The initial velocityof a particle along x axis is u at t=0 x=0 and its acceleration is given by a =2x then whats the correct equation for v^2= u^2 +2as

v² = u² + 2x²

Explanation:

v² = u² + 2as is only valid for constant acceleration.  Here, the acceleration is a function of position.  We can find the function of velocity by integrating.  Acceleration is the derivative of velocity with respect to time:

a = 2x

dv/dt = 2x

Apply chain rule:

dv/dt = dx/dt × dv/dx

dv/dt = v × dv/dx

Therefore:

v dv/dx = 2x

Separate the variables and integrate:

v dv = 2x dx

½ v² |ᵤᵛ = x² |₀ˣ

½ (v² − u²) = x²

v² − u² = 2x²

v² = u² + 2x²

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )