# Kinematics (Complex)

• May 30th 2008, 02:40 PM
Simplicity
Kinematics (Complex)
Question:
A particle moves on the positive x-axis. The particle is moving towards the origin $O$ when it passes through the point $A$, where $x=2a$, with speed $\sqrt{\left(\frac{k}{a}\right)}$, where $k$ is constant. Given that the particle experiences an acceleration $\frac{k}{2x^2} + \frac{k}{4a^2}$ in a direction away from $O$, show that it come instantaneously to rest at a point $B$, where $x=a$. Immediately the particle reaches B the acceleration changes to $\frac{k}{2x^2} - \frac{k}{4a^2}$ in a direction away from $O$. Show that the particle next comes instantaneously to rest at $A$.

My Problem with this question:
I have no idea on how to approach it. Please help. Thanks in advance.
• May 30th 2008, 03:00 PM
Mathstud28
Quote:

Originally Posted by Air
Question:
A particle moves on the positive x-axis. The particle is moving towards the origin $O$ when it passes through the point $A$, where $x=2a$, with speed $\sqrt{\left(\frac{k}{a}\right)}$, where $k$ is constant. Given that the particle experiences an acceleration $\frac{k}{2x^2} + \frac{k}{4a^2}$ in a direction away from $O$, show that it come instantaneously to rest at a point $B$, where $x=a$. Immediately the particle reaches B the acceleration changes to $\frac{k}{2x^2} - \frac{k}{4a^2}$ in a direction away from $O$. Show that the particle next comes instantaneously to rest at $A$.

My Problem with this question:
I have no idea on how to approach it. Please help. Thanks in advance.

HAve you tried taking the derivative and realizing that there is a max or min when v(t)=0?