A particle located at "x=0" at time "t=0" starts moving along the positive "x" - direction with a velocity "v" that varies as "v=alpha sqrt(x)".The displacement "(x)" of the particle varies with time as "(alpha" is constant)

A particle located at x = 0 at time t = 0 , starts moving along with the positive x-direction with a velocity 'v' that varies as v = a sqrt(x) . The displacement of the particle varies with time as

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity that varies as v=psqrtx . The displacement of the particle varies with time as (where, p is constant)

A particle located at position x=0, at time t=0, starts moving along the positive x-direction with a velocity v^2=alpha x (where alpha is a positive constant). The displacement of particle is proportional to

If the displacement of a particle varies with time as sqrt x = t+ 3

If the displacement of a particle varies with time as sqrt(x) = t + 7 . The -

If the velocity "v" of a particle varies as the square of its displacement "x" then the acceleration varies as

" If the velocity v of a particle varies as the square of its displacement "x" then the acceleration varies as "

The displacement to particle is zero at t=0 and is z at t= t . It starts moving in the positive x-direction with a velocity which varies, v= k sqrt x , wher (k) is a constant. Find the relation for variation of velocity with time.

The dispkacement of particle is zero at t= 0 and it is x , at t-t . It starts moving in the positive x-direction with a velocity which varices as v= k sqrt x , where (k) is a constant. Show that velocity is directly proprtional to time.

Velocity of a particle varies with time as v=4t. Calculate the displacement of particle between t=2 to t=4 sec.