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

Step-by-step solution

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 = dV dt = dV dx dx dt = V 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 x = 1 or x = 4 Now, find the derivative of velocity with respect to x : dV dx = 2x - 5 At x = 1 , dV dx = 2(1) - 5 = -3 At x = 4 , dV dx = 2(4) - 5 = 3 Substituting V = 0 and dV dx into the accel