Consider a particle moving along the positive direction of X-axis. The velocity of the particle is given by `v = alphasqrt(x)` (`alpha` is a positive constant). At time t = 0, if the particle is located at x = 0, the time dependence of the velocity and the acceleration of the particle are respectively 

Options

A. `(alpha^(2))/(2)t" and "(alpha^(2))/(2)`
B. `alpha^(2)t" and "alpha^(2)`
C. `(alpha)/(2)t" and "(alpha)/(2)`
D. `(alpha^(2))/(4)t" and "(alpha^(2))/(4)`

Answer

A

Explanation

Given, `v=alpha sqrt(x)` and at x = 0, t = 0 To find velocity and acceleration in terms of time, we need to find x as function of t.
`because " "(dx)/(dt)=alpha sqrt(x)`
`:." "(dx)/(sqrt(x))=alpha" dt"`
On integrating both sides, we get
`2sqrt(x)=alpha t+C`
Using, `" "x=0" at "t=0impliesC=0`
`:." "sqrt(x)=(alpha)/(2)t`
Hence, velocity, `v=(alpha^(2))/(2)t" "(because v=alpha sqrt(x))`
and acceleration, `a=(dv)/(dt)=(alpha^(2))/(2)`