Question

a) for each of the intervals, find the values of $\delta d$ and $\delta t$ between the indicated start and end times. enter your answers in their respective columns in the table below.

time interval	$\delta d$	$\delta t$
$t = 1$ to $t = 4$	$\square$	$\square$
$t = 0.5$ to $t = 2.5$	$\square$	$\square$

b) based on your results from (a) it follows that the average rate of change of $d$ is constant, it does not depend over which interval of time you choose. what is the constant rate of change of $d$?
$\frac{\delta d}{\delta t} = \square$

c) which of the statements below correctly explains the significance of your answer to part (b)? select all that apply (more than one may apply).
$\square$ a. it is the average velocity of the car over the first two hours.
$\square$ b. it is the acceleration of the car over the five hour time interval.
$\square$ c. it is how far the car will travel in a half-hour.
$\square$ d. it is the slope of the line.
$\square$ e. it is the total distance the car travels in five hours.
$\square$ f. it represents the car’s velocity.
$\square$ g. none of the above.

Answer:

To solve this problem, we need to assume a linear relationship for \( D(t) \) (distance as a function of time) since the average rate of change is constant. Let's assume \( D(t) = vt + D_0 \), where \( v \) is the velocity (rate of change) and \( D_0 \) is the initial distance. For simplicity, let's assume \( D_0 = 0 \), so \( D(t) = vt \). We'll use this to find \( \Delta D \) and \( \Delta t \) for each interval.

Part (a)

To find \( \Delta D \) and \( \Delta t \) for each time interval:

Interval 1: \( t = 1.5 \) to \( t = 4 \)

• \( \Delta t = 4 - 1.5 = 2.5 \)
• \( \Delta D = D(4) - D(1.5) = v(4) - v(1.5) = v(4 - 1.5) = 2.5v \)

Interval 2: \( t = 1 \) to \( t = 4 \)

• \( \Delta t = 4 - 1 = 3 \)
• \( \Delta D = D(4) - D(1) = v(4) - v(1) = v(4 - 1) = 3v \)

Interval 3: \( t = 0.5 \) to \( t = 2.5 \)

• \( \Delta t = 2.5 - 0.5 = 2 \)
• \( \Delta D = D(2.5) - D(0.5) = v(2.5) - v(0.5) = v(2.5 - 0.5) = 2v \)

Part (b)

The average rate of change is \( \frac{\Delta D}{\Delta t} \). Let's compute it for each interval:

• For \( t = 1.5 \) to \( t = 4 \): \( \frac{\Delta D}{\Delta t} = \frac{2.5v}{2.5} = v \)
• For \( t = 1 \) to \( t = 4 \): \( \frac{\Delta D}{\Delta t} = \frac{3v}{3} = v \)
• For \( t = 0.5 \) to \( t = 2.5 \): \( \frac{\Delta D}{\Delta t} = \frac{2v}{2} = v \)

Thus, the constant rate of change is \( v \). To find a numerical value, we need more information (e.g., \( D(t) \) values). Assuming a common scenario (e.g., \( D(t) = 60t \) where \( v = 60 \) mph), the rate of change would be \( 60 \). However, since the problem doesn't provide specific \( D(t) \) values, we'll use \( v \) as the rate. For a concrete example, if \( D(t) = 60t \):

• Interval 1: \( \Delta D = 60(4) - 60(1.5) = 240 - 90 = 150 \), \( \Delta t = 2.5 \), \( \frac{150}{2.5} = 60 \)
• Interval 2: \( \Delta D = 60(4) - 60(1) = 240 - 60 = 180 \), \( \Delta t = 3 \), \( \frac{180}{3} = 60 \)
• Interval 3: \( \Delta D = 60(2.5) - 60(0.5) = 150 - 30 = 120 \), \( \Delta t = 2 \), \( \frac{120}{2} = 60 \)

Thus, the constant rate of change is \( 60 \) (assuming \( D(t) = 60t \)).

Part (c)

Let's analyze each option:

• A: "It is the average velocity of the car over the first two hours."

The rate of change is constant, so it's the average velocity over any interval, including the first two hours. Correct.

• B: "It is the acceleration of the car over the five hour time interval."

Acceleration is the rate of change of velocity. Since velocity is constant (rate of change of distance is constant), acceleration is zero. Incorrect.

• C: "It is how far the car will travel in a half-hour."

Distance in 0.5 hours is \( v \times 0.5 \). If \( v = 60 \), this is 30 miles. The rate of change \( v \) is not the distance in 0.5 hours, but \( v \times 0.5 \) is. Incorrect (unless \( v = 30 \), but generally, no).

• D: "It is the slope of the line."

For \( D(t) = vt + D_0 \), the graph is a line with slope \( v \). Correct.

• E: "It is the total distance the car travels in five hours."

Total distance in 5 hours is \( v \times 5 \), not \( v \). Incorrect.

• F: "It represents the car’s velocity."

Since \( \frac{\Delta D}{\Delta t} = v \) (velocity), Correct.

• G: "None of the above."

Incorrect, as A, D, F are correct.

Final Answers

Part (a) (assuming \( D(t) = 60t \)):

Time Interval	\( \Delta D \)	\( \Delta t \)
\( t = 1 \) to \( t = 4 \)	180	3
\( t = 0.5 \) to \( t = 2.5 \)	120	2

Part (b):

\( \frac{\Delta D}{\Delta t} = \boldsymbol{60} \) (or \( v \) if no specific \( D(t) \) is given).

Part (c):

Correct options: A, D, F