A car A is traveling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B which is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of `20 "km/h"^2`. After what distance and time will B catch up with A

Explanation

Suppose the car B catches up the car A in t hours.
The distance travelled by the car A moving with a velocity of 60 km `hr^(-1)` in time t hours is given by `S_1` -60 tkm.
For motion of car B
The distance travelled by the car B moving with an intrial velocity u = 70 km `hr^(-1)` and decelerated at the rate of 20 km `hr^(-2)` in time t hour is given by
` s_2 = u t + at ^2 = 70 t + 1/2 (-20) t^2 , s_2 = 70 t -10 t^2 (or) S_1 -s_1 = 2.5 km`
` therefore 70 t - 10 t^2 + 2.5 =0 or 4t^2 - 4t +1=0 (or) ( 2t -1)^2 =0 i.e, 2t -1=0 , t=1//2 = 0.5 hr^(-1)`
Using relative velocity concept,
` u= 70-60 = 10 km hr^(-1) , v =60-60 =0 , a =-20 km hr^(-1)`
`t=( v-u) /(a) = (0-10)/(-20) = 0.5 hr^(-1) ,` since `S_2 = 70 t - 10 t^2`
putting value of t we get `S_2 = 70 xx 0.5 - 10 xx (0.5 )^2 = 35-3.5 therefore S_2 = 32.5 km`