Problem 19 - Applied Calculus

Question

With \(t\) in seconds, the velocity of an object is \(v(t)=\) \(10+8 t-t^{2} \mathrm{m} / \mathrm{sec}\). (a) Represent the distance traveled during the first 5 seconds as a definite integral and as an area. (b) Estimate the distance traveled by the object during the first 5 seconds by estimating the area. (c) Calculate the distance traveled.

Step-by-Step Solution

Answer

The distance traveled is approximately 108.33 meters.

Step 1: Understand the velocity function

The velocity function given is \( v(t) = 10 + 8t - t^2 \) in meters per second. This function describes how the velocity of the object changes over time.

Step 2: Set up the definite integral for distance

The distance traveled by the object from time \( t = a \) to \( t = b \) is the definite integral of the velocity function over those limits. Therefore, for the first 5 seconds (\( t = 0 \) to \( t = 5 \)), the distance is represented as \[ \int_{0}^{5} (10 + 8t - t^2) \, dt. \]

Step 3: Interpret the integral as an area

The definite integral \( \int_{0}^{5} (10 + 8t - t^2) \, dt \) represents the net area under the curve of the velocity function from \( t = 0 \) to \( t = 5 \). This area corresponds to the total distance traveled by the object over this time period.

Step 4: Estimate the area

To estimate the area under the curve, you could use numerical estimation methods such as the trapezoidal rule or left/right endpoint approximations. However, providing a precise estimate without computation tools simply involves noting that the function follows a parabolic curve with openings downwards from \( t = 0 \) to \( t = 5 \). It is symmetrical around \( t = 4 \) due to the quadratic term.

Step 5: Calculate the definite integral

First, find the antiderivative of \( 10 + 8t - t^2 \). The antiderivative is \( 10t + 4t^2 - \frac{t^3}{3} \). Compute this from \( t = 0 \) to \( t = 5 \):1. \( 10(5) + 4(5)^2 - \frac{(5)^3}{3} \)2. \( 50 + 100 - \frac{125}{3} \)3. Simplify: \( 150 - \frac{125}{3} = 150 - 41.67 \approx 108.33. \)Thus, the distance traveled is approximately \( 108.33 \) meters.

Key Concepts

Velocity Function

A velocity function is a mathematical representation of how an object's speed and direction change over time. In our example, the velocity function is given by:

• \( v(t) = 10 + 8t - t^2 \)

This polynomial function describes the object's velocity in meters per second as it varies based on the time \( t \), measured in seconds.
Let's break down this function:

• The term \( 10 \) represents the initial velocity at \( t = 0 \). This is the speed at which the object starts its motion.
• The term \( 8t \) shows how the velocity increases linearly with time. Essentially, the object speeds up by 8 meters per second for each additional second.
• The term \( -t^2 \) indicates a decrease in velocity over time, due to its parabolic shape that curves downwards. This part of the function reduces the speed as time increases, accounting for factors like friction or air resistance which typically decelerate an object.

Understanding the velocity function aids in analyzing how objects move through space over time, which is essential in physics and engineering.

Distance Estimation

Estimating the distance traveled by an object involves calculating the area under the curve of its velocity function over a given time interval. This concept relates to the real-world scenario of determining how far an object has moved based on its speed changes.
To visualize this, imagine the velocity function graph from \( t = 0 \) to \( t = 5 \). The area under this curve—bounded by the time axis and the curve itself—corresponds to the distance traveled by the object during the first 5 seconds:

• One common method to estimate this area is the trapezoidal rule, where the curve is approximated as a series of trapezoids rather than precise curves. This provides an easier calculation of the approximate area.
• You can also use left and right endpoint approximations. This involves summing up areas of rectangles under the curve, either starting from the left or right side of the interval.

While these methods provide an estimated area, understanding them is crucial as they lead up to more precise calculations using integrals.

Antiderivative Calculation

Calculating an antiderivative, also known as finding the indefinite integral, is a primary step in determining the exact area under the curve of a velocity function. For the given problem, we calculate the antiderivative of the velocity function \( v(t) = 10 + 8t - t^2 \).
The antiderivative is found by reversing the differentiation process:

• The antiderivative of a constant, such as 10, becomes \( 10t \).
• The antiderivative of \( 8t \) becomes \( 4t^2 \) because you add one to the exponent and divide by the new exponent.
• The antiderivative of \(-t^2\) is \(-\frac{t^3}{3} \).

Combining these results gives us the full antiderivative expression:

• \( 10t + 4t^2 - \frac{t^3}{3} \)

By evaluating this antiderivative from \( t = 0 \) to \( t = 5 \) and subtracting, we can accurately calculate the total distance covered:

• Perform the calculation \( \left[ 10(5) + 4(5)^2 - \frac{(5)^3}{3} \right] - \left[ 10(0) + 4(0)^2 - \frac{(0)^3}{3} \right] \)
• This results in a distance of approximately 108.33 meters.

Mastering antiderivative calculation is crucial for precise distance assessments in physics and calculus.