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Position, velocity, acceleration and speed on a line; displacement against distance travelled; and reading derivatives and integrals in the units of the situation they came from.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can move between position, velocity and acceleration by differentiating and integrating, say which way an object is going and whether it is speeding up, and tell displacement from distance travelled. You can also read any derivative or integral in the units of its context — deciding from those units alone whether a number is a rate, an accumulated total, or the rate at which a rate is changing.
You can differentiate and integrate polynomials, and you know that integrating a rate gives an accumulated change. Motion on a line is where those two facts are at their most concrete — and where the signs carry meaning that is easy to lose.
Position $s(t)$: where the object is, with a sign for direction.
Velocity $v = s'$: how fast and which way.
Speed $|v|$: how fast, with no direction.
Acceleration $a = v' = s''$.
Displacement: $\int v$, a signed change in position.
Distance travelled: $\int |v|$, which is never smaller.
For an object moving along a line, three functions describe everything, and they form a chain:
$$s(t) \xrightarrow{\;\text{differentiate}\;} v(t) \xrightarrow{\;\text{differentiate}\;} a(t)$$
Each step down differentiates; each step up integrates, and needs a constant that an initial condition supplies. That is why a motion problem always comes with a starting position or a starting velocity: without one, the integral names a family rather than a motion.
Signs carry the direction. Positive velocity is one way along the line, negative the other. Speed is $|v|$, and it is speed that 'speeding up' refers to — so the object speeds up when $v$ and $a$ have the same sign, and slows down when they have opposite signs, regardless of which direction it is travelling.
Displacement against distance. $\int_a^b v\,dt$ is the signed change in position: time spent going backwards subtracts. $\int_a^b |v|\,dt$ is the distance actually travelled, and it is never smaller. To compute it, find where $v$ changes sign, integrate over each stretch separately, and add the sizes.
Turning round happens where $v$ changes sign — so solve $v = 0$ and then check that the sign really crosses rather than touching.
Another way: picture
Picture a bead on a wire, and a separate graph of its velocity against time. Where the velocity graph is above the axis the bead slides right; below, it slides left. The area above minus the area below is where the bead ends up; the two areas added is how far it has actually travelled.
Another way: steps
| $v$ | $a$ | Moving | Speed |
|---|---|---|---|
| $+$ | $+$ | Forwards | Increasing |
| $+$ | $-$ | Forwards | Decreasing |
| $-$ | $-$ | Backwards | Increasing |
| $-$ | $+$ | Backwards | Decreasing |
Rows two and three are the ones that catch people. A negative acceleration speeds an object up when it is already going backwards.
"Negative acceleration means slowing down." It means accelerating in the negative direction. An object moving backwards with negative acceleration is speeding up. The rule is about the signs agreeing: same signs, speeding up; opposite signs, slowing down.
"Displacement and distance are the same." They agree only when the velocity never changes sign. A particle that goes out three metres and comes back has travelled six and been displaced zero.
"Velocity zero means it turned round." Only if the velocity changes sign there. A velocity that touches zero and returns the way it came is a momentary pause, not a reversal.
$v(t) = t - 3$ m/s on $[0, 5]$: the velocity is negative before $t = 3$ and positive after.
Find the sign change.
Displacement $= \int_0^5 (t-3)dt = \left[\tfrac{t^2}{2} - 3t\right]_0^5 = -2.5$ m.
Signed.
Distance $= |{-4.5}| + |2| = 6.5$ m, splitting at $t = 3$.
Sizes added.
$v(2) = -6$ m/s and $a(2) = -4$ m/s².
Both negative.
Same signs, so the speed is increasing: the object is moving backwards faster and faster.
Calling that 'decelerating' because $a < 0$ would be exactly wrong.
$v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$.
Differentiate and factor.
Zero at $t = 1$ and $t = 3$.
And $v$ changes sign at both, so the particle turns round twice — the distance travelled must be computed in three pieces.
A particle's position on a line is $s(t) = 3t^3 - 6t$ metres at time $t$ seconds. Find its velocity at $t = 1$.
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At one instant a particle has velocity $-2$ m/s and acceleration $5$ m/s². Is it speeding up or slowing down?
A particle's velocity is $v(t) = 12t^2$ m/s. Find its displacement over the first $5$ seconds.
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You can differentiate and integrate, and you have used both on quantities with units. This section is about nothing else: saying what a derivative or an integral means in the situation it came from, which is the part an examiner and an employer both care about most.
Rate: a derivative, in output units per input unit.
Accumulation: an integral of a rate, back in the output's units.
Marginal: the derivative of a total — the cost or benefit of one more.
Linearisation: current value plus rate times elapsed time.
Second derivative in context: whether the rate itself is rising or falling.
Every derivative and every integral in an applied problem has units, and reading them settles what the number is.
A derivative's units are the output's over the input's. If $C(x)$ is the cost in pounds of $x$ tonnes, then $C'(x)$ is in pounds per tonne — the marginal cost, the cost of roughly one more tonne. If $P(t)$ is a population in thousands and $t$ is years, $P'(t)$ is thousands per year.
An integral's units are the integrand's times the variable's. So $\int_a^b C'(x)\,dx$ is in pounds: the cost of going from $a$ tonnes to $b$. Notice that this is the fundamental theorem stated in units — integrating a marginal cost gives a total cost.
The second derivative says whether the rate is changing. $P'' < 0$ with $P' > 0$ is a quantity still growing, but more slowly: the two derivatives answer different questions, and conflating them is how a report about slowing inflation becomes a report about falling prices.
Linear approximation is the everyday use. Current value plus rate times elapsed time, $f(a) + f'(a)\Delta t$ — the estimate people make constantly without calling it calculus, reliable over a short interval and wrong in a direction the concavity predicts.
And an integral gives a change, never a total. Whatever was there at the start has to be added separately.
Another way: picture
Picture two dials on a dashboard: the speedometer and the odometer. The speedometer is the derivative and the odometer is the integral, and they are never in the same units. Every applied question in this section is asking which of the two dials you have been shown.
Another way: steps
| You have | You want | Operation | Units |
|---|---|---|---|
| Total $Q(t)$ | Rate | Differentiate | $Q$'s units per unit of $t$ |
| Rate $R(t)$ | Change | Integrate | $R$'s units $\times$ $t$'s |
| Rate $R(t)$ | Total | Integrate, then add the start | $Q$'s units |
| Rate $R(t)$ | Is the rate rising? | Differentiate | per unit of $t$, twice |
The third row is the one with an extra step, and it is the step that gets left out.
Reporting a rate as a quantity. $P'(5) = 20$ does not mean twenty of anything exist. It means the quantity is growing at twenty per unit of time.
"Growth is slowing" read as "it is shrinking." The first is about the second derivative and the second about the first. A quantity can grow while its growth slows for years.
Forgetting the starting amount. $\int_a^b R(t)\,dt$ is the change over that interval. The total at the end is that plus whatever was there at the start, and the missing initial value is the commonest single lost mark in this topic.
$C(x)$ pounds for $x$ items, $C'(200) = 14$.
Given.
Units: pounds per item. So the 201st item costs about £14.
Read the units first.
$\int_{200}^{250} C'(x)\,dx$ is in pounds: the cost of items 201 to 250.
Integrating a marginal gives a total.
Water enters at $R(t)$ litres/min; the tank starts with 100 litres.
$\int_0^{30} R\,dt = 240$ means 240 litres were added in half an hour.
The change.
So the tank holds $100 + 240 = 340$ litres.
The start, added.
Units: litres per hour, and the sign is negative.
So at $t = 3$ the tank is losing fuel at 8 litres an hour.
And $\int_3^5 F'(t)\,dt$ would be in litres — the change over those two hours, not the amount remaining.
$C(x)$ is the cost in pounds of producing $x$ tonnes. Give the units of $C'(x)$, and then the units of $\int_0^{4} C'(x)\,dx$.
a; b
A city's population $P(t)$ is measured in thousands, $t$ in years, and $P''(8) = -2$. What does that say?
A hospital's oxygen store holds $16$ cylinders at midnight. Cylinders are consumed at $3$ an hour for $6$ hours, and a delivery of $12$ arrives at the end. How many are in store then?
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Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A particle has velocity $v(t) = 2t^2 - 4t$ m/s for $t \ge 0$. At what positive time does it change direction?
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A tank holds $33$ litres and is filling at $6$ litres a minute. Using only that rate, estimate the volume $4$ minutes later.
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You can analyse motion and interpret rates in context. Without looking: when is an object slowing down, and what has to be added to $\int_a^b R(t)\,dt$ to get a total rather than a change?
9. Your turn: $s(t) = t^3 - 6t^2 + 9t$. When is the particle at rest?, step 3
20. Your turn: $F(t)$ is fuel in litres, $t$ in hours, $F'(3) = -8$, step 3