91Ó°ÊÓ

Given data is $F\left(x\right)=-2m/x^5,v=-1,x=1$, at $t=0$.

The definition of velocity is the rate of change of distance. i.e.$v\left(x\right)=\frac{dx}{dt}$. It is a vector quantity.

From the given information, noticing the given equation:

$m\frac{d^{2}x}{d{t}^2}=F\left(x\right)$ ……. (1)

With denoting differentiation w.r.t the time variable, can be cast into the more convenient form by multiplying it's both sides with$v\left(x\right)=\frac{dx}{dt}$:

$m\frac{d^{2}x}{d{t}^2}\frac{dx}{dt}=F\left(x\right)\frac{dx}{dt}$

This form can be simplified by noting that the first fraction on the LHS of the previous equation is a second time derivative of position, a first-time derivative of velocity, so inserting$\frac{dv}{dt}=\frac{d}{dt}\frac{dx}{dt}=\frac{d^{2}x}{d{t}^2}$into it:

$m\frac{dv}{dt}\frac{dx}{dt}=F\left(x\right)\frac{dx}{dt}$

Further simplifying the expression by exploiting the definition of velocity$v\left(x\right)≡\frac{dx}{dt}$:

$mv\frac{dv}{dt}\frac{dx}{dt}=F\left(x\right)\frac{dx}{dt}$

After multiplying both sides of the previous equation by$dt$ to separate the integration integral, to obtain:

$mvdv=F\left(x\right)dt$

Now integrate it by using the table integral$∫xdx=\frac{1}{2}{x}^2+$const. to demonstrate the convenient form of the equation (1):

$\frac{1}{2}m{v}^2\left(x\right)=∫F\left(x\right)dx+\text{const.}$ ……. (2)

By given the function$F\left(x\right)=-2m/x^5,v=-1,x=1$find the term on LHS of the above equation:

$∫F\left(x\right)dx=-∫\frac{2m}{x^5}dx=\frac{1}{2}\frac{m}{x^4}+\text{const}$

Since$∫\frac{1}{x^5}dx=-\frac{1}{4}\frac{1}{x^4}+$const. inserting this result into equation (2) and denoting the overall integration constant as $A$, to obtain:

$\frac{1}{2}m{v}^2=\frac{1}{2}\frac{m}{x^4}+A$

Integrate it by using the table integral $∫xdx=\frac{1}{2}{x}^2+$const to demonstrate the convenient form of the equation (1):

$\frac{1}{2}m{v}^2\left(x\right)=∫F\left(x\right)dx+\text{const.}$

By given the function$F\left(x\right)=m/x^3,v=0,x=1$find the term on LHS of the above equation:

$∫F\left(x\right)dx=∫\frac{m}{x^3}dx=-\frac{1}{2}\frac{m}{x^2}+\text{const}$

Since$∫\frac{1}{x^5}dx=-\frac{1}{4}\frac{1}{x^4}+$const. inserting this result into equation (2) and denoting the overall integration constant as$A$, to obtain:

$\frac{1}{2}m{v}^2=\frac{1}{2}\frac{m}{x^4}+A$

The initial condition evaluates the previous equation at time$t=0$to obtain the constant$A$

$$\begin{gathered} \frac{1}{2}m\left[v\right(t=0){]}^2=\frac{1}{2}\frac{m}{\left|x{(t=0)}^4\right|}+A \\ \frac{1}{2}m=\frac{1}{2}m+A \\ A=0 \end{gathered}$$

Rewrite the equation as:

$\frac{1}{2}m{v}^2=\frac{1}{2}\frac{m}{x^4}$

After diving through by$\frac{m}{2}$as:

$v^2=\frac{1}{x^4}$

Rewrite the above expression using$v\left(x\right)=\frac{dx}{dt}$and taking the square root as:

$\frac{dx}{dt}=Â\pm \frac{1}{x^2}$

It is possible to choose the appropriate sign, since$v\left(x\right)=\frac{dx}{dt}$is negative at time$t=0$and$x^2$square is positive. So that if we separate the equation to obtain:

$∫x^{2}dx=-∫dt$

Both side of the previous equation are easily integrated using the table integral

$$\begin{gathered} ∫x^{n}dx=\frac{1}{n+1}{x}^{n+1}+C \\ \frac{1}{3}{x}^3=-t+A \end{gathered}$$

With the integration constant labeled by $A$. Exploiting the initial condition $x(t=0)=1$to obtain a numerical expression for $A$:

$$\begin{gathered} \frac{1}{3}=0+A \\ A=\frac{1}{3} \end{gathered}$$

Insert it into the solution to arrive at the expression:

$$\begin{gathered} \frac{1}{3}{x}^3=-t+\frac{1}{3} \\ {x}^3=1-3t \\ x=(1-3t{)}^{\frac{1}{3}} \end{gathered}$$

Thus, the given function's solution is $x=(1-3t{)}^{\frac{1}{3}}$.