Question 12. [Maximum mark: 8]
A tank of water initially contains 400 litres. Water is leaking from the tank such that after
10 minutes there are 324 litres remaining in the tank.
The volume of water, V litres, remaining in the tank after t minutes, can be modelled by the differential equation
\(\frac{dV}{dt}= -k\sqrt{V}\), where k is a constant.
(a)Show that = \((20-\frac{1}{5})^{2}\) . [6]
(b)Find the time taken for the tank to empty. [2]
▶️Answer/Explanation
(a) \(\frac{dv}{dt}-kv^{\frac{1}{2}}\) use of separation of variables \(\Rightarrow \int v ^{\frac{1}{2}}dv = \int -k dt 2v ^{\frac{1}{2}}= – kt (+c)\) considering initial conditions 40 = c \(\sqrt[2]{324}=-10k+40 \Rightarrow k = 0.4 \sqrt[2]{v}= -04t+ 40 \Rightarrow \sqrt{v}\)=20-0.2t Note: Award A1 for any correct intermediate step that leads to the AG. \(\Rightarrow =(20-\frac{t}{5})^{2}\) Note: Do not award the final A1 if the AG line is not stated. (b) 0 = \((20-\frac{t}{5})^{2}\) ⇒ t = 100 minutes
Question
A body is moving in a straight line. When it is \(s\) metres from a fixed point O on the line its velocity, \(v\), is given by \(v = – \frac{1}{{{s^2}}},{\text{ }}s > 0\).
Find the acceleration of the body when it is 50 cm from O.
▶️Answer/Explanation
Markscheme
\(\frac{{{\text{d}}v}}{{{\text{d}}s}} = 2{s^{ – 3}}\) M1A1
Note: Award M1 for \(2{s^{ – 3}}\) and A1 for the whole expression.
\(a = v\frac{{{\text{d}}v}}{{{\text{d}}s}}\) (M1)
\(a = – \frac{1}{{{s^2}}} \times \frac{2}{{{s^3}}}\left( { = – \frac{2}{{{s^5}}}} \right)\) (A1)
when \(s = \frac{1}{2},{\text{ }}a = – \frac{2}{{{{(0.5)}^5}}}{\text{ }}( = – 64){\text{ (m}}{{\text{s}}^{ – 2}})\) M1A1
Note: M1 is for the substitution of 0.5 into their equation for acceleration.
Award M1A0 if \(s = 50\) is substituted into the correct equation.
[6 marks]