The acceleration of a car is $\frac{1}{{40}}(60 - v){\text{ m}}{{\text{s}}^{ - 2}}$, when its velocity is $v{\text{ m}}{{\text{s}}^{ - 2}}$. Given the car starts from rest, find the velocity of the car after 30 seconds.

METHOD 1

$\frac{{{\text{d}}v}}{{{\text{d}}t}} = \frac{1}{{40}}(60 - v)$     (M1)

attempting to separate variables $\int {\frac{{{\text{d}}v}}{{60 - v}} = \int {\frac{{{\text{d}}t}}{{40}}} }$     M1

$- \ln (60 - v) = \frac{t}{{40}} + c$     A1

$c = - \ln 60$ (or equivalent)     A1

attempting to solve for v when t = 30     (M1)

$v = 60 - 60{e^{ - \frac{3}{4}}}$

$v = 31.7{\text{ (m}}{{\text{s}}^{ - 1}})$     A1

METHOD 2

$\frac{{{\text{d}}v}}{{{\text{d}}t}} = \frac{1}{{40}}(60 - v)$     (M1)

$\frac{{{\text{d}}t}}{{{\text{d}}v}} = \frac{{40}}{{60 - v}}$ (or equivalent)     M1

$\int_0^{{v_f}} {\frac{{40}}{{60 - v}}{\text{d}}v = 30}$ where ${v_f}$ is the velocity of the car after 30 seconds.     A1A1

attempting to solve $\int_0^{{v_f}} {\frac{{40}}{{60 - v}}{\text{d}}v = 30}$ for ${v_f}$     (M1)

$v = 31.7{\text{ (m}}{{\text{s}}^{ - 1}})$     A1

[6 marks]

Most candidates experienced difficulties with this question. A large number of candidates did not attempt to separate the variables and instead either attempted to integrate with respect to v or employed constant acceleration formulae. Candidates that did separate the variables and attempted to integrate both sides either made a sign error, omitted the constant of integration or found an incorrect value for this constant. Almost all candidates were not aware that this question could be solved readily on a GDC.