Car A and car B start moving simultaneously in the same direction along the line joining them. Car A moves 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 overtake car B.

Options

A. `2.5 s`
B. `2 s`
C. `3 s`
D. `3.5 s`

Answer

A

Explanation

Given, `u_(A)=0, u_(B)=1 ms^(-1), a_(A) = 4 ms^(-2)` and `a_(B) = 0`
Assuming car B to be rest, we have
`u_(AB)=u_(A)=0-1=-1 ms^(-1)`
`a_(AB)=a_(A)-a_(B)=4-0=4 ms^(-2)`
Now, the problem can be assumed in simplified from as follows

Substituting the proper values in equation
`s=ut+(1)/(2)at^(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 `-2 s`
Ignoring the negative value, the desired time is `2.5 s`
Note : The above problem can also be solved without using the concept of relative motion as under.
At the time when A overtakes B, `S_(A)=S_(B)+10`
`therefore (1)/(2)xx4xxt^(2)=1xxt+10` or `2t^(2)-t-10=0`
Which on solving gives `t = 2.5 s` and `-2 s`, the same as we found above. As per my opinion, this approach (by taking values) is more suitable in case of two body problem in one dimensional motion. Let us see one more example in support of it.