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.