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Chapter 2: Problem 22

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

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

Step by step solution

01

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 \(t = \frac{\ln{8}}{3\alpha}\).
02

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 \(v(t) = \frac{3}{4}\alpha x_0 e^{3\alpha t}\).
03

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 \(a(t) = \frac{9}{4}\alpha^2 x_0 e^{3\alpha t}\).
04

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\(^{-1}\).

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