Answer: \(V=50\); acceleration \(=-1\); exact acceleration \(=\frac{13}{3}\); no change of direction because \(v\gt0\) for \(t\gt0\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use \(v=ds/dt\) and \(a=dv/dt\), then substitute the requested value of time.

(a)(i) The distance travelled is the area under the velocity-time graph.

From \(0\) to \(30\), the area is

\(30\times10=300.\)

From \(30\) to \(70\), the area is a trapezium:

\(\frac12(10+V)\times40.\)

From \(70\) to \(120\), the area is a triangle:

\(\frac12\times50\times V.\)

The total distance is \(2750\), so

\(300+\frac12(10+V)40+\frac12(50V)=2750.\)

Thus

\(300+200+20V+25V=2750.\)

So

\(45V=2250,\)

and hence

\(V=50.\)

(a)(ii) For \(70\lt t\lt120\), acceleration is the gradient of the line segment:

\(\frac{0-50}{120-70}=-1.\)

So the acceleration is \(-1\text{ m s}^{-2}\).

(b)(i) The velocity is

\(v=t(t^2+5)^{1/2}.\)

Differentiate using the product rule:

\(\frac{\mathrm dv}{\mathrm dt}=(t^2+5)^{1/2}+t\cdot\frac12(t^2+5)^{-1/2}(2t).\)

So

\(\frac{\mathrm dv}{\mathrm dt}=(t^2+5)^{1/2}+\frac{t^2}{(t^2+5)^{1/2}} =\frac{2t^2+5}{\sqrt{t^2+5}}.\)

When \(t=2\),

\(\frac{\mathrm dv}{\mathrm dt}=\frac{2(2^2)+5}{\sqrt{2^2+5}}=\frac{13}{3}.\)

(b)(ii) For \(t\gt0\), both \(t\) and \(\sqrt{t^2+5}\) are positive. Therefore

\(v=t\sqrt{t^2+5}\gt0.\)

So the velocity does not change sign, and the particle does not change direction.

This completes the solution and gives the required result.