Question

The velocity of a particle on the \(x\) axis, \(x \geq 0\), is always numerically equal to the square root of its displacement \(x\). If \(v=0\) when \(x=0\), find \(x\) as a function of \(t .\) Show that the given conditions are satisfied if the particle remains at the origin for any arbitrary length of time \(t_{0}\), and then moves away; find \(x\) for \(t>t_{0}\) for this case.

Short answer

\( x \) as a function of \( t \) is \( x = \frac{t^{2}}{4} \), and for \( t > t_{0} \), \( x = \frac{(t - t_{0})^{2}}{4} \).

Step-by-step solution

Understand the Given Information

The problem states that the velocity of a particle is equal to the square root of its displacement on the x-axis, where the velocity is given by: \( v = \frac{dx}{dt} = \sqrt{x} \). It is also given that \( v = 0 \) when \( x = 0 \).

Set Up the Differential Equation

We can start with the relationship between velocity and displacement: \( \frac{dx}{dt} = \sqrt{x} \).

Separate and Integrate

To solve for \( x \) as a function of \( t \), separate the variables and integrate both sides: \( \int \frac{dx}{\sqrt{x}} = \int dt \). Using the substitution \( u = x^{1/2} \) simplifies the integral.

Perform the Integral

Performing the integral on both sides: \( 2\(\sqrt{x}\) = t + C \), where \( C \) is the integration constant. Since at \( t = 0 \), \( x = 0 \), it follows that \( C = 0 \).

Solve for \( x \)

Solving for \( x \), we get: \( 2\(\sqrt{x}\) = t \) yields \( x = (\frac{t}{2})^{2} = \frac{t^2}{4} \).

Consider the Particle at Rest for Arbitrary \( t_0 \)

To show the particle can remain at the origin for a time \( t_{0} \), we start by assuming \( x = 0 \) for \( t \leq t_{0} \). For \( t > t_{0} \), the equation must match our earlier solution.

Define \( x \) for \( t > t_{0} \)

For \( t > t_{0} \), reset the clock so the equation remains continuous: \( \frac{dx}{dt} = \sqrt{x} \) should be valid for \( t - t_{0} \). Then, \( x = \frac{(t - t_{0})^{2}}{4} \).

Key concepts

Particle Motion

Understanding particle motion is essential in different areas of physics. It describes how an object moves in space over time, including its path, velocity, and acceleration. In this exercise, we are analyzing a particle moving along the x-axis.
The particle's motion is characterized by its velocity, which in this case, is directly related to its displacement. This unique scenario allows us to use differential equations to determine the particle's position over time.
Let's break down the process step by step, starting with the given relationship between velocity and displacement and moving through the mathematical steps to find the position function, \(x(t)\).

Velocity-Displacement Relationship

The relationship between velocity and displacement is our starting point. Here, it's given that the velocity, \(v\), of the particle is the square root of its displacement, \(x\). This is expressed as:
\( v = \frac{dx}{dt} = \sqrt{x} \).
This equation tells us that as the particle moves, its speed at any point is the square root of its current position on the x-axis. It's a special case because typically, velocity is generally a function of both displacement and time. Here, however, we have a direct dependency on displacement alone.
To solve for the displacement, we need to convert this relationship into a form that allows integration. We can do this by separating the variables \(dx\) and \(dt\) on either side of the equation.

Variable Separation

Variable separation is a crucial technique in solving differential equations. It involves rearranging the equation so that all terms involving one variable (in this case, \(x\)) are on one side, and all terms involving the other variable (time \(t\)) are on the other.
In our scenario, starting with:
\( \frac{dx}{dt} = \sqrt{x} \),
we can separate variables and integrate both sides. This means rewriting the equation as:
\( \int \frac{dx}{\sqrt{x}} = \int dt \).
The integral on the left simplifies using the substitution \( u = \sqrt{x} \), to:
\(2\sqrt{x} = t + C \),
where \(C\) is the constant of integration. Using the initial condition \(v=0\) when \(x=0\), we find \(C=0\). So, solving for \(x\), we get:
\( 2\sqrt{x} = t \implies x = \left(\frac{t}{2}\right)^2 = \frac{t^2}{4} \).
For \( t > t_0 \), to account for the particle starting from rest and then moving, we adjust to:\( x = \frac{(t - t_{0})^{2}}{4} \), ensuring continuity of motion.