A particle moves along a straight line so that after t seconds its displacement s , in metres, satisfies the equation \({s^2} + s - 2t = 0\) . Find, in terms of s , expressions for its velocity and its acceleration.

\(2s\frac{{{\text{d}}s}}{{{\text{d}}t}} + \frac{{{\text{d}}s}}{{{\text{d}}t}} - 2 = 0\)     M1A1

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

EITHER

\(a = \frac{{{\text{d}}v}}{{{\text{d}}t}} = \frac{{{\text{d}}v}}{{{\text{d}}s}}\frac{{{\text{d}}s}}{{{\text{d}}t}}\)     (M1)

\(\frac{{{\text{d}}v}}{{{\text{d}}s}} = \frac{{ - 4}}{{{{(2s + 1)}^2}}}\)     (A1)

\(a = \frac{{ - 4}}{{{{(2s + 1)}^2}}}\frac{{{\text{d}}s}}{{{\text{d}}t}}\)

OR

\(2{\left( {\frac{{{\text{d}}s}}{{{\text{d}}t}}} \right)^2} + 2s\frac{{{{\text{d}}^2}s}}{{{\text{d}}{t^2}}} + \frac{{{{\text{d}}^2}s}}{{{\text{d}}{t^2}}} = 0\)     (M1)

\(\frac{{{{\text{d}}^2}s}}{{\underbrace {{\text{d}}{t^2}}_a}} = \frac{{ - 2{{\left( {\frac{{{\text{d}}s}}{{{\text{d}}t}}} \right)}^2}}}{{2s + 1}}\)     (A1)

THEN

\(a = \frac{{ - 8}}{{{{(2s + 1)}^3}}}\)     A1

[6 marks]

Despite the fact that many candidates were able to calculate the speed of the particle, many of them failed to calculate the acceleration. Implicit differentiation turned out to be challenging in this exercise showing in many cases a lack of understanding of independent/dependent variables. Very often candidates did not use the chain rule or implicit differentiation when attempting to find the acceleration. It was not uncommon to see candidates trying to differentiate implicitly with respect to t rather than s, but getting the variables muddled.