Question

a motorcycle traffic police starts from rest at point A two seconds after a car speeding at the constant rate of 120 km/h passes point A (where speed limit is 100 km/h only) if the policeman accelerates at the rate of 6 m/s^2 until he reaches the speed of 150 km/h and maintains constant speed thereafter. calculate the distance, s, from point A to the point at which he overtakes the speeding car

Answer 1

120km/hr = 33.333 m/s
so, at time t, the car has gone 33.333 t meters

how long does the cop take to achieve 150km/hr? That's 41.666 m/s, so at 6m/s^2 it takes 6.9444 seconds.

During the 6.9444 seconds, he travels 1/2 (6)(6.9444)^2 = 144.67 m

So, the cop overtakes the car when

33.333t = 144.67 + 41.666(t-2-6.9444)
t = 27.362

check:
car(27.362) = 33.333*27.362 = 912m
cop(25.362) = 144.47+41.666*18.418 = 912m

so, the cop overtakes the car 912m from A.

Answer 2

120km/h=33.33m/s, 150km/h=41.67m/s
For motorcycle:
v^2=u^2+2*a*s1 (u=0 since at t=2 motorcycle is at rest)
s1=v^2/2a=41.67^2/(2*6)=144.68m
v=u+at1 (u=0)
t1=v/a=41.67/6=6.95s
s2=v(t-6.95) (1)
For car:
s3=v*t (2)
s3=s1+s2
33.33t=144.68+41.67(t-6.95)
t=17.38
and s3=s1+s2 => 579.3=579.3

so the cop overtakes the car 579.3 from A.