Question

The motion of a set of particles moving along the x‑axis is governed by the differential equation $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{,}$ where $\mathbf{x}\left(\mathbf{t}\right)$ denotes the position at time t of the particle.

⦁ If a particle is located at $\mathit{x}\mathbf{=}\mathbf{1}$ when $\mathit{t}\mathbf{=}\mathbf{1}$ , what is its velocity at this time?

⦁ Show that the acceleration of a particle is given by $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}\mathbf{.}$

⦁ If a particle is located at $\mathit{x}\mathbf{=}\mathbf{2}$when $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}$, can it reach the location $\mathit{x}\mathbf{=}\mathbf{1}$at any later time?

[Hint: ${\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{(}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{xt}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}\mathbf{.}$]

Short answer

⦁ 7

⦁ $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}$

⦁ No

Step-by-step solution

1(a): Finding the velocity by putting  x=1 and t=2  in  dxdt

Given, $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{,}$

Substituting $x=1$ and $t=2$ , one gets,

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{2}}^{\mathbf{3}}\mathbf{-}{\mathbf{1}}^{\mathbf{3}} \\ \mathbf{=}\mathbf{8}\mathbf{-}\mathbf{1} \\ \mathbf{=}\mathbf{7} \end{gathered}$$

i.e., Velocity at time $t=2$ and position $x=1$is 7.

Hence, the velocity is 7.

2(b): Differentiating  dxdt with respect to t.

$\frac{\mathbf{dx}}{\mathbf{dt}}$represents velocity while $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}$gives the acceleration of the particle.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{,} \\ \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{.}\frac{\mathbf{dx}}{\mathbf{dt}} \\ \mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{(}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{)} \\ \mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}} \end{gathered}$$

Hence, it is shown that the acceleration of a particle is given by $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}\mathbf{.}$

 3(c): Analyzing the position of particle at different time

From the graph, it is visible that the particle is stabilized about the velocity $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{1}$

Now, if $x=1$ and $t=2.5$, then $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{>}\mathbf{0}$, i.e., the particle moves away from $x=1$.

Thus, the position of the particle keeps on increasing and can reach x = 1 only once.