Answer: distance \(110\); \(v=(2t+4)^{1/2}+5\); \(s=\frac13(2t+4)^{3/2}+5t-51\).

(a)(i) The displacement rises from \(0\) to \(50\), so the runner travels \(50\) m in the first part.

It then falls from \(50\) to \(-10\), so the runner travels a further \(60\) m.

Hence the total distance travelled is

\(50+60=110\text{ m}.\)

(a)(ii) Velocity is the gradient of the displacement-time graph.

From \(t=0\) to \(t=10\),

\(v=\frac{50-0}{10-0}=5.\)

From \(t=10\) to \(t=40\),

\(v=\frac{-10-50}{40-10}=-2.\)

So the velocity-time graph is a horizontal line at \(v=5\) for \(0\leq t\leq10\), then a horizontal line at \(v=-2\) for \(10\leq t\leq40\).

(b)(i) Since acceleration is \(\frac{dv}{dt}\),

\(\frac{dv}{dt}=(2t+4)^{-1/2}.\)

Integrate:

\(v=(2t+4)^{1/2}+C.\)

Given \(v=9\) when \(t=6\),

\(9=(16)^{1/2}+C=4+C.\)

So \(C=5\). Therefore

\(v=(2t+4)^{1/2}+5.\)

(b)(ii) Since \(v=\frac{ds}{dt}\), integrate \(v\):

\(s=\int\left((2t+4)^{1/2}+5\right)\,dt.\)

This gives

\(s=\frac13(2t+4)^{3/2}+5t+D.\)

Given \(s=\frac13\) when \(t=6\),

\(\frac13=\frac13(16)^{3/2}+30+D.\)

Since \((16)^{3/2}=64\),

\(\frac13=\frac{64}{3}+30+D.\)

So

\(D=-51.\)

Hence

\(s=\frac13(2t+4)^{3/2}+5t-51.\)