# 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 meters. Find the magnitude of acceleration of the particle when the velocity of particle is zero?

## A: $0 \frac{m}{s} ^ 2$ B: $2 \frac{m}{s} ^ 2$ C: $3 \frac{m}{s} ^ 2$ D: None of the above

Aug 8, 2017

A

#### Explanation:

Given velocity
v=x^2−5x+4
Acceleration $a \equiv \frac{\mathrm{dv}}{\mathrm{dt}}$

:.a=d/dt(x^2−5x+4)
=>a=(2x(dx)/dt−5(dx)/dt)

We also know that $\frac{\mathrm{dx}}{\mathrm{dt}} \equiv v$

=>a=(2x −5)v

at $v = 0$ above equation becomes
$a = 0$