The acceleration in ms⁻² of a particle moving in a straight line at time \(t\) seconds, \(t \geqslant 0\) , is given by the formula \(a = - \frac{1}{2}v\). When \(t = 0\) , the velocity is \(40\) ms⁻¹ .

Find an expression for \(v\) in terms of \(t\) .

\(\frac{{{\text{d}}v}}{{{\text{d}}t}} = - \frac{1}{2}v\)     A1

\(\int {\frac{{{\text{d}}v}}{v}}  = \int { - \frac{1}{2}} {\text{d}}t\)     (A1)

\(\ln v = - \frac{1}{2}t + c\)     (A1)

\(v = {{\text{e}}^{ - \frac{1}{2}t + c}}\)   \(\left( { = A{{\text{e}}^{ - \frac{1}{2}t}}} \right)\)     (A1)

\(t = 0\), \(v = 40\), so \(A = 40\)     M1

\({v = 40{{\text{e}}^{ - \frac{1}{2}t}}}\)   (or equivalent)     A1

[6 marks]

This was a poorly answered question which linked the topic of kinematics with that of first order differential equations. Many candidates seemed unaware that the acceleration is the time derivative of the velocity. This was often followed by a failure to recognize a separable differential equation and/or integration with respect to the wrong variable.