A particle move a distance `x` in time `t` according to equation `x = (t + 5)^-1`. The acceleration of particle is alphaortional to.

Options

A. `("velocity")^(3//2)`
B. `("distance")^(2)`
C. `("distance")^(-2)`
D. `("velocity")^(2//3)`

Answer

A

Explanation

Given, `x=(1)/(t+5)`
Differentiating both sides w.r.t. t, we get
`(dx)/(dt)=-(1)/((t-5)^(2))xx1`
`therefore` Velocity, `upsilon =(dx)/(dt)=-(1)/((t+5)^(2))` ….(i)
Again differentiating both sides w.r.t. t, we get
`(d upsilon)/(dt)=(1xx2)/((t+5)^(3))(1)=(2)/((t+5)^(3))` ....(ii)
Now, from Eqs. (i) and (ii), we get
`(d upsilon)/(dt)=-2upsilonxx(1)/((t+5))`
`=(-2upsilon)(sqrt(upsilon))=-2(upsilon)^(3//2)`
`therefore` Acceleration, `a=(d upsilon)/(dt)=-2upsilon^(3//2)rArr a prop ("velocity")^(3//2)`