Answer: \(V=48\) and the acceleration at \(t=35\) is \(-\frac{16}{5}\text{ m s}^{-2}\).

Use the given velocity, acceleration or displacement relationship consistently. Remember that acceleration is the derivative of velocity, and displacement is obtained by integrating velocity.

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

The graph is made up of four simple regions:

\(\frac12(10)(10)+10(10)+\frac12(10)(10+V)+\frac12(15)(V)=800.\)

Simplify:

\(50+100+50+5V+\frac{15V}{2}=800.\)

So

\(200+\frac{25V}{2}=800.\)

Hence

\(\frac{25V}{2}=600,\)

and

\(V=48.\)

(b) When \(t=35\), the particle is on the straight-line section from \(t=30\) to \(t=45\). The acceleration is the gradient of this section:

\(\frac{0-48}{45-30}=-\frac{48}{15}=-\frac{16}{5}.\)

So the acceleration is

\(-\frac{16}{5}\text{ m s}^{-2}.\)