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Two particles A and B are initially 40 m...

Two particles `A` and `B` are initially `40 m`apart, `A` is behind `B`. Particle `A` is moving with uniform velocity of `10 m s^(-1)` towards `B`. Particle `B` starts moving away from `A` with constant acceleration of `2 m s^(-1)`.
The time for which there is a minimum distance between the two is .

A

2 s

B

4 s

C

5 s

D

6 s

Text Solution

Verified by Experts

The correct Answer is:
C

`s_(1)=vt`
`s_(2)=lo +1//2at^(2)=40+1//2xx2xxt^(2)=40+t^(2)`
`Deltas=40+t^(2)-10t`
`(Deltas)/(Deltat)=2t-10=0,t=5sec`
`(d^(2)s)/(dt^(2))=2("constant")`
so only one minimum in this function is at
t=5 sec
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