Answer: (a) \(t=\frac13\) and \(t=5\); (d)(i) acceleration \(=6t-16\).

Answer: (a) \(t=\frac13\) and \(t=5\); (d)(i) acceleration \(=6t-16\).

First expand the displacement:

\(s=(t+2)(t-5)^2=(t+2)(t^2-10t+25).\)

So

\(s=t^3-8t^2+5t+50.\)

The velocity is

\(v=\frac{ds}{dt}=3t^2-16t+5.\)

For zero velocity, solve

\(3t^2-16t+5=0.\)

Factorising gives

\((3t-1)(t-5)=0.\)

Hence

\(t=\frac13\quad\text{or}\quad t=5.\)

For the displacement-time graph,

\(s(0)=2(25)=50,\)

so it meets the displacement axis at \((0,50)\). Also, \(s=0\) when \(t=5\), so it meets the time axis at \((5,0)\). The graph has a local maximum at \(t=\frac13\) and a local minimum at \((5,0)\).

For the velocity-time graph,

\(v=3t^2-16t+5.\)

It meets the velocity axis at \((0,5)\), and it meets the time axis at

\(\left(\frac13,0\right)\quad\text{and}\quad(5,0).\)

The acceleration is

\(\frac{dv}{dt}=6t-16.\)

So the acceleration-time graph is a straight line with vertical-axis intercept \((0,-16)\). It crosses the time axis when

\(6t-16=0,\)

so

\(t=\frac83.\)

Therefore the acceleration graph crosses the time axis at \(\left(\frac83,0\right)\).