Velocity and acceleration

At rest, direction, speed

In 19 of 20 recent exams
2% of the marks
Paper 1 or 2
1 to 5 marks

How it is set in the exam

Paper 2 (calculator) mostly (15 of 19 exam versions), Paper 1 (no calculator) in 4; 1–5 marks per part, usually 2, and nearly always the first parts of a kinematics question. Very common (19 of 20 exam versions). The velocity is given in 15 of them, the displacement in 4.

Spot it in the exam

Exam phrases to spot (6)

“find the velocity of the particle when $t = 2$ ”
“find an expression for the acceleration”
“find the times when the particle is at rest”
“find the value of $t$ at which the particle changes direction”
“find the acceleration when the particle changes direction”
“find the values of $t$ for which the particle is moving in the positive direction”

In disguise (8)

only $s(t)$ is given: differentiate first. "At rest" is where the graph of $s$ is flat ( $v = s'(t) = 0$ ). It is a maximum or minimum of $s$ only if the particle also turns; markschemes accept the minimum of $s$ when it does
"passes through $O$ " or "is at $O$ ", with $s(t)$ given: $s(t) = 0$ . If it starts at $O$ , $t = 0$ is a root, but not an answer when the question asks when it passes through $O$ or is next (or again) at $O$ : keeping it there scores
"moving towards $O$ " or "away from $O$ ": the signs of $s$ and $v$
"speeding up" or "slowing down": the signs of $v$ and $a$
a given speed, such as "the speed is $4$ m s $^{-1}$ ": $v = 4$ or $v = -4$
"sketch the graph of $v$ for $0 \le t \le 3$ ": the marks are for the zeros of $v$ , its shape, its turning point and its ends
two particles "have the same velocity": $v_A(t) = v_B(t)$
a velocity–time graph, with the acceleration or the turns asked for: the gradient of the graph, and where it crosses the axis (its areas belong to Displacement, position and distance travelled)

Variants

$v$ given ( exact, on the GDC); $s$ given (differentiate first); $a(t)$ asked for as an expression (product or chain rule); the velocity or the acceleration equal to a given non-zero value (every solution in the interval, and no others)

Looks like it, but isn't (5)

the greatest or least velocity, speed, acceleration or displacement: Greatest or least velocity, speed, acceleration or displacement on an interval
a displacement, a distance travelled or a position from $v$ or $a$ , or "at $O$ " when only $v$ or $a$ is given: Displacement, position and distance travelled
a velocity from a given acceleration (integrate up, with $v(0)$ ): Displacement, position and distance travelled
the velocity in the long run: a limit, Rate of change in context
the rate of change of a quantity that isn't a position (a population, a temperature): Rate of change in context

Worked example

At rest twice, but only one turn

Paper 1 (no calculator)

Question

A particle moves in a straight line. Its velocity $v$ m s $^{-1}$ at time $t$ seconds is $v(t) = (t - 2)^2(t - 5)$ , for $0 \le t \le 6$ .
(a) Find the times at which the particle is at rest. [2]
(b) Find the value of $t$ at which the particle changes direction. Justify your answer. [3]
(c) Show that the acceleration of the particle is $a(t) = 3(t - 2)(t - 4)$ . [3]
(d) Find the values of $t$ for which the particle is speeding up. [4]

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See it move: A particle and its velocity graph

$t = 1.00$ : $v = -4.0$ , $a = 9.0$ , $s = -10.8$ . Slowing down: $v$ is negative and $a$ is positive.
$v(0) = -20$ : the particle sets off backwards.
At $t = 2$ the graph touches 0 and goes back down: the particle stops for an instant and carries on the same way.
At $t = 5$ the graph crosses 0: the particle turns.
Speeding up where $v$ and $a$ share a sign: $2 < t < 4$ and $5 < t \le 6$ .

Steps

Given what?
$s$ , $v$ or $a$ . Differentiate down to what you need: $v = s'(t)$ , $a = v'(t)$ . On a value at one time can come from the GDC (gradient at a point). A condition on $a$ over an interval () can be read from the graph of the derivative of $v$ .
Translate the wording
into a condition, with the table in From exam wording to a condition.
Solve on the given interval only.
: factorise, or solve exactly. : zeros and intersections, with the equation and the interval written down. Give every solution in the interval, and no others.
Sign questions
(direction, towards or away from $O$ , speeding up): a sign table, cut at the zeros of every function involved ( $s$ and $v$ , or $v$ and $a$ ).
Answer with units:
m s $^{-1}$ for $v$ , m s $^{-2}$ for $a$ , and intervals as inequalities in $t$ . At a zero of $v$ or $a$ , strict and non-strict inequalities are both accepted; a time outside the interval never is.

Common mistakes

Extra times
a zero outside the interval (or at $t < 0$ ) left in the answer scores , even with the right times there too.
The start left in the answer
when the particle starts at $O$ , $s(t) = 0$ also gives $t = 0$ . For "when is it at $O$ (again)", $t = 0$ is the start: leaving it in the answer scores (Example 1.4(c)).
Every zero of $v$ taken as a change of direction
a squared factor touches $0$ without changing sign (Example 1.1).

More mistakes (6)

"At rest" found only as a maximum or minimum of $s$
a flat point of the graph of $s$ that is not a maximum or minimum is still a moment at rest (Example 1.1, $t = 2$ : $v$ touches $0$ there without changing sign, so the graph of $s$ has a stationary point of inflexion there, not a turning point).
"Towards $O$ " as $v < 0$
, and "away" as $v > 0$ : wrong whenever $s < 0$ (Example 1.2, $7 < t \le 8$ ).
"Speeding up" as $a > 0$
it is $v$ and $a$ with the same sign (Example 1.1, $2 < t < 4$ ).
Speed $k$ as $v = k$ only
the other time is $v = -k$ (Example 1.4(e)).
$0$ for the acceleration when the particle turns
it is $v$ that is $0$ there. The acceleration at that time is $v'(t)$ (Example 1.3(b)).
Degree mode
on with trig: ().

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