Question

A particle moves uniformly with speed v along a parabolic path y = kx^(2), where k is a positive constant. Find the acceleration of the particle at the point x = 0.

Answer 1

Therefore the acceleration of the particle at the point x=0 is 2k.

Explanation:

Velocity: The ratio of distance to time.

$Velocity=\frac{Distance}{time}$

Acceleration: The ratio of difference between of initial velocity and final velocity to the time.

$Acceleration =\frac{\textrm{Change of velocity}}{Time}$

Rules of Derivatives:

• $\frac{d}{dx}(ax^b)=a\frac{d}{dx}(x^b)$

• $\frac{d}{dx}(x^n)=nx^{n-1}$  
• $\frac{dx}{dx}=1$

Given that, a particle moves uniformly along a parabolic path

$y=kx^2$

To find the acceleration, we have to find out the second order derivative of the path.

It means $\texrm{Acceleration}=a=\frac{d^2y}{dx^2}$

∴$y=kx^2$

Differentiating with respect to x

$v= \frac{dy}{dx}= \frac{d}{dx}(kx^2)$

           $=k\frac{d}{dx}(x^2)$

            =2kx

Again differentiating with respect to x

$a=\frac{d^2y}{dx^2}=\frac{d}{dx}(2kx)$

           $=2k\frac{d}{dx}(x)$

           $=2k$

$\therefore[\frac{d^2y}{dx^2}]_{x=0}=2k$

Therefore the acceleration of the particle is 2k.