A body travelling with uniform acceleration crosses two point `A` and `B` with velocities `20 m s^-1` and `30 m s^-1` respectively. The speed of the body at the mid-point of `A` and `B` is.

Options

A. `25.5 ms ^(-1)`
B. `25 ms ^(-1)`
C. `24 ms ^(-1)`
D. `10sqrt6 ms ^(-1)`

Answer

A

Explanation

Let x be the distance between two points A and B and O is the mid-point of AB . Let a be the uniform acceleration of the body and v be the velocity of the body at point O respectively.
Using `v^(2) -u^(2)=2as`
According to given problem
`(v.)^(2)-(20)^(2) =2a(x)/(2) " "...(i)`
and `(30)^(2) -(v.)^(2) =2a(x)/(2) " "...(ii)`
Equating equations (i) and (ii) , we get
`(v.)^(2) -(20)^(2) =(30)^(2) -(v.)^(2)`
or `2(v.)^(2) = (30)^(2) +(20)^(2)`
or `v.=sqrt(((30)^(2)+(20)^2)/2)=sqrt((900+400)/2)=sqrt(650)=25.5ms^(-1)`