The velocity of a particle moving along \(x\)-axis is given as \(V = x^2 - 5x + 4\) (in m/s) where \(x\) denotes the \(x\)-coordinate of the particle in metres. The magnitude of the acceleration of the particle when the velocity of the particle zero is

Given the velocity of the particle as a function of position:

\(V = x^2 - 5x + 4\)

The acceleration \(a\) of a particle is given by the chain rule:

\(a = \dfrac{dV}{dt} = \dfrac{dV}{dx} \dfrac{dx}{dt} = V \dfrac{dV}{dx}\)

We are asked to find the acceleration when the velocity is zero (\(V = 0\)).

First, find the positions where \(V = 0\):

\(x^2 - 5x + 4 = 0\)

\((x - 1)(x - 4) = 0 \Rightarrow x = 1 \text{ or } x = 4\)

Now, find the derivative of velocity with respect to \(x\):

\(\dfrac{dV}{dx} = 2x - 5\)

At \(x = 1\), \(\dfrac{dV}{dx} = 2(1) - 5 = -3\)

At \(x = 4\), \(\dfrac{dV}{dx} = 2(4) - 5 = 3\)

Substituting \(V = 0\) and \(\dfrac{dV}{dx}\) into the acceleration formula for either position:

\(a = 0 \times (\pm 3) = 0\)

Thus, the magnitude of the acceleration when the velocity is zero is \(0\).

Answer: Zero