# please solve both A and B part

Initial velocity, u = 12 m/s

rate at which velocity decreases = acceleration, a = - 0.5 m/s

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(a) Let the time at which the particle comes to rest be t.

Then, using the kinematic relation for motion under uniform acceleration

$v=u+at\phantom{\rule{0ex}{0ex}}0=12-0.5t\phantom{\rule{0ex}{0ex}}t=\frac{12}{0.5}=\frac{120}{5}=24s$

(b) Let the distance covered before coming to rest be x.

Then, using another kinematic relation for motion under uniform acceleration

$x=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}x=12\times 24-\frac{1}{2}\times 0.5\times {24}^{2}=288-144=144m$

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