A particle moves along `x`-axis in positive direction. Its acceleration `'a'` is given as `a=cx+d`, where `x` denotes the `x`- coordinate of particle, `c` and `d` are positive constants. For velocity position graph of particle to be of type as shown in figure, the value of speed of particle at `x=0` should be

Options

A. `sqrt((2d^(2))/c)`
B. `sqrt((8d^(2)/c)`
C. `sqrt((4d^(2))/c)`
D. `sqrt((d^(2))/c)`

Answer

D

Explanation

`a=V(dv)/(dx)=cx+d`
Let at `x=0 v=u`
`:.int_(u)^(v) vdv=int_(0)^(x)(cx+d) dx`
or `v^(2)=cx^(2)+2dx+u^(2)`
`v` shall be linear function of `x` if `cx^(2)+2dx+u^(2)` is perfect square