Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

The velocity function for an object is given. Assuming that the object is at the origin at time \(t=0\), find the position at time \(t=4\). $$ v(t)=\left\\{\begin{array}{ll} t / 2 & \text { if } 0 \leq t \leq 2 \\ 1 & \text { if } 2

Step-by-Step Solution

Answer

The position at time \(t=4\) is 3 units.

Step 1: Understand the velocity function

Recognize that the velocity function is piecewise with different expressions depending on the interval of time. For the interval \(0 \leq t \leq 2\), the velocity \(v(t) = \frac{t}{2}\). For the interval \(2 < t \leq 4\), the velocity \(v(t) = 1\).

Step 2: Calculate position from \(t=0\) to \(t=2\)

The position function is the integral of the velocity function. Integrate \(v(t) = \frac{t}{2}\) from \(t=0\) to \(t=2\). \[x(t) = \int_{0}^{2} \frac{t}{2} \, dt = \left. \left( \frac{1}{2} \cdot \frac{t^2}{2} \right) \right|_{0}^{2} = \left. \frac{t^2}{4} \right|_{0}^{2} = \frac{4}{4} - 0 = 1\]At \(t=2\), the position \(x(2) = 1\).

Step 3: Calculate position from \(t=2\) to \(t=4\)

Continue with the position calculation by integrating \(v(t) = 1\) from \(t=2\) to \(t=4\). The initial position at \(t=2\) is \(1\). \[x(t) = 1 + \int_{2}^{4} 1 \, dt = 1 \left. \, t \, \right|_{2}^{4} = 1 + (4 - 2) = 1 + 2 = 3\]Thus, at \(t=4\), the position \(x(4) = 3\).

Key Concepts

Piecewise Functions

Piecewise functions are mathematical functions that have distinct expressions over different intervals in their domain. This means that the function changes its behavior depending on the value of the input variable.
For example, in the given exercise, the velocity function is defined as:

• \(v(t) = \frac{t}{2}\) for \(0 \leq t \leq 2\)
• \(v(t) = 1\) for \(2 < t \leq 4\)

This setup tells us that from time zero to two seconds, the velocity depends on time \(t\) in the form of \(\frac{t}{2}\).
However, once the time passes two seconds but does not exceed four, the velocity remains constant at 1 unit.
Understanding piecewise functions is crucial because it helps in modeling real-world phenomena where the condition or situation changes in intervals. Knowing the function's expression in each range allows you to solve problems more effectively by focusing on one interval at a time.

Velocity and Position

Velocity and position are fundamental concepts in both physics and calculus. They are closely related, as velocity is essentially the rate of change of position with respect to time.
The velocity function provides information about the speed and direction of an object's movement.
In this problem, we have been given a piecewise velocity function. Our task is to find out how far the object has traveled at a specific time, which involves using integration to determine the position function.

• For \(0 \leq t \leq 2\), the velocity \(v(t) = \frac{t}{2}\) tells us that velocity is increasing over time.
• For \(2 < t \leq 4\), \(v(t) = 1\) indicates a constant velocity.

Using integrals, which calculate the area under the velocity-time graph, we find the total distance traveled.
For instance, by integrating the first part of the piecewise function, we determine the position at \(t=2\), and then continue its calculation using the second part to find the position at \(t=4\).
Thus, understanding velocity and its relation to position through integration is key to solving these types of problems.

Integration Methods

Integration methods help us find the antiderivative or the area under a curve described by a function.
This is particularly useful when calculating the position of an object, as it's the cumulative effect of the velocity over time.
In the given exercise, we employed the following integration techniques for each piece of the piecewise velocity function:

• For \(v(t) = \frac{t}{2}\), we used the power rule for integration. Calculating the integral of \(\frac{t}{2}\) from \(t=0\) to \(t=2\), we found that the position at \(t=2\) is 1 unit.
• For \(v(t) = 1\), integration is straightforward. We integrated the constant function from \(t=2\) to \(t=4\), starting with the position from the previous interval. This gave us the final position at \(t=4\) as 3 units.

The integration technique involves breaking the problem into manageable parts, especially when dealing with piecewise functions.
Each section is analyzed separately, and their solutions are combined to provide the final outcome.
This method solidifies our understanding of calculus by giving us practical applications of integrals and their computation.