A particle moves along a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ - 1}}\)at time t seconds is given by \(v = 6{{\rm{e}}^{3t}} + 4\) . When \(t = 0\) , the displacement, s, of the particle is 7 metres. Find an expression for s in terms of t.

evidence of anti-differentiation     (M1)

e.g. \(s = \int {(6{{\rm{e}}^{3x}} + 4)} {\rm{d}}x\)

\(s = 2{{\rm{e}}^{3t}} + 4t + C\)     A2A1

substituting \(t = 0\) ,    (M1)

\(7 = 2 + C\)     A1

\(C = 5\)

\(s = 2{{\rm{e}}^{3t}} + 4t + 5\)    A1     N3

[7 marks]

There were a number of completely correct solutions to this question. However, there were many who did not know the relationship between velocity and position. Many students differentiated rather than integrated and those who did integrate often had difficulty with the term involving e. Many who integrated correctly neglected the C or made \(C = 7\) .