A body is moving through a liquid so that its acceleration can be expressed as

\[\left( { - \frac{{{v^2}}}{{200}} - 32} \right){\text{m}}{{\text{s}}^{ - 2}},\]

where \(v{\text{ m}}{{\text{s}}^{ - 1}}\) is the velocity of the body at time t seconds.

The initial velocity of the body was known to be \(40{\text{ m}}{{\text{s}}^{ - 1}}\).

(a)     Show that the time taken, T seconds, for the body to slow to \(V{\text{ m}}{{\text{s}}^{ - 1}}\) is given by

\[T = 200\int_V^{40} {\frac{1}{{{v^2} + {{80}^2}}}{\text{d}}v.} \]

(b)     (i)     Explain why acceleration can be expressed as \(v\frac{{{\text{d}}v}}{{{\text{d}}s}}\), where s is displacement, in metres, of the body at time t seconds.

  (ii)     Hence find a similar integral to that shown in part (a) for the distance, S metres, travelled as the body slows to \(V{\text{ m}}{{\text{s}}^{ - 1}}\).

(c)     Hence, using parts (a) and (b), find the distance travelled and the time taken until the body momentarily comes to rest.

(a)     \(\frac{{{\text{d}}v}}{{{\text{d}}t}} = - \frac{{{v^2}}}{{200}} - 32\left( { = \frac{{ - {v^2} - 6400}}{{200}}} \right)\)     (M1)

\(\int_0^T {{\text{d}}t = \int_{40}^V { - \frac{{200}}{{{v^2} + {{80}^2}}}{\text{d}}v} } \)     M1A1A1

\(T = 200\int_V^{40} {\frac{1}{{{v^2} + {{80}^2}}}{\text{d}}v} \)     AG

[4 marks]

(b)     (i)     \(a = \frac{{{\text{d}}v}}{{{\text{d}}t}} = \frac{{{\text{d}}v}}{{{\text{d}}s}} \times \frac{{{\text{d}}s}}{{{\text{d}}t}}\)     R1

\( = v\frac{{{\text{d}}v}}{{{\text{d}}s}}\)     AG

(ii)     \(v\frac{{{\text{d}}v}}{{{\text{d}}s}} = \frac{{ - {v^2} - {{80}^2}}}{{200}}\)     (M1)

\(\int_0^S {{\text{d}}s = \int_{40}^V -{\frac{{200v}}{{{v^2} + {{80}^2}}}{\text{d}}v} } \)     M1A1A1

\(\int_0^S {{\text{d}}s = \int_V^{40} {\frac{{200v}}{{{v^2} + {{80}^2}}}{\text{d}}v} } \)     M1

\(S = 200\int_V^{40} {\frac{v}{{{v^2} + {{80}^2}}}{\text{d}}v} \)     A1

[7 marks]

(c)     letting V = 0     (M1)

distance \( = 200\int_0^{40} {\frac{v}{{{v^2} + {{80}^2}}}{\text{d}}v = 22.3{\text{ metres}}} \)     A1

time \( = 200\int_0^{40} {\frac{1}{{{v^2} + {{80}^2}}}{\text{d}}v = 1.16{\text{ seconds}}} \)     A1

[3 marks]

Total [14 marks]

Many students failed to understand the problem as one of solving differential equations. In addition there were many problems seen in finding the end points for the definite integrals. Part (b) (i) should have been a simple point having used the chain rule, but it seemed that many students had not seen this, even though it is clearly in the syllabus.