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

a) To find the time when the particle's position reaches \(2x_0\), we solve the equation \(2x_0 = \frac{1}{4}x_0 e^{3\alpha t}\) for t: The solution is: $$t = \frac{\ln{8}}{3\alpha}$$ b) The speed of the particle as a function of time is: $$v(t) = \frac{3}{4}\alpha x_0 e^{3\alpha t}$$ c) The acceleration of the particle as a function of time is: $$a(t) = \frac{9}{4}\alpha^2x_0 e^{3\alpha t}$$ d) The SI units for \(\alpha\) is s\(^{-1}\).

Step-by-step solution

a) Finding the time when the particle's position reaches \(2x_0\)

We start by setting \(x(t) = 2x_0\) and solve for the time, \(t\): $$2x_0 = \frac{1}{4}x_0 e^{3\alpha t}$$ Divide both sides by \(x_0\): $$2 = \frac{1}{4}e^{3\alpha t}$$ Now, multiply both sides by 4 to get: $$8 = e^{3\alpha t}$$ Take the natural logarithm of both sides: $$\ln{8} = 3\alpha t$$ Solve for \(t\): $$t = \frac{\ln{8}}{3\alpha}$$

b) Finding the speed as a function of time

Take the first derivative of the position function with respect to time to get the speed function, \(v(t)\): $$v(t) = \frac{d}{dt} \left( \frac{1}{4}x_0 e^{3\alpha t}\right) = \frac{3}{4}\alpha x_0 e^{3\alpha t}$$

c) Finding the acceleration as a function of time

Take the first derivative of the speed function with respect to time to get the acceleration function, \(a(t)\): $$a(t) = \frac{d}{dt} \left( \frac{3}{4}\alpha x_0 e^{3\alpha t}\right) = \frac{9}{4}\alpha^2x_0 e^{3\alpha t}$$

d) Determining the SI units for \(\alpha\)

In order to determine the SI units for \(\alpha\), we can analyze the units of the given position equation, which is expressed in terms of \(\alpha\). The position's unit is length and time's unit is seconds. So, we have: $$[\text{length}] =\frac{1}{4} [\text{length}] \times e^{[\alpha] \times [\text{time}]}$$ The units of \(\text{length}\) on both sides of the equation cancel. Then, to find the unit of \(\alpha\), we analyze the exponent in the exponential function: $$[\alpha] \times [\text{time}] = [\text{unit-less}]$$ Since the unit of time is seconds (s), the unit of \(\alpha\) must be s\(^{-1}\) in order to cancel the time unit and result in a unit-less value for the exponent. Therefore, \(\alpha\) has SI units of s\(^{-1}\).

Key concepts

Position-Time Relationship

Understanding the position-time relationship in particle kinematics is crucial for predicting a particle's location at any given moment. It describes how the position of a particle changes with time. In our exercise, the position of a particle is given by the function
\(x(t)=\frac{1}{4} x_{0} e^{3 \alpha t},\) where \(x(t)\) is the particle's position at time \(t\), \(x_{0}\) is the initial position, and \(\alpha\) is a positive constant. To find at what time the particle is at \(2 x_{0}\), you set \(x(t)\) equal to \(2 x_{0}\) and solve for \(t\), yielding \(t = \frac{\ln{8}}{3\alpha}\) after logarithmic manipulation. Understanding this relationship allows us to map out the trajectory of the particle over time.

Derivative Velocity

Derivative velocity refers to the rate of change of position with respect to time, which is the definition of speed or velocity in physics. To find the speed as a function of time for our particle, we take the first derivative of the position function, \(x(t)\), with respect to time. This process gives us
\(v(t) = \frac{d}{dt} \left( \frac{1}{4}x_0 e^{3\alpha t}\right) = \frac{3}{4}\alpha x_0 e^{3\alpha t}\). The resulting function, \(v(t)\), illustrates how the particle's speed changes over time. The ability to calculate velocity through differentiation is a powerful tool in both theoretical and applied physics, facilitating the prediction of a particle's future speed based on its current motion.

Acceleration-Time Relationship

The acceleration-time relationship in particle kinematics demonstrates how a particle's acceleration varies with time. Acceleration is the second derivative of the position function or the first derivative of the velocity function with respect to time. In the given problem, we derive the acceleration equation by differentiating the velocity function, which results in
\(a(t) = \frac{d}{dt} \left( \frac{3}{4}\alpha x_0 e^{3\alpha t}\right) = \frac{9}{4}\alpha^2 x_0 e^{3\alpha t}\). This equation, \(a(t)\), shows the acceleration of the particle as a continuous function of time. This relationship is particularly important when analyzing the dynamics of a particle under various forces since acceleration is directly linked to force by Newton's second law of motion.

SI Units

In science, particularly physics, the International System of Units (SI) provides a standard for measuring and expressing quantities. Understanding SI units is essential for accurately describing physical phenomena. In this exercise, we're asked to determine the SI units for the constant \(\alpha\). SI units must remain consistent throughout equations to ensure the correctness of physical laws. Since \(\alpha\) is found within the exponent in the position function for a unit-less result, its units, when combined with time (in seconds), must cancel out. Thus, we deduce that the units of \(\alpha\) are reciprocal seconds, or s\(^{-1}\). This SI unit of \(\alpha\) reflects its role in influencing the rate of exponential change in the particle's position over time.