The given situation of entire motion of motor bike is shown in the figure.


If time taken by the motorbike to reach from point $A$ to $B$ is $t_{A B}$, then for $A$ to $B$,
$v_A=0, v_B=10 \mathrm{~ms}^{-1}, a=0.5 \mathrm{~ms}^{-2}$
$\therefore$ From equation, $v=u+a t$
$\begin{aligned}
v_B & =v_A+a t_{A B} \Rightarrow 10=0+0.5 t_{A B} \\
\Rightarrow \quad t_{A B} & =\frac{10}{0.5}=20 \mathrm{~s}
\end{aligned}$
Since, motorbike moves from point $B$ to $C$ with constant velocity. Hence, time taken to travel distance $B C$ is given as
$t_{B C}=\frac{B C}{v_B}=\frac{10 \mathrm{~km}}{10 \mathrm{~ms}^{-1}}=\frac{10000 \mathrm{~m}}{10 \mathrm{~ms}^{-1}}=1000 \mathrm{~s}$
For motion of motorbike from point $C$ to $D$, time $=t_{C D}, v_B=10 \mathrm{~ms}^{-1}, v_f=0, a^{\prime}=0.2 \mathrm{~ms}^{-2}$
From equation, $v=u-a t$,
$\begin{gathered}
v_f=v_B-a^{\prime} t_{C D} \\
0=10-0.2 \times t_{C D} \Rightarrow t_{C D}=50 \mathrm{~s} \\
\therefore \text { Total time }=t_{A B}+t_{B C}+t_{C D} \\
=20 \mathrm{~s}+1000 \mathrm{~s}+50 \mathrm{~s}=1070 \mathrm{~s}
\end{gathered}$