Question

Displacement from velocity The following functions describe the velocity of a car (in \(\mathrm{mi} / \mathrm{hr}\) ) moving along a straight highway for a 3-hr interval. In each case, find the function that gives the displacement of the car over the interval \([0, t]\), where \(0 \leq t \leq 3\) (Check your book to see figure) $$v(t)=\left\\{\begin{array}{ll}30 & \text { if } 0 \leq t \leq 2 \\\50 & \text { if } 2< t \leq 2.5 \\\44 & \text { if } 2.5< t \leq 3\end{array}\right.$$

Short answer

Answer: The displacement of the car at time t can be found using the following piecewise function: $$s(t)=\left\\{\begin{array}{ll}30t & \text { if } 0 \leq t \leq 2 \\\50t - 40 & \text { if } 2< t \leq 2.5 \\\44t - 31 & \text { if } 2.5< t \leq 3\end{array}\right.$$

Step-by-step solution

Understand the given function

The given function is a piecewise function defining the velocity of the car in mi/hr for a 3-hour interval. The function has 3 intervals: 1. \(0 \leq t \leq 2\): The car moves at a constant velocity of 30 mi/hr. 2. \(2 < t \leq 2.5\): The car accelerates to 50 mi/hr. 3. \(2.5 < t \leq 3\): The car decelerates to 44 mi/hr.

Integrate each part of the function

Since the velocity function is given, we have to integrate it in order to obtain the displacement function. For the first interval: \(0 \leq t \leq 2\) $$\int_0^t 30 \, dt = 30t$$ For the second interval: \(2 < t \leq 2.5\) $$\int_2^t 50 \, dt = 50(t-2) + 30(2) = 50t - 40$$ For the third interval: \(2.5 < t \leq 3\) $$\int_{2.5}^t 44 \, dt = 44(t-2.5) + 50(0.5) + 30(2) = 44t - 31$$

Form the piecewise function for displacement

Using the results from the previous step, we can write the displacement function as: Displacement function: $$s(t)=\left\\{\begin{array}{ll}30t & \text { if } 0 \leq t \leq 2 \\\50t - 40 & \text { if } 2< t \leq 2.5 \\\44t - 31 & \text { if } 2.5< t \leq 3\end{array}\right.$$ The obtained function provides the car's displacement over the given time intervals.

Key concepts

Velocity Function

A velocity function, such as the one provided in the problem, is a mathematical expression that describes how fast an object is moving with respect to time. It's measured in units of distance over time, like miles per hour (mi/hr) in this example. In simpler terms, it tells us how quickly the position of the car is changing. In the given scenario, the velocity function is piecewise. This means it consists of different expressions depending on the time interval. Understanding these intervals is crucial:

• From 0 to 2 hours, the car travels at 30 mi/hr. It's a constant speed, so the velocity does not change.
• Between 2 and 2.5 hours, the velocity jumps to 50 mi/hr, indicating the car accelerates.
• Finally, from 2.5 to 3 hours, the car decelerates to 44 mi/hr.

The velocity function gives us precise control over how we understand the car’s movement and allows us to mathematically calculate other related aspects, like displacement.

Piecewise Function

A piecewise function is essentially a collection of functions, each defined over a specific interval. Think of it like a story in segments; each part tells a piece of the whole picture. For the car's velocity, the function is divided into three distinct intervals, each with its own constant velocity value. To work with piecewise functions, it's important to:

• Identify the different segments: Each piece of the function corresponds to a specific time range in our example.
• Apply the correct function: Use the appropriate mathematical expression for each interval when working with calculations like integration.
• Understand transitions: Knowing when the function switches between segments helps in correctly piecing together the overall story of the problem.

Managing piecewise functions helps us break down more complex problems into smaller, manageable sections, much like solving a puzzle!

Integration of Functions

Integration is a fundamental concept in calculus, used to find a function's antiderivative. In the context of this problem, integrating the velocity function helps us determine the displacement, which shows the overall distance the car has traveled. For each segment of the piecewise velocity function, we perform separate integrations:

• The first interval is integrated to obtain a simple formula: \( \int_0^t 30 \, dt = 30t \). This result signifies that in 2 hours, the car would have traveled a total of 60 miles at that constant speed.
• For the second interval, we integrate and adjust: \( \int_2^t 50 \, dt = 50(t-2) + 60 \). Here, adjustments account for the starting velocity.
• The last section gives us \( \int_{2.5}^t 44 \, dt = 44(t-2.5) + 50(0.5) + 60 \), completing the car's journey with a decrease in speed.

Thus, integration links the car's changing velocities into a complete picture of its movement, transforming the piecewise velocity function into a cohesive story of displacement over time.