2024 H2 Maths Paper 1 Q11 — The Parachutist

The model, livespeed against time

Velocity Distance fallen Terminal velocity Landing

Part (a)(i)

1 mark

Use $v=\frac{dx}{dt}$ to write down a differential equation in $v$ and $t$ .

Approach

One mark, one substitution — and the rest of the question rests on it. The model is given as a second-order equation in $x$ . Since $v$ is itself $\frac{dx}{dt}$ , differentiating $v$ gives the second derivative of $x$ , and the whole thing drops to first order in $v$ .

The model given is

\frac{d^{2}x}{dt^{2}}+\alpha\frac{dx}{dt}=9.8.

Since $v=\frac{dx}{dt}$ , differentiating both sides with respect to $t$ gives $\frac{dv}{dt}=\frac{d^{2}x}{dt^{2}}$ , so

$\therefore\ \dfrac{dv}{dt}+\alpha v=9.8$ .

What the equation says, in three moments. Gravity contributes a constant 9.8; the resistance term

\alpha v

grows with speed and eats into it. When the two balance the acceleration is zero, and the speed stops changing.

Part (a)(ii)

1 mark

Given that $\frac{dv}{dt}=5$ when $v=2.4$ , find the value of $\alpha$ .

The equation from part (i) holds at every instant, so it holds at that one:

\begin{aligned}5+\alpha(2.4)&=9.8\\2.4\alpha&=4.8\\\alpha&=2\end{aligned}

To check your answer:

With $\alpha=2$ the model reads $\frac{dv}{dt}=9.8-2v$ . Putting $v=2.4$ into the right-hand side gives $9.8-4.8=5$ , the rate that was given.

$\therefore\ \alpha=2$ .

Note

$\frac{dv}{dt}=9.8-\alpha v$ is a straight line in $v$ , and the data give one point on it. That is all a single mark is asking for: not a solved differential equation, just one substitution into an equation that is true at every moment of the fall.

Part (b)

5 marks

Given that the initial velocity is zero, show that $v=A(1-e^{Bt})$ , and find $A$ and $B$ .

Approach

The variables separate: every $v$ can be gathered on one side and every $t$ on the other. The one place marks go missing is the constant — it has to be carried through the exponentiation and only then fixed by $v=0$ at $t=0$ , not applied to the logarithmic line. Note too that $9.8-2v\gt0$ throughout the fall, so the modulus signs come off without a case to consider.

With $\alpha=2$ the differential equation is $\frac{dv}{dt}=9.8-2v$ . Separating the variables:

\begin{aligned}\int\frac{1}{9.8-2v}\,dv&=\int 1\,dt\\[2pt]-\frac{1}{2}\ln|9.8-2v|&=t+c\\[2pt]\ln|9.8-2v|&=-2t+c'\end{aligned}

Exponentiating both sides and writing $K=\pm e^{c'}$ :

9.8-2v=Ke^{-2t}

The parachutist starts from rest, so $v=0$ when $t=0$ :

9.8-0=K\quad\Longrightarrow\quad K=9.8

Substituting back and making $v$ the subject:

\begin{aligned}9.8-2v&=9.8e^{-2t}\\2v&=9.8\left(1-e^{-2t}\right)\\v&=4.9\left(1-e^{-2t}\right)\ \ (\text{shown})\end{aligned}

To check your answer:

At $t=0$ , $v=4.9(1-1)=0$ , matching the initial condition. Differentiating, $\frac{dv}{dt}=9.8e^{-2t}$ ; and $9.8-2v=9.8-9.8\left(1-e^{-2t}\right)=9.8e^{-2t}$ as well, so the differential equation holds at every $t$ , not just at the start.

$\therefore\ A=4.9$ and $B=-2$ .

Note — what $A$ and $B$ mean

$A=4.9$ is the speed the parachutist approaches but never reaches: put $\frac{dv}{dt}=0$ in the model and $v=\frac{9.8}{\alpha}=4.9$ . $B=-2$ sets how fast he gets there — after 2.3 seconds he is already within 1% of it. Drag the time slider and watch the gap close.

Part (c)

5 marks

The helicopter is 2000 m above the ground. Find the time taken for the parachutist to reach the ground.

Thought process

Velocity is the rate of change of distance, so the distance fallen is the integral of $v$ . Set that equal to 2000 and solve for $t$ — except the equation that comes out cannot be solved exactly, which is the point of the part. The exponential term has to be shown to be negligible before it is dropped.

Since $v=\frac{dx}{dt}$ , integrating the velocity gives the distance fallen:

\begin{aligned}x&=\int 4.9\left(1-e^{-2t}\right)dt\\[2pt]&=4.9\left(t+\frac{1}{2}e^{-2t}\right)+c\end{aligned}

He starts at the helicopter, so $x=0$ when $t=0$ :

0=4.9\left(0+\frac{1}{2}\right)+c\quad\Longrightarrow\quad c=-\frac{4.9}{2}

x=4.9\left(t+\frac{1}{2}e^{-2t}-\frac{1}{2}\right)

He reaches the ground when he has fallen 2000 m:

\begin{aligned}4.9\left(t+\frac{1}{2}e^{-2t}-\frac{1}{2}\right)&=2000\\[2pt]t+\frac{1}{2}e^{-2t}-\frac{1}{2}&=\frac{20000}{49}\end{aligned}

The fall takes hundreds of seconds, so $e^{-2t}$ is utterly negligible beside the other terms and may be dropped:

t=\frac{20000}{49}+\frac{1}{2}=\frac{40049}{98}

To check your answer:

$\frac{40049}{98}=408.66\ldots$ , and at that value $e^{-2t}=e^{-817}$ , a number with more than three hundred zeros after the decimal point. Solving the full equation numerically returns the same value to well over twenty decimal places, so nothing was lost by dropping the term.

$\therefore$ the parachutist reaches the ground after $\dfrac{40049}{98}=408.66$ seconds (2 d.p.), a little under 7 minutes.

Note — why the answer is so nearly a straight-line answer

After the first couple of seconds the parachutist is falling at a constant 4.9 m s⁻¹, so 2000 m ought to take $\frac{2000}{4.9}=\frac{20000}{49}$ seconds. The extra $\frac{1}{2}$ second is the whole cost of speeding up at the start: while accelerating he covered 2.45 m less than a body already at full speed would have. The model's entire history reduces to half a second.

What this question was really testing

Reducing the order. $v=\frac{dx}{dt}$ turns a second-order equation into a first-order one you can actually separate. One mark, and the other eleven depend on it.
Handling the constant properly. Exponentiate, let the constant become a multiplier, then apply the initial condition. Doing it in the other order is where solutions unravel.
Reading the answer form. "Show that $v=A(1-e^{Bt})$ " tells you the shape to aim for — and that $B$ is negative, since the speed settles rather than runs away.
Justifying an approximation. Dropping $e^{-2t}$ is not a shortcut; it is a claim about size that has to be stated. Examiners reward the sentence, not the silence.

A parachutist, a differential equation, and the speed he never quite reaches

Part (a)(i)

Part (a)(ii)

Part (b)

Part (c)

What this question was really testing

This is how we mark, and how we teach.