Question

The position of a particle as a function of time is given as \(x(t)=\frac{1}{4} x_{0} e^{3 \alpha t}\), where \(\alpha\) is a positive constant. a) At what time is the particle at \(2 x_{0}\) ? b) What is the speed of the particle as a function of time? c) What is the acceleration of the particle as a function of time? d) What are the SI units for \(\alpha\) ?

Short answer

Answer: The particle is at position \(2x_0\) when \(t = \frac{\ln{8}}{3\alpha}\).

Step-by-step solution

Find time when particle is at \(2x_0\)

We need to solve the equation \(x(t) = 2x_0\) for the time \(t\). Start by setting up the equation: $$\frac{1}{4}x_0 e^{3\alpha t} = 2x_0$$ Now, we will solve for \(t\): $$e^{3\alpha t} = 2 \times 4$$ $$3\alpha t = \ln{8}$$ $$t = \frac{\ln{8}}{3\alpha}$$ Therefore, the particle is at \(2x_0\) when \(t = \frac{\ln{8}}{3\alpha}\).

Find the speed of the particle as a function of time

The speed of the particle is given by the derivative of the position function with respect to time: $$v(t) = \frac{d}{dt} x(t) = \frac{d}{dt} \left(\frac{1}{4}x_0 e^{3\alpha t}\right)$$ Using the chain rule, we find: $$v(t) = \frac{1}{4}x_0 (3\alpha) e^{3\alpha t} = \frac{3}{4}\alpha x_0 e^{3\alpha t}$$ Hence, the speed of the particle as a function of time is \(v(t) = \frac{3}{4}\alpha x_0 e^{3\alpha t}\).

Find the acceleration of the particle as a function of time

The acceleration of the particle is given by the derivative of the speed function with respect to time: $$a(t) = \frac{d}{dt} v(t) = \frac{d}{dt} \left(\frac{3}{4}\alpha x_0 e^{3\alpha t}\right)$$ Again, using the chain rule, we find: $$a(t) = \frac{3}{4}\alpha x_0 (3\alpha) e^{3\alpha t} = \frac{9}{4}\alpha^2 x_0 e^{3\alpha t}$$ Hence, the acceleration of the particle as a function of time is \(a(t) = \frac{9}{4}\alpha^2 x_0 e^{3\alpha t}\).

Find the SI units for \(\alpha\)

Since we know that \(x(t)\) represents position, its unit is meters (m). The time variable \(t\) has the unit of seconds (s). Let's analyze the equation to determine the units for \(\alpha\): $$x(t) = \frac{1}{4}x_0 e^{3\alpha t}$$ From this equation, we see that \(\alpha\) must be inside an exponential function, which is dimensionless. Therefore, we need to find the units for \(\alpha\) that make \(\alpha t\) dimensionless. Since the unit of \(t\) is seconds (s), the unit of \(\alpha\) must be the reciprocal of the unit of time, which is inverse seconds or s\(^{-1}\). Thus, the SI units for \(\alpha\) are s\(^{-1}\).

Key concepts

Particle Dynamics

Particle dynamics refers to the study of the motion of particles and how forces affect this motion. It is a fundamental part of classical mechanics. In this exercise, we deal with a particle whose position changes over time following an exponential function. Dynamics help us to understand more than just the particle's current position—it also involves calculating the particle's speed (velocity) and acceleration. For exponential motion, the particle exhibits rapid growth or decay in position, velocity, and acceleration values.

• This exponential property is indicative of systems where change happens rapidly after a certain point, similar to population growth or radioactive decay.
• Understanding these changes helps in predicting the future position and behavior of the particle.

Position Function

In this exercise, the position function of the particle is given by \(x(t) = \frac{1}{4} x_{0} e^{3 \alpha t}\). The position function expresses the location of the particle at any time \(t\). Here, \(x_0\) refers to the initial position, and \(e^{3 \alpha t}\) describes how the position changes exponentially over time.

• The function starts at \(\frac{1}{4} x_0\) and grows quickly as time progresses due to the exponential term \(e^{3 \alpha t}\).
• The exponent \(3 \alpha t\) suggests that the speed of change in the position is influenced significantly by the constant \(\alpha\).

To find at what time the particle is at a specific position, like \(2x_0\), we solve for \(t\) by setting the position function equal to this desired position.

Kinematic Equations

Kinematic equations describe the motion details of objects through formulas that relate position, velocity, and acceleration with time. In this problem, we derived both speed and acceleration functions of the particle from the provided position function.

• The velocity function derived is \( v(t) = \frac{3}{4}\alpha x_0 e^{3\alpha t} \), showcasing that velocity also changes exponentially with time.
• Similarly, the acceleration is \( a(t) = \frac{9}{4}\alpha^2 x_0 e^{3\alpha t} \), meaning acceleration also follows an exponential growth.

These kinematic expressions are crucial as they give detailed insights into how fast and how much the motion is changing at any point in time.

SI Units

The International System of Units (SI) provides standard units for measuring physical quantities. It's essential to identify the correct SI units to understand the magnitude and dimensions of different variables. For our exercise, the position \(x(t)\) is measured in meters \((\text{m})\) and time \(t\) in seconds \((\text{s})\).
When determining the units for \(\alpha\), it must ensure that the exponential argument \(3\alpha t\) is dimensionless.

• Given that \(t\) has units of seconds, \(\alpha\) must have units of inverse seconds \(\text{s}^{-1}\).

Making sure \(\alpha\) is in s\(^{-1}\) ensures that our mathematical expressions comply with dimensional analysis, keeping the equations consistent and physically meaningful.