Question

The speed of a particle on the \(x\) axis, \(x \geq 0\), is always numerically equal to the square root of its displacement \(x\). If \(x=0\) when \(t=0\), 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

The displacement is \( x = \frac{t^2}{4} \). For \( t > t_0 \), \( x = \frac{(t - t_0)^2}{4} \).

Step-by-step solution

Understand the Problem

The problem states that the speed of a particle, which is the derivative of displacement with respect to time, is equal to the square root of its displacement. We need to find the relationship between displacement, x, and time, t.

Set Up the Equation

Given that speed is the square root of displacement, we can write the following differential equation: \ \( \frac{dx}{dt} = \sqrt{x} \).

Separate Variables and Integrate

We separate the variables to integrate: \ \( \frac{dx}{\sqrt{x}} = dt \). Then, integrate both sides: \ \( 2\sqrt{x} = t + C \) where C is the constant of integration.

Apply Initial Conditions

Given that at \( t = 0, x = 0 \), substitute these values in: \ \( 2\sqrt{0} = 0 + C \). Therefore, C = 0. Our equation simplifies to: \ \( 2\sqrt{x} = t \).

Solve for x

Square both sides to solve for x: \ \( x = \frac{t^2}{4} \). This is the function of x in terms of t.

Consider the Case When Particle Remains at Origin

If the particle remains at the origin for a time \( t_0 \), it means that x = 0 until \( t = t_0 \).

Find x for \( t > t_0 \)

For \( t > t_0 \), we restart the motion with \( x = \frac{(t - t_0)^2}{4} \).

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In this exercise, we know the speed of a particle and how it relates to its position over time. The task is to find a function that describes the particle's displacement, which means we need to solve the differential equation given by the problem. Differential equations can be tricky because they often involve functions and their rates of change. In our case, we start with the relationship \( \frac{dx}{dt} = \sqrt{x} \). The goal is to find the function \( x(t) \), which describes the displacement of the particle over time.

Initial Conditions

Initial conditions specify the values of the function and its derivatives at a particular point, which is essential for solving differential equations uniquely. For our problem, the initial condition is given as \( x = 0 \) when \( t = 0 \). This means at time zero, the displacement of the particle is also zero. When we solve for the integration constant, initial conditions help us determine its exact value. Applying the initial condition in our integration, we find that the constant C is equal to zero, which simplifies our equation to \( 2\sqrt{x} = t \). This is crucial because without it, we wouldn't know the precise behavior of our particle at the start.

Separation of Variables

Separation of variables is a method used to solve differential equations. This method involves rearranging the equation so that each variable and its differential are on different sides of the equation. In our case, we start with \( \frac{dx}{dt} = \sqrt{x} \). To separate the variables, we rewrite this as \( \frac{dx}{\sqrt{x}} = dt \). By doing this, we prepare the equation for integration. Separation of variables transforms a complex differential equation into simpler integrals, making it easier to solve. Once variables are separated, we simply integrate each side with respect to its variable.

Integration

Integration is a fundamental concept in calculus, used to find the original function from its derivative. Once we separate the variables in our equation, we need to integrate both sides to solve for \( x(t) \). On the left side, we integrate \( \frac{dx}{\sqrt{x}} \), which equals \( 2\sqrt{x} \). On the right side, integrating \( dt \) gives us \( t + C \). Therefore, our integrated equation is \( 2\sqrt{x} = t + C \). Applying the initial condition, we determine that \( C = 0 \), simplifying it to \( 2\sqrt{x} = t \). Finally, solving for \( x \) by squaring both sides results in \( x = \frac{t^2}{4} \). This solution describes the displacement of our particle as it moves along the x-axis over time. If the particle starts moving after a certain time \( t_{0} \), the solution adjusts to account for this delay, yielding \( x = \frac{(t - t_{0})^2}{4} \).