In Problem 11, find $v\left(x\right)$if$v=0,x=1,$at$t=0$. Then write an integral for$t\left(x\right)$.

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

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

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

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:

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

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

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

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 $\int xdx=\frac{1}{2}{x}^2+$const. to demonstrate the convenient form of the equation (1):

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

……. (2)

By given the function $F\left(x\right)=-2m/x^5,x=1,t=0$find the term on LHS of the above equation:

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

Since $\int \frac{1}{x^5}dx=-\frac{1}{4}\frac{1}{x^4}+$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$A$

Rewrite the equation as:

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

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

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

Taking the square root of the previous expression, obtained a solution for $v\left(x\right)$, i.e. for velocity as a function of the function of the position:

$v\left(x\right)=\pm \sqrt{\frac{1}{x^4}-1}$

Exploiting the definition of velocity $v\equiv \frac{dx}{dt}$,to find that:

The initial moment $t=0$to some arbitrary constant, defined by position $x$and time $t$:

Using the table integral$\int dx=x+$const. multiplying both sides of the previous equation by$\pm$yield a final a final expression for$t\left(x\right)$:

$t\left(x\right)=\pm {\int }_1^x\frac{dy}{\sqrt{\frac{1}{y^4}-1}}$

Thus, given function's solution isrole="math" localid="1664366427306" $t\left(x\right)=\pm {\int }_1^x\frac{dy}{\sqrt{\frac{1}{y^4}-1}}$ .