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30 July, 07:05

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

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  1. 30 July, 07:27
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    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
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