Displacement, position and distance travelled

In 17 of 20 recent exams
3% of the marks
Paper 1 or 2
1 to 7 marks

How it is set in the exam

Paper 2 (calculator) mostly (12 of 17 exam versions), Paper 1 (no calculator) in 5; 1–7 marks per part, often the longest parts of the question. Very common (17 of 20 exam versions). On Paper 2 (calculator) the distance is $\int \lvert v \rvert$ on the GDC (7 versions); on Paper 1 (no calculator) it is split at the zeros of $v$ by hand. In 4 versions the displacement $s(t)$ is given, and the distance comes from its turning points.

Spot it in the exam

Exam phrases to spot (5)

“find the displacement of the particle from $t = 1$ to $t = 4$ ”
“find the total distance travelled in the first 5 seconds”
“find an expression for $s(t)$ ”
“does the particle return to its starting point?”
“find the time at which the particles meet”

In disguise (6)

the acceleration is given, with an initial velocity: integrate twice, with one constant each time
a velocity–time graph made of straight lines: the areas of triangles, rectangles and trapezia
"during the fifth second": $4 \le t \le 5$
"passes through $O$ " when only $v$ (or $a$ ) is given: find $s$ first ( $+c$ and $s(0)$ ), then solve $s = 0$
two particles: "when do they meet?", "the distance between them": both positions measured from the same $O$ , when only $v(t)$ or $a(t)$ is given (with both positions given as formulas, Solve graphically)
a distance travelled is given and the time is unknown: the time is the upper limit, $\int_0^T \lvert v \rvert\,dt = D$

Variants

a value of the displacement read from a given $s(t)$ , such as the initial displacement $s(0)$ or the position at a given time (a 1–2 mark part: substitute); $v$ changes sign inside the interval (split, or $\lvert v \rvert$ on the GDC); $s(t)$ given, so there is nothing to integrate: the distance is the sum of the sizes of the changes in $s$ between the start, the times where $v = 0$ (the turning points of $s$ ) and the end, or $\int \lvert s'(t) \rvert\,dt$ on the GDC; a periodic velocity (the same distance in every period); the $+c$ route or $s(0) + \int_0^t v$

Looks like it, but isn't (6)

"at $O$ " when $s(t)$ is given: solve $s(t) = 0$ directly, Velocity and acceleration: at rest, direction, speed
an amount that isn't a position, from a given rate: Total change from a rate (a change is signed; a distance never is:
an area between a curve and the $x$ -axis, with no motion: Area between a curve and the x-axis
the greatest displacement, or the greatest distance from $O$ : Greatest or least velocity, speed, acceleration or displacement on an interval
$f$ from $f'$ and a point, with no motion: Find f from f′ and a point
both positions are already given as functions of $t$ , so nothing is integrated, and the question is when the objects meet: set the positions equal, or find the zeros of the gap, Solve graphically: intersections, how many solutions, above or below a level

Worked example

long Section B question: split, return, and the distance while speeding up

Paper 1 (no calculator)

Question

A particle starts at $O$ and moves in a straight line with velocity $v(t) = 3t^2 - 15t + 12$ m s $^{-1}$ , for $0 \le t \le 5$ .
(a) Find the displacement of the particle from $O$ when $t = 5$ . [3]
(b) Find the total distance travelled by the particle. [5]
(c) Does the particle ever return to its starting point after it sets off? Give a reason for your answer. [3]
(d) Find the total distance travelled while the particle is speeding up. [5]

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See it move: Area above counts forwards, area below counts back

$t = 0.60$ : $s = 4.72$ m from $O$ , and the distance so far is 4.72 m. Moving forwards.
Area above the axis counts forwards and area below counts back, so $s(t)$ is one integral: $s(5) = -\frac{5}{2}$ m.
The distance adds the sizes: $v = 0$ at $t = 1$ and $t = 4$ splits it into $\frac{11}{2} + \frac{27}{2} + \frac{11}{2} = \frac{49}{2}$ m.
Back at $O$ at $t \approx 2.31$ , though $s(5) \ne 0$ : a net displacement of 0 is the wrong test.
Speeding up where $v$ and $a$ share a sign, $1 < t < \frac{5}{2}$ and $4 < t \le 5$ : $\frac{49}{4}$ m of the journey.

Which method?

a distance, $v$ changes sign
(by hand)
find the zeros of $v$ ; add the sizes of the changes in $s$ between consecutive times: the start, each zero, the end
(GDC)
$\int_{t_1}^{t_2} \lvert v(t) \rvert\,dt$ , written in full
a distance, $v$ doesn't change sign
(by hand)
$\lvert s(t_2) - s(t_1) \rvert$
(GDC)
the same, or $\int_{t_1}^{t_2} \lvert v(t) \rvert\,dt$
a distance, $s(t)$ given
(by hand)
add the sizes of the changes in $s$ between consecutive times: the start, each turning point of $s$ , the end
(GDC)
the same, with the turning points from maximum or minimum of a graph; or $\int_{t_1}^{t_2} \lvert s'(t) \rvert\,dt$
the position at time $t$
(by hand)
$\int v\,dt$ with $+c$ , then the starting position; or $s(t) = s(0) + \int_0^t v(u)\,du$ (the letter inside the integral is $u$ only so that it isn't confused with the upper limit $t$ )
(GDC)
$s(0) + \int_0^{T} v\,dt$ for the position at a time $T$ ; a formula still needs the algebra
a $v$ – $t$ graph of straight lines
(by hand)
areas of shapes, counted negative below the axis for a displacement
(GDC)
the same
the time to travel a distance $D$
(by hand)
$\lvert s(T) - s(0) \rvert = D$ if $v$ doesn't change sign; if it does, add the distance up to each turn, then solve for the rest: $\lvert s(T) - s(t_{\text{turn}}) \rvert = D - (\text{distance so far})$ , or triangle areas on a straight-line $v$ – $t$ graph
(GDC)
$\int_0^T \lvert v(t) \rvert\,dt = D$ (definite integral with an unknown limit)

Asked for	(by hand)	(GDC)
a distance, $v$ changes sign	find the zeros of $v$ ; add the sizes of the changes in $s$ between consecutive times: the start, each zero, the end	$\int_{t_1}^{t_2} \lvert v(t) \rvert\,dt$ , written in full
a distance, $v$ doesn't change sign	$\lvert s(t_2) - s(t_1) \rvert$	the same, or $\int_{t_1}^{t_2} \lvert v(t) \rvert\,dt$
a distance, $s(t)$ given	add the sizes of the changes in $s$ between consecutive times: the start, each turning point of $s$ , the end	the same, with the turning points from maximum or minimum of a graph; or $\int_{t_1}^{t_2} \lvert s'(t) \rvert\,dt$
the position at time $t$	$\int v\,dt$ with $+c$ , then the starting position; or $s(t) = s(0) + \int_0^t v(u)\,du$ (the letter inside the integral is $u$ only so that it isn't confused with the upper limit $t$ )	$s(0) + \int_0^{T} v\,dt$ for the position at a time $T$ ; a formula still needs the algebra
a $v$ – $t$ graph of straight lines	areas of shapes, counted negative below the axis for a displacement	the same
the time to travel a distance $D$	$\lvert s(T) - s(0) \rvert = D$ if $v$ doesn't change sign; if it does, add the distance up to each turn, then solve for the rest: $\lvert s(T) - s(t_{\text{turn}}) \rvert = D - (\text{distance so far})$ , or triangle areas on a straight-line $v$ – $t$ graph	$\int_0^T \lvert v(t) \rvert\,dt = D$ (definite integral with an unknown limit)

Steps

What is asked for?
A position (a formula or a value), a displacement (signed), a distance travelled (never negative), or a time (it returns, it meets another particle, it has travelled a given distance).
Position:
integrate $v$ with $+c$ and use the starting position, or $s(t) = s(0) + \int_0^t v(u)\,du$ ( $u$ is just the time variable renamed, as in the Which method? table). $a$ given: integrate twice, with one constant each time, fixed by $v(0)$ and then by $s(0)$ . $s$ given: there is nothing to integrate.
Displacement from $t_1$ to $t_2$ :
$\int_{t_1}^{t_2} v\,dt = s(t_2) - s(t_1)$ . Keep the sign.
Distance:
find the zeros of $v$ in the interval first. : split there and add the sizes. : $\int_{t_1}^{t_2} \lvert v \rvert\,dt$ (definite integral). A straight-line $v$ – $t$ graph: areas of shapes.
Times.
Returns to its start at $t_1$ : $s(t) = s(t_1)$ for some later $t$ in the interval (a sign change of $s(t) - s(t_1)$ proves it), then solve if the time is asked for. Meets: $s_A(t) = s_B(t)$ , both from the same $O$ . Has travelled $D$ : $\int_0^T \lvert v \rvert\,dt = D$ .
Sense-check:
the distance is at least the size of the displacement, and equal to it only if $v$ never changes sign. A particle's distance can't be less than a gap it had to close on its own. For two particles that meet, their two distances together can't be less than the gap between them. A distance that is less loses the final mark ().

Common mistakes

Distance as displacement
when $v$ changes sign: find the zeros of $v$ first, then split (or use $\lvert v \rvert$ ).
The modulus outside the integral
$\left\lvert \int_{t_1}^{t_2} v\,dt \right\rvert$ is the size of the displacement. Markschemes accept it for a distance only when $v$ doesn't change sign (Example 3.3(c)).
A turn outside the interval counted
for the distance from $t = 1$ to $t = 5$ , only the zeros of $v$ between $1$ and $5$ split it (Example 3.3).

More mistakes (9)

A negative displacement made positive
(). A displacement keeps its sign; only a distance is positive.
"Returns to its start" tested by the net displacement being $0$
the question asks whether $s(t) = s(t_1)$ at some later time in the interval. Show that $s(t) - s(t_1)$ takes values of opposite signs, or that after its first turn the distance back (up to its next turn, or to the end) is at least the distance out (Example 3.1(c)).
Substituting the starting value before writing $+c$ ,
or never writing it: the method mark needs the constant to be there.
"In the 5th second" as $0 \le t \le 5$
it is $4 \le t \le 5$ .
A GDC value with no integral written down
on the written integral earns the method mark. Include the $dt$ : markschemes have condoned it when it is missing, but don't rely on that.
Rounded limits, or a rounded one-period distance, reused later
(Example 3.4(c)): keep full accuracy (keep full-accuracy values).
Two particles measured from different points
both positions must be from the same $O$ .
A distance less than a gap that had to be closed
one particle closing a gap on its own travels at least the gap, and two particles that meet travel at least the gap between them together (Example 3.5). Markschemes withhold the final for a distance that fails this, even under follow-through. A distance is also never less than the size of the displacement.
Degree mode
on with trig: () (Examples 3.3 and 3.4).

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