Answer: (a) \(w=18.4\), (b)(i) \(t=4.5\), (b)(ii) \(4\lt t\lt 5\), (b)(iii) \(\frac{59}{2}\) m.

Answer: (a) \(w=18.4\), (b)(i) \(t=4.5\), (b)(ii) \(4\lt t\lt 5\), (b)(iii) \(\frac{59}{2}\) m.

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

For the first \(12\) seconds, the distance is

\(12w.\)

During the next \(10\) seconds, the velocity decreases uniformly from \(w\) to \(w-14\), so the distance is

\(\frac12(w+(w-14))\cdot10.\)

For the final \(28\) seconds, the distance is

\(28(w-14).\)

The total distance is \(458\), so

\(12w+\frac12(w+w-14)\cdot10+28(w-14)=458.\)

Simplify:

\(12w+10w-70+28w-392=458.\)

Thus

\(50w-462=458,\)

so

\(50w=920.\)

Therefore

\(w=18.4.\)

(b)(i) The velocity is

\(v=(t-4)(t-5)=t^2-9t+20.\)

Acceleration is \(\frac{dv}{dt}\), so

\(a=2t-9.\)

Set \(a=0\):

\(2t-9=0.\)

Hence

\(t=4.5.\)

(b)(ii) The velocity is negative when

\((t-4)(t-5)\lt 0.\)

This happens between the two roots, so

\(4\lt t\lt 5.\)

(b)(iii) The velocity is positive from \(0\) to \(4\), and negative from \(4\) to \(5\). Distance travelled is the integral of speed, so

An antiderivative is

\(\frac{t^3}{3}-\frac{9t^2}{2}+20t.\)

From \(0\) to \(4\), the distance is

\(\frac{64}{3}-72+80=\frac{88}{3}.\)

From \(4\) to \(5\), the signed displacement is

So the distance on this interval is \(\frac16\). Therefore the total distance is