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Car A and car B start moving simultaneou...

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.

A

`2.5s`

B

`3.5s`

C

`-2s`

D

none

Text Solution

Verified by Experts

The correct Answer is:
A
Given, `u_A=0, u_B=1 m//s^2, a_(A)=4 m//s^(2)` and `a_B=0`
Assuming car B to be at rest, we have
`u_(AB)=u_A-u_B=0-1=-1 m//s`
`a_(AB)=a_A-a_B=4-0=4 m//s^2`
Now, the problem can be assumed in simplified from as shown below.

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+-sqrt81)/4`
`(1+-9)/4`
or `t=2.5s and -2s`
Ignoring the negative value, the desired time is `2.5 s.`
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