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Two cars A and B are racing along straig...

Two cars `A` and` B` are racing along straight line. Car `A` is leading, such that their relative velocity is directly proportional to the distance between the two cars. When the lead of car `A` is `l_(1)=10m` , its running `10m//s` faster than car `B`. Determine the time car `A` will take to increase its lead to `l_(2)=20m` from car `B`.

Text Solution

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The correct Answer is:
B, C

As given
`(V_(A)-V_(B))x_(A)-x_(B)(V_(A)-V_(B))=K(x_(A)-x_(B))`
when `x_(A)-x_(B) =10`
We have `V_(A)-V_(B) =10`
We get
`10 = K10 rArr K=1`
`rArr V_(A)-V_(B)=(x_(A)-x_(B))`....(A)
Now let
`x_(A) -x_(B) =y` ....(B)
On differentiating with respect to 't' on both side.
`rArr (dx_(A))/(dt)-(dx_(B))/(dt)=(dy)/(dt)`
`V_(A)-V_(B)=(dy)/(dt)` .....(C)
Using (A),(B),(C)
We get `(dy)/(dt) =y` Here `y` represent sepration between two cars
`rArr underset(10)overset(20)(int) (dy)/(y)=underset(0) overset(t)(int) dt`
`rArr [log_(e)y]_(10)^(20)=t`
`t=(log_(e)2)sec`.
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