A particle initially at rest starts moving from reference point x=0 along x-axis, with velocity v that varies as v=4 √x m / s. The acceleration of the particle is __________ ms⁻².

Solution

Physics sir · 4 steps

Velocity is given as a function of x, so use a = v dv/dx and the x cancels out.

1. Right formula. Here v depends on position, not time. So write acceleration as a = dv/dt = dv/dx·dx/dt = vdv/dx.
2. Differentiate. With v = 4√x, we get dv/dx = 4 × 1/2√x = 2/√x.

   a = vdv/dx = 4√x × 2/√x = 8 m s⁻²

3. Quick check. Squaring gives v² = 16x. Starting from rest at x = 0, uniform acceleration gives v² = 2ax, so 2a = 16 and a = 8 m s⁻², a constant.
4. So. The acceleration of the particle is 8 m s⁻², the same at every point.

The common slip: 16. Matching v² = 16x with v² = ax and forgetting the factor 2 in v² = 2ax gives 16, not 8.

The idea to keep. When velocity is given in terms of position, use a = vdv/dx. If v² comes out proportional to x, the acceleration is constant and equals half the coefficient.

Answer from the NTA final answer key. Solution by PrepWiser.