2017 H2 Maths Paper 1 Q11 — Worked Solution

Part (i)(a)

1 mark

\frac{dv}{dt} = c

Part (i)(b)

3 marks

Approach

With $\frac{dv}{dt} = c$ constant, integrating directly gives $v$ as a linear function of $t$ ; the initial condition and the value after $2.5\ s$ then pin down the two constants.

\frac{dv}{dt} = c

\int dv = \int c\, dt

v = ct + D

Since $v = 4$ when $t = 0$ , $D = 4$ , so $v = ct + 4$ .

Using the given value $v = 29$ when $t = 2.5$ :

29 = 2.5c + 4

c = 10

$\therefore$ $v = 10t + 4$ , with $c = 10$ .

Part (ii)

5 marks

Approach

The new rate of change is $\frac{dv}{dt} = 10 - kv$ , a first-order equation with $v$ and $t$ separable. We integrate both sides, using the initial condition $v = 0$ at $t = 0$ to fix the constant, then rearrange to make $v$ the subject.

\frac{dv}{dt} = 10 - kv

Separating variables,

\int\frac{1}{10 - kv}\, dv = \int dt

- \frac{1}{k}\ln|10 - kv| = t + D

Since $v = 0$ when $t = 0$ :

- \frac{1}{k}\ln 10 = D

Substituting back:

- \frac{1}{k}\ln|10 - kv| = t - \frac{1}{k}\ln 10

Multiplying throughout by $- k$ :

\ln|10 - kv| = - kt + \ln 10

\ln\left| \frac{10 - kv}{10} \right| = - kt

Note: while the terminal velocity has not yet been reached, $v < \frac{10}{k}$ , so $10 - kv > 0$ and the modulus can be dropped.

\frac{10 - kv}{10} = e^{- kt}

10 - kv = 10e^{- kt}

v = \frac{10}{k}\left( 1 - e^{- kt} \right)

To check your answer: at

t = 0

,

v = \frac{10}{k}(1 - 1) = 0

, which matches the given initial condition. ✓

$\therefore$ $v = \frac{10}{k}\left( 1 - e^{- kt} \right)$ .

Part (iii)

4 marks

Approach

The terminal velocity is the limit of $v$ as $t \rightarrow \infty$ . As $t \rightarrow \infty$ , $e^{- kt} \rightarrow 0$ (since $k > 0$ ), so this limit is $\frac{10}{k}$ , which lets us find $k$ from the given terminal velocity before solving for $t$ .

As $t \rightarrow \infty$ , $e^{- kt} \rightarrow 0$ , so the terminal velocity is

\lim_{t \rightarrow \infty}v = \frac{10}{k}

Since the terminal velocity is $40\ m\, s^{- 1}$ :

\frac{10}{k} = 40

k = \frac{1}{4}

So $v = 40\left( 1 - e^{- t/4} \right)$ . We require $t$ such that $v = 0.9(40) = 36$ :

36 = 40\left( 1 - e^{- t/4} \right)

1 - e^{- t/4} = 0.9

e^{- t/4} = 0.1

- \frac{t}{4} = \ln 0.1

t = - 4\ln 0.1 = 4\ln 10

$\therefore$ $t = 4\ln 10 = 9.21\ s$ (3 s.f.).

2017 H2 Maths Paper 1, Question 11

Part (i)(a)

Part (i)(b)

Part (ii)

Part (iii)

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