Problem 12 - Applied Calculus

Question

The velocity of a car (in miles per hour) is given by \(v(t)=40 t-10 t^{2}\), where \(t\) is in hours. (a) Write a definite integral for the distance the car travels during the first three hours. (b) Sketch a graph of velocity against time and represent the distance traveled during the first three hours as an area on your graph. (c) Use a computer or calculator to find this distance.

Step-by-Step Solution

Answer

Distance traveled is 60 miles.

Step 1: Understanding the Problem

To find the distance a car travels, we integrate the velocity function over the given time period. Here, the velocity function is given as \(v(t) = 40t - 10t^2\). We need to integrate this function from \(t = 0\) to \(t = 3\).

Step 2: Writing the Definite Integral

The distance traveled is the definite integral of the velocity function over the time interval \([0, 3]\). So, we set up the integral as follows:\[\int_{0}^{3} (40t - 10t^2) \, dt.\]

Step 3: Setting Up the Graph

To sketch the graph of velocity against time, plot the equation \(v(t) = 40t - 10t^2\). Note that it is a downward-facing parabola due to the negative \(t^2\) term. The parabola starts at the origin (\(t = 0, v = 0\)), rises to a maximum, and then falls. The area under this curve from \(t = 0\) to \(t = 3\) represents the distance traveled.

Step 4: Calculating the Integral

To calculate the definite integral, use any integration tool such as a computer or calculator. Solve the integral:\[\int (40t - 10t^2) \, dt = 20t^2 - \frac{10}{3}t^3 + C.\]Evaluate this from 0 to 3:\[[20(3)^2 - \frac{10}{3}(3)^3] - [20(0)^2 - \frac{10}{3}(0)^3] = 60.\]Thus, the car travels 60 miles in the first 3 hours.

Key Concepts

Velocity Function

A velocity function describes how an object's speed changes with respect to time. In our problem, the car's velocity function is given by \(v(t) = 40t - 10t^2\). This quadratic function lets us understand how quickly the car is moving at any given time \(t\). Here,

• At \(t = 0\), the velocity is \(0\) mph, meaning the car starts from rest.
• As time increases, the velocity increases initially, reaches a peak, and then decreases due to the \(-10t^2\) term.

Velocity functions are essential for understanding movement in a mathematical sense. By representing real-world motion with an equation, we can perform calculations, like integrating to find the distance traveled.

Distance Traveled

The distance traveled by an object moving with a varying velocity can be determined by integrating the velocity function over a specified time interval. For this exercise, we're interested in the car's distance covered during the first three hours. The formula to calculate this is the definite integral: \[\int_{0}^{3} (40t - 10t^2) \, dt\] By evaluating this integral, we determine the total distance. It represents the sum of all infinitesimally small distances the car travels at each moment within the given time frame. Distance is crucial in many real-world applications, from trip planning to analyzing motion in physics.

Parabola Graph

In mathematics, a parabola is a curve where any point is at an equal distance from a fixed point (focus) and a fixed line (directrix). Our velocity function \(v(t) = 40t - 10t^2\) forms a downward-facing parabola when plotted against time. To visualize:

• Start at \(t = 0\) where velocity is \(0\), mark this origin point on the graph.
• As \(t\) increases, the parabola rises to a peak where the velocity is highest, then descends.
• The vertex of this parabola indicates maximum velocity.

This parabolic shape resonates with real-world scenarios like a car accelerating to a peak speed and then decelerating. Understanding the graphical representation of functions helps in visualizing and analyzing the physical scenarios they depict.

Integration

Integration is a calculus technique used to find the area under a curve, which translates to solving many practical problems, like computing the distance from a velocity function. Here, we need to integrate \(40t - 10t^2\) over the interval \([0, 3]\). To proceed:

• Integrate the function: \( \int (40t - 10t^2) \, dt = 20t^2 - \frac{10}{3}t^3 + C \)
• Evaluate this at the bounds 3 and 0: \([20(3)^2 - \frac{10}{3}(3)^3]\) minus \([20(0)^2 - \frac{10}{3}(0)^3]\).

As a result, we find that the car travels 60 miles within the first 3 hours. Integration is powerful, transforming complex function descriptions of movement, growth, or change into actionable insights.