please solve both A and B part

Solution,
Initial velocity, u = 12 m/s
rate at which velocity decreases = acceleration, a =  - 0.5 m/s²
(a) Let the time at which the particle comes to rest be t.
Then, using the kinematic relation for motion under uniform acceleration
$$\begin{gathered} v=u+at \\ 0=12-0.5t \\ t=\frac{12}{0.5}=\frac{120}{5}=24s \end{gathered}$$

(b) Let the distance covered before coming to rest be x.
Then, using another kinematic relation for motion under uniform acceleration
$$\begin{gathered} x=ut+\frac{1}{2}a{t}^2 \\ x=12\times 24-\frac{1}{2}\times 0.5\times {24}^2=288-144=144 m \end{gathered}$$