2024 H2 Mathematics Paper 1 Question 11

Differential Equations (DEs)

Answers

(a)

$\frac{d v}{d t} + α v = 9.8.$

$α = 2.$

$A = 4.9,$ $B = - 2.$

$409 s .$

(a)

(i)

\begin{array}{lrlr} v & = \frac{d x}{d t} \\ \frac{d v}{d t} & = \frac{d^{2} x}{d t^{2}} \end{array}

Substituting $(1)$ and $(2)$ into the given differential equation,

\begin{aligned} \frac{d v}{d t} + α v & = 9.8 ∎ \end{aligned}

(ii)

Since $\frac{d v}{d t} = 5$ when $v = 2.4,$

\begin{aligned} 5 + 2.4 α & = 9.8 \\ 2.4 α & = 4.8 \\ α & = 2 ∎ \end{aligned}

(b)

\begin{aligned} \frac{d v}{d t} + 2 v & = 9.8 \\ \frac{d v}{d t} & = 9.8 - 2 v \\ \frac{1}{9.8 - 2 v} \frac{d v}{d t} & = 1 \\ \int \frac{1}{9.8 - 2 v} d v & = \int 1 d t \\ - \frac{\ln | 9.8 - 2 v |}{2} & = t + C \\ \ln | 9.8 - 2 v | & = - 2 t + C^{'} \\ | 9.8 - 2 v | & = e^{- 2 t + C^{'}} \\ 9.8 - 2 v & = A e^{- 2 t} \end{aligned}

When $t = 0,$ $v = 0$

\begin{aligned} A & = 9.8 \end{aligned}

\begin{aligned} 9.8 - 2 v & = 9.8 e^{- 2 t} \\ 2 v & = 9.8 - 9.8 e^{- 2 t} \\ v & = 4.9 (1 - e^{- 2 t}) ∎ \end{aligned}

(c)

Since $v = \frac{d x}{d t},$

\begin{aligned} \frac{d x}{d t} & = 4.9 (1 - e^{- 2 t}) \\ \int 1 d x & = \int 4.9 (1 - e^{- 2 t}) d t \\ x & = 4.9 (t + \frac{e^{- 2 t}}{2}) + C \end{aligned}

When $t = 0,$ $x = 0$

\begin{matrix} 0 = 4.9 (0 + \frac{e^{- 2 (0)}}{2}) + C \\ C = - 2.45 \end{matrix}

x = 4.9 (t + \frac{e^{- 2 t}}{2}) - 2.45

When $x = 2000,$

2000 = 4.9 (t + \frac{e^{- 2 t}}{2}) - 2.45

Solving with a GC,

t = 409 s ∎