Acceleration + velocity calculus question

Question

Explanation for Calculus question needed The acceleration $a\,\text{m/s}^2$ of a particle $P$ moving in a straight line is given by $a = 3(1-x^2)$ , where $x$ metres is the displacement of the particle to the right of the origin. Initially the particle is at the origin moving with a velocity of $4\,\text{m/s}$. Show that the velocity $v\,\text{m/s}$ of the particle is given by $v^2 = 16+6x-2x^3$.

Answer 1

You have a differential equation $$\ddot{x}=3(1-\frac{1}{x^2})$$

You have initial conditions:

$$\begin{equation} x(0)=0 \\ \dot{x}(0)=4 \end{equation}$$

Now you only need to solve it and then invert the relation between t and x to obtain speed at each coordinate.

Answer 2

we know that; $a=\frac{dv}{dt}$

$a= \frac{dv}{dt}\cdot\frac{dx}{dx}$

$a= \frac{dv}{dx}\cdot v$ $\qquad$as $\frac{dx}{dt} =v$

$a\cdot\,dx=v\cdot dv$

integrate on both sides;

$\int a\,dx=\int v\,dv$

$\int (3-3x^2)\,dx=\int v\,dv$

$3x-x^3+C = \frac{v^2}2$

$v^2 =6x-2x^3+C$

given thatat origin velocity is $4\,ms^{-1}$

$16=0+C$

$C=16$

therefore the equation is ;

$v^2=6x-2x^3+16$

Answer 3

Since $F=ma$ we know that the force acting on the particle as a function of its mass $m$ and its displacement $x$ is

$F=m(3-3x^2)$

We also know that the change in energy of the particle is the integral of force with respect to distance, so

$E(x) - E(0) = \int_{0}^x m(3-3x^2) dx = m(3x - x^3)$

where $E(x)$ is the energy of the particle at displacement $x$. But this energy is all kinetic energy (no other forms of energy are mentioned in the question) so

$E(x) = \frac{1}{2}mv^2$

and, since we know that $v=4$ when $x=0$, we also have

$E(0) = 8m$

Now you have everything you need to solve the question.

(Update: the comments below are correct - the purely kinematic approach is simpler and does not require the assumption of constant mass)

Answer 4

Another try:

$\dfrac{dv}{dt} = a = 3(1-x^2).$

$\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt} =v\dfrac{dv}{dx};$

Hence:

$v\dfrac{dv}{dx}= 3(1-x^2).$

Separating variables:

$(v)dv = 3(1-x^2)dx$.

Integrating:

$(1/2)v^2 =3x - x^3 +C;$

For $x=0$: $v (x=0) =4$.

$C= (1/2)(16) =8.$

Finally :

$v^2= 6x -2x^3 +16.$