Car A and car B start moving simultaneously in the same direction along the line joining them. Car A with a constant acceleration a = 4 `m//s^2` , while car B moves with a constant velocity v = 1 m/s. At time t = 0, car A is 10 m behind car B. Find the time when car A overtakes car B.

Explanation

Given : `u_A-u_B=1m//s,a_A=4m//s^2` and `a_B=0`
Assuming car B to be at rest, we have
`u_(AB)=u_A-u_B=0-1=-1m//s`
`a_(AB)=a_A-a_B=4-0=4m//s^2`
Now, the problem can be assumed in simplified form as follow :

Substituting the proper values in equation

`s=ut+1/2at^2`
We get `10=-t+1/2(4)(t)^2`
or `2t^2-t-10=0`
or `t=(1+-sqrt(1+80))/4=(1+-sqrt(81))/(4)=(1+-9)/(4)` or t=2.5s
and -2s
Ignoring the negative value, the desired time is 2.5s.