Question

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=2 e^{-t / 6} ; v(0)=1, s(0)=0$$

Short answer

Question: Given the acceleration function \(a(t) = 2 e^{-t/6}\) of an object moving along a line with initial conditions \(v(0) = 1\) and \(s(0) = 0\), find the position function, \(s(t)\). Answer: The position function is \(s(t) = -12e^{-t/6} + 13t + 12\).

Step-by-step solution

Integrate the acceleration function

First, we need to integrate the acceleration function, \(a(t)\), to find the velocity function, \(v(t)\). We have: $$v(t) = \int a(t) dt = \int 2 e^{-t/6} dt$$ To integrate the exponential function, we can use substitution: Let \(u = -\frac{t}{6}\). Thus, \(du = -\frac{1}{6} dt\), and \(dt = -6 du\). We get: $$v(t) = \int 2e^u (-6 du)$$ $$v(t) = -12 \int e^u du$$ Now, integrate with respect to \(u\): $$v(t) = -12e^u + C_1$$ Substitute back the variable \(u = -\frac{t}{6}\): $$v(t) = -12e^{-t/6} + C_1$$

Apply the initial velocity condition

Now, apply the initial velocity condition \(v(0) = 1\) to find the constant of integration \(C_1\). We get: $$1 = -12e^{-0/6} + C_1$$ $$1 = -12 + C_1$$ $$C_1 = 13$$ So, the velocity function is: $$v(t) = -12e^{-t/6} + 13$$

Integrate the velocity function

Next, we need to integrate the velocity function, \(v(t)\), to find the position function, \(s(t)\). We have: $$s(t) = \int v(t) dt = \int (-12e^{-t/6} + 13) dt$$ Separate the integral: $$s(t) = \int -12e^{-t/6} dt + \int 13 dt$$ The integral of the first part is the same as before, and the integral of the second part is a simple linear function: $$s(t) = -12e^{-t/6} + 13t + C_2$$

Apply the initial position condition

Finally, apply the initial position condition \(s(0) = 0\) to find the constant of integration \(C_2\). We get: $$0 = -12e^{-0/6} + 13\cdot 0 + C_2$$ $$0 = -12 + C_2$$ $$C_2 = 12$$ Thus, the position function is: $$s(t) = -12e^{-t/6} + 13t + 12$$

Key concepts

Understanding the Acceleration Function

The acceleration function, denoted as \(a(t)\), represents how the speed of an object changes over time as it moves along a line. When given an acceleration function like \(a(t) = 2e^{-t/6}\), it describes how rapidly the velocity of the object increases or decreases.

• In our exercise, the exponential term \(e^{-t/6}\) suggests that the acceleration decreases over time.
• The factor of 2 scales this decrease, affecting the rate at which velocity will change over time.

Understanding the decay of acceleration is crucial for integrating and finding both the velocity function and the position function. This step helps us understand how the initial strength of acceleration diminishes, impacting the object's speed and position.

Finding the Velocity Function

The velocity function, \(v(t)\), is critical as it indicates the speed and direction of the object's motion along a path. By integrating the acceleration function, we can find this velocity function:

• In our problem, we integrate \(a(t)\) to find \(v(t) = \int 2e^{-t/6} dt\). This involves using substitution, a common integration technique.
• We replace \(t\) with a different variable \(u\), simplifying the integral to something more manageable.
• The integration results in \(v(t) = -12e^{-t/6} + C_1\).

To determine the constant \(C_1\), we apply the initial velocity condition \(v(0) = 1\). Solving for \(C_1\) ensures our velocity function accurately reflects the initial state of the object's motion. This gives us a complete velocity function: \(v(t) = -12e^{-t/6} + 13\). This expression defines how the body's speed changes based on initial conditions and exponential decay.

The Role of Initial Conditions

Initial conditions are the starting values that help us solve for unknown constants in our calculated functions. In physics, these are critical to finding precise solutions that match real-world scenarios.

• For this exercise, we had two initial conditions: \(v(0) = 1\) and \(s(0) = 0\).
• The initial velocity \(v(0) = 1\) was used to solve for the constant \(C_1\) in the velocity function \(v(t)\).
• Similarly, the initial position \(s(0) = 0\) allows us to find \(C_2\) when integrating \(v(t)\) to get \(s(t)\).

These conditions aren't just arbitrary numbers; they define where the object starts in terms of location and speed. With these constants determined, the position function that describes the object's pathway over time is given as: \(s(t) = -12e^{-t/6} + 13t + 12\). This full position function provides both the amplitude and direction of travel, completely specifying the object's future position.