Problem 52 - Calculus Early Transcendentals

Question

Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of _____. $$v(t)=2 t+6, \text { for } 0 \leq t \leq 8$$

Step-by-Step Solution

Answer

Answer: 14 units per unit of time.

Step 1: Integrate the velocity function

Integrate the function v(t) = 2t + 6 with respect to t over the interval [0, 8]. This will give us the distance traveled by the object. $$ D = \int_0^8 (2t + 6) dt $$

Step 2: Solve the integral

We'll now solve the integral using the power rule for integration. $$ \begin{aligned} D &= \int_0^8 (2t + 6) dt \\ &= \left[ t^2 + 6t \right]_0^8 \end{aligned} $$

Step 3: Calculate the distance traveled

Plug in the limits of integration to find the distance traveled. $$ \begin{aligned} D &= \left[ t^2 + 6t \right]_0^8 \\ &= (8^2 + 6(8)) - (0^2 + 6(0)) \\ &= 64 + 48 \\ &= 112 \end{aligned} $$ The distance traveled by the object is 112 units.

Step 4: Find the constant velocity for the same distance traveled

Now we want to find the constant velocity (V) that would result in the same distance traveled over the time period [0, 8]. Using the formula distance = velocity × time: $$ 112 = V \times 8 $$ Divide both sides by 8 to find the constant velocity: $$ V = \frac{112}{8} $$ $$ V = 14 $$ The same distance could have been traveled over the given time period at a constant velocity of 14 units per unit of time.

Key Concepts

Understanding the Velocity Function

When we're dealing with motion, the velocity function is a crucial tool. It describes how an object's velocity changes over time. In this exercise, the velocity function is given as \( v(t) = 2t + 6 \). This equation tells us how fast and in which direction the object is moving at any given time \( t \). Here:

• The term \( 2t \) represents a velocity that increases as time passes. As \( t \) increases, \( 2t \) increases, showing that the object is accelerating.
• The constant term \( 6 \) represents an initial velocity at \( t = 0 \), which means at the start, the object is already moving at 6 units of speed.
• The key to understanding velocity functions lies in recognizing how changes in \( t \) affect the overall velocity. For this specific function, as time increases, so does the velocity, indicating a linear acceleration.

By integrating this function over a given interval, we can determine the total distance traveled.

Exploring Constant Velocity

In physics, a constant velocity means that an object moves at the same speed and in the same direction. Unlike our original velocity function, which shows acceleration, constant velocity does not change with time. It is a straight line on a velocity vs. time graph.

In our exercise, after calculating the distance traveled using the velocity function, we wish to find an equivalent constant velocity. This is a velocity at which the same total distance would be covered over the same period, without any acceleration or deceleration.

• The idea is to simplify the motion so that instead of accelerative movement, you have unchanging, consistent movement.
• This concept is useful for understanding the average motion or energy in a system over time.
• To determine this constant velocity, you would ensure the total distance and time remain unchanged.

This exploration teaches us the relationship between time, distance, and varying versus unvarying rates of movement.

Distance Calculation Using Integration

Calculating distance when given a velocity function involves integration, a fundamental concept in calculus. Integration allows us to sum up infinite, tiny increments of distance over time to reach the total distance traveled.

In our exercise, we integrated the velocity function \( v(t) = 2t + 6 \) over the time interval [0, 8].

• The integral \( \int_0^8 (2t + 6) dt \) represents the area under the curve of the velocity function on a graph, which corresponds to the total distance.
• Solving this gives us \( D = t^2 + 6t \) evaluated from 0 to 8, resulting in a distance of 112 units.
• This method effectively translates the changing speed over time into a fixed, quantifiable distance.

Thus, integration not only provides a way to compute distance but also connects the dots between instantaneous velocities and overall displacement.