When the particle crosses the origin, would its speed (velocity) be finite?
Relevant equations:
| Symbolic | Word |
|---|---|
| \(F=ma\) | \(\text{force}=\text{mass}\times\text{acceleration}\) |
| \(E=Fd\) | \(\text{energy transferred}=\text{force}\times\text{distance}\) |
| \(E_k=\frac{1}{2}mv^2\) | \(\text{kinetic energy}=\frac{1}{2}\text{mass}\times\text{velocity}^2\) |
Since the force on the particle is not constant, the total (kinetic) energy gained by the particle can be calculated with
$$ \begin{align*} E_k&=\int^5_0Fdr\\ &=\int^5_0\frac{1}{r^2}dr\\ &=\infty \end{align*} $$
Calculating the corresponing velocity of this kinetic energy, it’s obvious that \(v=\infty\).
Express \(r\) as a function of \(t\).
My current best attempt only goes up to writing an equation down representing the situation, but I was not able to solve it, I don’t even know if this equation is the simplest way to represent the situation.
Expressing \(r\) and \(v\) as functions of \(t\): $$ \begin{align*} r(t)&=r(t-dt)-v(t)dt\\ v(t)&=v(t-dt)+\frac{1}{r(t)^2}dt \end{align*} $$
With the starting condition \(r(0)=5\) and \(v(0)=0\), as \(dt\to 0\), the functions become more and more accurate.
However, I could not find a way to write it as a non-iterative formula, hence its unsolved.