Question

The motion of a set of particles moving along the x-axis is governed by the differential equation dx/dt=t^3-x^3 where x(t) denotes the position at time t of the particle. a. If a particle is located at x=1 when t=2, what is its velocity at this time? b. Show that the acceleration of a particle is given by d^2x/dt^2=3t^2-3t^3x^2+3x^5 c. If a particle is located at x=2 when t=2.5, can it reach the location x=1 at any later time? (hint: t^3-X^3=(t-x)(t^2+xt+x^2)

Answer 1

$\dfrac{\mathrm dx}{\mathrm dt}$ gives the velocity at any point $(t,x)$, so you need only plug these values in:

$\dfrac{\mathrm dx}{\mathrm dt}\bigg|_{(t,x)=(2,1)}=2^3-1^3=7$

Acceleration can be determined by differentiating the velocity function:

$\dfrac{\mathrm d^2x}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{\mathrm dx}{\mathrm dt}\right]$
$=\dfrac{\mathrm d}{\mathrm dt}[t^3-x^3]$
$=3t^2-3x^2\dfrac{\mathrm dx}{\mathrm dt}$
$=3t^2-3x^2(t^3-x^3)$
$=3t^2-3x^2t^3+3x^5$

In order for any particle to reach the point $(t,x)=(2.5,2)$, we require that $x(t)$ be decreasing for some value of $t>2.5$. In other words, $\dfrac{\mathrm dx}{\mathrm dt}$ must be negative for some $t>2.5$. This basically amounts to solving the inequality,

$\dfrac{\mathrm dx}{\mathrm dt}=t^3-x^3=(t-x)(t^2+xt+x^2)<0$

Note that the left hand side is negative when either $t-x<0$ or $t^2+xt+x^2<0$, but not both.

When $x=1$, in the first case you have $t-1<0\implies t<1$. But this isn't possible, since $t$ (time) must always be growing in the positive direction.

Meanwhile, in the second case, you have

$t^2+t+1=\left(t+\dfrac12\right)^2+\dfrac34>0$

for all $t$, which means $\dfrac{\mathrm dx}{\mathrm dt}$ can never be negative, which in turn means $x(t)$ must be increasing. Therefore it's not possible for any particle to reach a position of $x=1$ for any $t>2.5$.

Answer 2

The velocity at this time is 7 meters per second.

The acceleration of the particle is $\rm \dfrac{d^2x}{dt^2} = 3t^2-3t^3x^2+3x^5$.

It's not possible for any particle to reach a position of x =1 for any t >2.5.

Given that

The motion of a set of particles moving along the x-axis is governed by the differential equation;

$\rm \dfrac{dx}{dt}=t^3-x^3$

Where x(t) denotes the position at time t of the particle.

According to the question

The motion of a set of particles moving along the x-axis is governed by the differential equation;

$\rm \dfrac{dx}{dt}=t^3-x^3$

1. If a particle is located at x=1 when t=2, what is its velocity at this time is,

$\rm \rm \dfrac{dx}{dt}=t^3-x^3\\\\\rm \dfrac{dx}{dt}=2^3-1^3\\\\\rm \dfrac{dx}{dt}=8-1\\\\\rm \dfrac{dx}{dt}=7$

The velocity at this time is 7 meters per second.

2. Show that the acceleration of a particle is given by,

The acceleration of the particle is given by,

$\rm \dfrac{d^2x}{dt^2} = {3t^2-3x^2}{\dfrac{dx}{dt}}\\\\\dfrac{d^2x}{dt^2} = {3t^2-3x^2} \times (t^3-x^3)\\\\\dfrac{d^2x}{dt^2} = 3t^2-3t^3x^2+3x^5$

3. If a particle is located at x=2 when t=2.5, can it reach the location x =1 at any later time.

it's not possible for any particle to reach a position of x =1 for any t >2.5.

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