 16$ ; (b) $\tfrac{1}{4}\ \text{m s}^{-2}$ ; (c) $\dfrac{208}{3} - 16\ln 2$ m, about $58.2$ m.

The banner means every integral must be shown with its limits; a value typed straight from a calculator's integration key would score nothing.

In (c) each piece is integrated over its own interval, and $\ln 8 - \ln 4$ is combined to $\ln 2$ so that the answer is a single simplified expression.

The velocity is positive throughout, so $x$ is also the distance travelled.

Worked example 3: greatest speed, then an exact displacement

A particle moves along a straight line.

At time $t$ seconds, $t \geq 0$ , its velocity is $v\ \text{m s}^{-1}$ , where $v = \dfrac{8t}{t^2 + 4}$ .

When $t = 0$ the particle is at the point $O$ .

(a) Using calculus, show that the greatest speed of the particle is $2\ \text{m s}^{-1}$ . (5)
(b) Find the exact distance of the particle from $O$ when $t = 4$ . (4)

1. (a) $\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{8(t^2 + 4) - 8t(2t)}{(t^2 + 4)^2}$ M1 A1

2. $= \dfrac{8(4 - t^2)}{(t^2 + 4)^2} = 0$ when $t = 2$ ( $t \geq 0$ ) DM1

3. $\dfrac{\mathrm{d}v}{\mathrm{d}t} > 0$ for $0 \leq t < 2$ and $< 0$ for $t > 2$ , so $t = 2$ gives the greatest value A1

4. $v = \dfrac{16}{8} = 2\ \text{m s}^{-1}$ A1*

5. (b) $x = \displaystyle\int_0^4 \frac{8t}{t^2 + 4}\,\mathrm{d}t$ M1

6. $= \Big[4\ln(t^2 + 4)\Big]_0^4$ M1 A1

7. $= 4\ln 20 - 4\ln 4 = 4\ln 5$ m A1

Answer: (b) $4\ln 5$ m, about $6.44$ m.

In (a), "greatest" needs a reason: the sign of the acceleration changes from positive to negative at $t = 2$ , so the particle speeds up and then slows down.

For all $t \geq 0$ , $v \geq 0$ , so the speed is $v$ itself.

In (b), the integrand is a multiple of $\dfrac{f'(t)}{f(t)}$ , recognised from P3, so no substitution is needed.

Worked example 4: a start at $t = 1$, not $t = 0$

For $t \geq 1$ , a particle $P$ moving on the $x$ -axis has acceleration $\left(2 + \dfrac{4}{t^2}\right)\ \text{m s}^{-2}$ in the positive $x$ direction.

When $t = 1$ , $P$ is at the origin and has velocity $-6\ \text{m s}^{-1}$ .

(a) Find $v$ in terms of $t$ . (3)
(b) Find the value of $t$ at which $P$ is instantaneously at rest. (3)

1. (a) $\dfrac{\mathrm{d}v}{\mathrm{d}t} = 2 + 4t^{-2}$ , so $v = 2t - 4t^{-1} + C$ M1 A1

2. $t = 1$ , $v = -6$ : $-6 = 2 - 4 + C$ , so $C = -4$ and $v = 2t - \dfrac{4}{t} - 4$ A1

3. (b) $2t - \dfrac{4}{t} - 4 = 0$ , so $2t^2 - 4t - 4 = 0$ , $t^2 - 2t - 2 = 0$ M1

4. $t = \dfrac{2 \pm \sqrt{4 + 8}}{2} = 1 \pm \sqrt{3}$ DM1

5. $t \geq 1$ , so $t = 1 + \sqrt{3}$ A1

Answer: (a) $v = 2t - \dfrac{4}{t} - 4$ ; (b) $t = 1 + \sqrt{3}$ , about $2.73$ .

The condition is given at $t = 1$ , and using $v = -6$ at $t = 0$ would give a different constant (and $\dfrac{4}{t}$ is not even defined at $t = 0$ ).

The negative root is rejected because the motion only starts at $t = 1$ .

Before $t = 1 + \sqrt{3}$ the particle moves in the negative direction; after it, the positive acceleration has turned it round.

Exam tip

When a calculus kinematics question carries the banner about calculator technology, show the derivative or the integral with its limits on the page, then substitute.

The M3 scheme for such a question gave no credit to a number without the calculus shown.

When a stem names the method ("Using calculus"), follow it: a graph or a table of values scores nothing.

Common mistakes

Using a condition at the wrong time. The tempting habit from M1 is "at $t = 0$ , $x = 0$ ". Fix: read the stem for the time it gives (such as $t = 1$ ) and substitute that.

Leaving the unknown constant in the answer. When the velocity is given in pieces, the constant comes from continuity at the changeover. Fix: set the two formulae equal at that time first.

Integrating a piecewise velocity in one go. Fix: integrate each formula over its own interval and add.

Displacement mistaken for distance. A single integral across a change of direction lets the backward motion cancel the forward motion. Fix: find where $v = 0$ and add the sizes of the separate areas.

Uncombined $\ln$ terms. Fix: use the laws of logarithms, so $16\ln 8 - 16\ln 4$ becomes $16\ln 2$ , and give a single simplified term where the question asks for an exact value.

Forgetting the constant of integration. Fix: write $+\,C$ on every indefinite integral and find it immediately from the given condition.

Check your understanding

1. A particle $P$ moves along the positive $x$ -axis. At time $t$ seconds, $t \geq 0$ , the acceleration of $P$ is $3\sqrt{t + 1}\ \text{m s}^{-2}$ in the positive $x$ direction. When $t = 0$ , $P$ is at rest at the origin $O$ .
(a) Find the velocity of $P$ in terms of $t$ . (3)
(b) Show that the distance of $P$ from $O$ when $t = 3$ is $18.8$ m. (4) Total [7]

2. Understanding check. A particle moves along a straight line from rest at a fixed point $O$ . At time $t$ seconds its velocity is $v\ \text{m s}^{-1}$ , where $v = 3t^2$ for $0 \leq t < 2$ and $v = k - \dfrac{24}{t^2}$ for $t \geq 2$ , and $k$ is a constant. The velocity is a continuous function of $t$ .
(a) Find the value of $k$ . (3)
(b) Find the distance travelled by the particle between $t = 0$ and $t = 6$ . (5)
(c) Show that the speed of the particle never reaches $18\ \text{m s}^{-1}$ , giving a reason in words. (3) Total [11]

3. A particle $P$ moves along the $x$ -axis. At time $t$ seconds, $t \geq 0$ , the velocity of $P$ is $4\sin\left(\dfrac{\pi t}{4}\right)\ \text{m s}^{-1}$ in the positive $x$ direction. When $t = 0$ , $P$ is at the origin $O$ .
(a) Find the exact acceleration of $P$ when $t = 1$ . (3)
(b) Find the exact total distance travelled by $P$ in the interval $0 \leq t \leq 6$ . (6)
(c) Show that $P$ is at the same point when $t = 2$ as when $t = 6$ . (3) Total [12]

4. For $t \geq 1$ , a particle $P$ moving along the $x$ -axis has acceleration $\dfrac{8}{t^3}\ \text{m s}^{-2}$ in the positive $x$ direction. When $t = 1$ , $P$ is at rest at the origin $O$ .
(a) Show that the velocity of $P$ is $4\left(1 - \dfrac{1}{t^2}\right)\ \text{m s}^{-1}$ . (3)
(b) Show that $x = \dfrac{4(t - 1)^2}{t}$ . (4)
(c) Find the value of $t$ when $P$ is $9$ m from $O$ . (3) Total [10]

5. [In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.] A particle moves along a straight line from a fixed point $O$ . At time $t$ seconds, $t \geq 0$ , its velocity is $v\ \text{m s}^{-1}$ , where $v = 20t\mathrm{e}^{-t/2}$ .
(a) Using calculus, show that the greatest speed of the particle is $\dfrac{40}{\mathrm{e}}\ \text{m s}^{-1}$ . (4)
(b) Find the exact distance travelled by the particle in the first $2$ seconds. (6) Total [10]

Answers

1. (a) $\dfrac{\mathrm{d}v}{\mathrm{d}t} = 3(t + 1)^{1/2}$ , $v = 2(t + 1)^{3/2} + C$ M1 A1; $t = 0$ , $v = 0$ gives $C = -2$ , so $v = 2(t + 1)^{3/2} - 2\ \text{m s}^{-1}$ A1. (b) $x = \displaystyle\int_0^3 \left(2(t + 1)^{3/2} - 2\right)\mathrm{d}t$ M1 $= \Big[\tfrac{4}{5}(t + 1)^{5/2} - 2t\Big]_0^3$ A1ft $= \left(\tfrac{128}{5} - 6\right) - \tfrac{4}{5}$ DM1 $= \tfrac{94}{5} = 18.8$ m A1*. The velocity is positive for $t > 0$ , so this is also the distance.

2. (a) At $t = 2$ : $3(4) = k - \dfrac{24}{4}$ M1 A1; $k = 18$ A1. (b) $\displaystyle\int_0^2 3t^2\,\mathrm{d}t = 8$ M1 A1; $\displaystyle\int_2^6 \left(18 - \frac{24}{t^2}\right)\mathrm{d}t = \Big[18t + \frac{24}{t}\Big]_2^6 = 112 - 48 = 64$ M1 A1; distance $= 72$ m A1. (c) For $t \geq 2$ , $v = 18 - \dfrac{24}{t^2}$ and $\dfrac{24}{t^2} > 0$ , so $v < 18$ M1; for $0 \leq t < 2$ , $v = 3t^2 < 12$ B1. So the speed is always below $18\ \text{m s}^{-1}$ : it approaches $18$ as $t$ grows but never reaches it A1*. The velocity is positive throughout, so the distance equals the displacement.

3. (a) $a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \pi\cos\left(\dfrac{\pi t}{4}\right)$ M1 A1; at $t = 1$ , $a = \dfrac{\pi\sqrt{2}}{2}\ \text{m s}^{-2}$ A1. (b) $v = 0$ when $\dfrac{\pi t}{4} = \pi$ , $t = 4$ B1; $\displaystyle\int_0^4 4\sin\left(\frac{\pi t}{4}\right)\mathrm{d}t = \Big[-\frac{16}{\pi}\cos\left(\frac{\pi t}{4}\right)\Big]_0^4 = \frac{32}{\pi}$ M1 A1; $\displaystyle\int_4^6 \ldots\,\mathrm{d}t = -\frac{16}{\pi}$ , size $\dfrac{16}{\pi}$ M1 A1; total distance $= \dfrac{48}{\pi}$ m A1. (c) $x = \dfrac{16}{\pi}\left(1 - \cos\dfrac{\pi t}{4}\right)$ M1; $t = 2$ : $x = \dfrac{16}{\pi}(1 - 0) = \dfrac{16}{\pi}$ ; $t = 6$ : $x = \dfrac{16}{\pi}\left(1 - \cos\dfrac{3\pi}{2}\right) = \dfrac{16}{\pi}$ A1; equal, so $P$ is at the same point: it passed that point moving forwards at $t = 2$ and returns to it moving backwards at $t = 6$ A1*.

4. (a) $\dfrac{\mathrm{d}v}{\mathrm{d}t} = 8t^{-3}$ , $v = -4t^{-2} + C$ M1 A1; $t = 1$ , $v = 0$ : $C = 4$ , so $v = 4 - \dfrac{4}{t^2} = 4\left(1 - \dfrac{1}{t^2}\right)$ A1*. (b) $x = 4t + \dfrac{4}{t} + D$ M1 A1; $t = 1$ , $x = 0$ : $D = -8$ M1; $x = \dfrac{4t^2 - 8t + 4}{t} = \dfrac{4(t - 1)^2}{t}$ A1*. (c) $4(t - 1)^2 = 9t$ , so $4t^2 - 17t + 4 = 0$ , $(4t - 1)(t - 4) = 0$ M1 A1; $t \geq 1$ , so $t = 4$ A1. The root $t = \tfrac{1}{4}$ is before the motion starts.

5. (a) $\dfrac{\mathrm{d}v}{\mathrm{d}t} = 20\mathrm{e}^{-t/2} - 10t\mathrm{e}^{-t/2} = 10\mathrm{e}^{-t/2}(2 - t)$ M1 A1; $= 0$ at $t = 2$ , positive before and negative after DM1; $v = 40\mathrm{e}^{-1} = \dfrac{40}{\mathrm{e}}\ \text{m s}^{-1}$ A1*. (b) $\displaystyle\int_0^2 20t\mathrm{e}^{-t/2}\,\mathrm{d}t$ by parts M1: $= \Big[-40t\mathrm{e}^{-t/2}\Big]_0^2 + \displaystyle\int_0^2 40\mathrm{e}^{-t/2}\,\mathrm{d}t$ A1 $= -80\mathrm{e}^{-1} + \Big[-80\mathrm{e}^{-t/2}\Big]_0^2$ M1 A1 $= -80\mathrm{e}^{-1} - 80\mathrm{e}^{-1} + 80$ DM1 $= 80 - \dfrac{160}{\mathrm{e}}$ m A1, about $21.1$ m. The velocity is never negative for $t \geq 0$ , so the displacement is the distance.

Acceleration and velocity as functions of time

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