Answer: (a)(i) \(20\text{ m s}^{-1}\), (ii) \(t=\dfrac{\pi}{4}\), (iii) \(-40\text{ m s}^{-2}\); (b)(i) \(V=35\), (ii) \(10\text{ m s}^{-1}\), (iii) \(t=27\).

(a)(i) The displacement is

\(x=10\sin2t-5.\)

Differentiate to find velocity:

\(v=\frac{dx}{dt}=20\cos2t.\)

When \(t=\pi\),

\(v=20\cos2\pi=20.\)

So the speed is

\(20\text{ m s}^{-1}.\)

(a)(ii) The particle is at rest when \(v=0\):

\(20\cos2t=0.\)

The first positive value occurs when

\(2t=\frac{\pi}{2}.\)

Hence

\(t=\frac{\pi}{4}.\)

(a)(iii) Acceleration is

\(a=\frac{dv}{dt}=-40\sin2t.\)

When \(t=\frac{\pi}{4}\),

\(a=-40\sin\frac{\pi}{2}=-40.\)

So the acceleration is

\(-40\text{ m s}^{-2}.\)

(b)(i) The particle accelerates at \(3.5\text{ m s}^{-2}\) for \(10\) s, starting from rest. Therefore

\(V=3.5\times10=35.\)

(b)(ii) From \(t=20\) to \(t=25\), the area under the velocity-time graph is \(112.5\). Let the velocity at \(t=25\) be \(u\). This part is a trapezium, so

\(112.5=\frac12(35+u)(5).\)

Thus

\(35+u=45,\)

and hence

\(u=10.\)

So the velocity when \(t=25\) is

\(10\text{ m s}^{-1}.\)

(b)(iii) During the \(5\) s from \(t=20\) to \(t=25\), the velocity decreases from \(35\) to \(10\). The deceleration is therefore

\(\frac{35-10}{5}=5\text{ m s}^{-2}.\)

At \(t=25\), the particle still has velocity \(10\text{ m s}^{-1}\). The extra time needed to come to rest is

\(\frac{10}{5}=2\text{ s}.\)

Therefore the particle comes to rest at

\(t=25+2=27.\)