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A particle moving in a straight line is acted upon by a force which works at a constant rate and chages ist melocity from (u and v ) over a distance x. Prove that the taken in it is
`3/2 (u+v)x/(u^(2)+v^(2)+uv)` .

A

`t=(3x(u+v))/(2(u^(2)+v^(2)+uv))`

B

`t=(3x(u-v))/(2(u^(2)+v^(2)+uv))`

C

`t=(3x(u+v))/(2(u^(2)+v^(2)-uv))`

D

`t=(3x(u+v))/(2(u^(2)-v^(2)+uv))`

Text Solution

Verified by Experts

The correct Answer is:
A
From work-energy principle, `W=DeltaKE`
`:. Pt=1/2m(v^(2)-u^(2)) (P="power")`
or `t=m/(2P)(v^(2)-u^(2))`
Further
`F.v=P`
`:. m.(dv)/(ds).v^(2)=P`
or `int_(u)^(v)v^(2)dv =P/mint_(0)^(x)`
`:. (v^(3)-u^(3) =(3P)/m.x`
or `m/P =(3x)/(v^(3)-u^(3)))`
Substituting in Eq. (i)
`t=(3x(u+v))/(2(u^(2)+v^(2)+uv))` Hence proved.
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