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 `60km hr^(-1) ` in time t hours is given by , `S_(1)=60t km`
For motion of car B
The distance travelled by the car B moving with an intrial velocity `u=70km hr^(-1)` and decelerated at the rate of `20km hr^(-2)` in time t hour is given by,
`S_(2)=ut+at1/2at^(2)=70t+1/2(-20)t^(2),S_(2)=70t-10t^(2)"(or)But" S_(2)-S_(1)=2.5km`
`:. 70t-10t^(2)-60t=2.5`
i.e., `10t^(2)-10t+2.5=0"or"4t^(2)-4t+1=0 "(or)" (2t-l)^(2)=0`i.e.,`2t-l=0, t=1//2=0.5 hr^(-1)`
using relative velocity concept,
`u=70-60=10km hr^(-1), v=60-60=0,a=-20km hr^(-1)`
`t=(v-u)/a=(0-10)/(-20)=0.5 hr^(-1), ` Since `S_(2)=70t-10t^(2)`
Putting value of t, we get , `S_(2)=70xx0.5-10xx(0.5)^(2)=35-3.5:. S_(2)=32.5km`