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
v=u+at0=12-0.5tt=120.5=1205=24s

(b) Let the distance covered before coming to rest be x.
Then, using another kinematic relation for motion under uniform acceleration
x=ut+12at2x=12×24-12×0.5×242=288-144=144 m

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