An iron rod of length `1.5m` lying on a horizontal table is pulled up from one end along a vertical line so as to move it with a constant velocity 3m/s, while the other end of the rod slides along the floor. After how much time the speed of the end sliding on the floor equals to the speed of the end being pulled up.

Options

A. `(1)/( 2 sqrt2) s`
B. `(1)/(sqrt2) s`
C. ` 3 sqrt2 s`
D. ` 1/4 s `

Answer

A

Explanation

Length of an iron rod, `l = 1.5 m`
Constant velocity,` V =3 m//s`
The diagram for given situation is given by

Since, velocity is constant, velocity of end P at point R is also `3 m//s`
After time 't', position of RS of rod is,
`(l -x)^(2) + (3t) ^(2) + = l ^(2)`
`(1.5 -x)^(2) + 9t ^(2) = (1.5) ^(2)`
`(1.5) ^(2) + x ^(2) - 2( 1.5) (x) + 9t^(2) = 2.25`
`2.25 + x ^(2) - 3x +_ 9t ^(2) = 2.25`
`x ^(2) - 3x + 9t ^(2) =0 " "...(1)`
Equation of motion of rod moving with constant velocity is,
`x = ut + 1/2 at ^(2)`
` x = ut + 0`
`x = 3t " "...(2)`
Substituting equation (2) in equation (1),
`(3t)^(2) - 3 (3t) + 9t ^(2) =0`
`9t ^(2) - 9t + 9t ^(2) =0`
`18t ^(2) - 9t =0`
` 2t -1 =0`
`2t -1 =0`
`therefore t = 1/2`