Question

Robbers in a car traveling at 20m/s pass a policeman on the motorcycle at rest. The policeman immediately starts chasing the robbers . The policeman accelerates at 3m per second square for 12s and there after travels at the constant velocity. Calculate the distance covered by the policeman before he overtake the car

Answer 1

s = ut + 0.5 at^2

2 = 2 u + 0.5*2^2*a
1 = u + 2 a.......................(1)`


2.00 + 2.20 = u ( 2 + 5) + 0.5 a ( 2+5)^2
4.20 = 7 u + 24.5 a.................(2)

mutiply (1) by 7, than subtract (2) from (1)
2.80 = - 10.5 a
a = - 0.2667 m /s^2
subtitute value of a in (1)

1= u + 2*(- 0.2667)
u = 1.5333 m/s initial velocity
v = u + at
= 1.5333 - 0.2667*7 = -0.3336 m/s................ans (your ans 2.22m/s seems to be not correct)

2.
In 12 s police man will travel s = ut + 0.5at^2 = 0 + 0.5*3*144 = 216 m
final velocity of police man after 12 s = v = u + at = 0 + 3*12 = 36 m/s

Suppose he over takes in 12 + t sec, in which he travels = 216 + 36 * t m

thr robbers will travel in the same time 20 * ( 12+t) m
216 + 36 t = 240 + 20 t
t = 24 / 16 = 1.5 s

distance traveled by policeman in 12 + 1.5 = 13.5 s
is 216 + 36* 1.5 = 270 m........................ans

I hope this helps you