(a) For the first \(10\text{ s}\),

\[ u=0,\quad a=0.5,\quad t=10. \]

So the speed after \(10\text{ s}\) is

\[ v_1=u+at=0+0.5(10)=5\text{ m s}^{-1}. \]

For the next part of the motion,

\[ u=5,\quad a=1.5,\quad s=13. \]

Using \(v^2=u^2+2as\),

\[ v^2=5^2+2(1.5)(13)=25+39=64. \]

Therefore

\[ v=8. \]

Answer: \(v=8\text{ m s}^{-1}\).

(b)(i) The distance travelled in the first \(10\text{ s}\) is

\[ s_1=\frac12(0.5)(10)^2=25\text{ m}. \]

The next distance is \(13\text{ m}\).

The distance travelled at constant speed is

\[ s_3=8(5)=40\text{ m}. \]

The final distance is \(28\text{ m}\).

Hence the total distance is

\[ 25+13+40+28=106. \]

Answer: \(106\text{ m}\).

(b)(ii) The time for the second part is

\[ t_2=\frac{8-5}{1.5}=2\text{ s}. \]

For the final part, the cyclist decelerates from \(8\text{ m s}^{-1}\) to rest over \(28\text{ m}\). The average speed during this part is

\[ \frac{8+0}{2}=4\text{ m s}^{-1}. \]

So the time for the final part is

\[ t_4=\frac{28}{4}=7\text{ s}. \]

Total time:

\[ 10+2+5+7=24\text{ s}. \]

Average speed:

\[ \frac{106}{24}=4.416\ldots \]

Answer: \(4.42\text{ m s}^{-1}\).