Use \(a=\dfrac{v-u}{t}\).

Here \(u=4\), \(v=10\), \(t=3\).

So \(a=\dfrac{10-4}{3}=2\text{ m s}^{-2}\).

Since it starts from rest, \(u=0\).

Use \(a=\dfrac{v-u}{t}=\dfrac{10-0}{4}=2.5\text{ m s}^{-2}\).

Use \(v=u+at\).

So \(12=3+6t\).

Hence \(9=6t\), so \(t=1.5\text{ s}\).

Use \(v=u+at\).

So \(v=4+3\times 5=19\text{ m s}^{-1}\).

Use \(v=u+at\).

So \(9=u+1.5\times 4=u+6\).

Hence \(u=3\text{ m s}^{-1}\).

Take deceleration as \(a=-2\text{ m s}^{-2}\).

Use \(v=u+at\).

So \(8=u+(-2)(3)=u-6\).

Hence \(u=14\text{ m s}^{-1}\).

First find the time using \(v=u+at\):

\(8=4+0.5t\), so \(t=8\text{ s}\).

Then use average velocity \(=\dfrac{u+v}{2}=\dfrac{4+8}{2}=6\text{ m s}^{-1}\).

Displacement \(=6\times 8=48\text{ m}\).

Use \(s=\dfrac{u+v}{2}t\).

So \(60=\dfrac{u+9}{2}\times 10\).

This gives \(60=5(u+9)\), so \(u+9=12\), hence \(u=3\text{ m s}^{-1}\).

Now use \(a=\dfrac{v-u}{t}=\dfrac{9-3}{10}=0.6\text{ m s}^{-2}\).

To use the standard equations of uniformly accelerated motion, we assume the athlete moves in a straight line with constant acceleration over the whole time interval.

Using \(t=\dfrac{v-u}{a}\):

Time from 0 to 1 is \(\dfrac{1}{a}\).

Time from 1 to 5 is \(\dfrac{4}{a}\).

Given that the second time is 1 s longer:

\(\dfrac{4}{a}=\dfrac{1}{a}+1\).

So \(\dfrac{3}{a}=1\), hence \(a=3\text{ m s}^{-2}\).

Use \(s=ut+\dfrac12at^2\).

Here \(s=100\), \(t=10\), \(a=-0.4\).

So \(100=10u+\dfrac12(-0.4)(10^2)=10u-20\).

Hence \(10u=120\), so \(u=12\text{ m s}^{-1}\).

Now use \(v=u+at=12+(-0.4)(10)=8\text{ m s}^{-1}\).

Use \(v^2=u^2+2as\).

So \(v^2=10^2+2(0.1)(80)=100+16=116\).

Thus \(v=\sqrt{116}\approx 10.77\text{ m s}^{-1}\).

This is below the safe limit of 11 m s^-1.

Therefore he should do nothing: braking would make him slower, and pedalling would make him less safe.