Question

1. determine the average rate of change, in mph, from 2 to 4 hours on the graph shown below.

Explanation:

Step1: Identify the points at t=2 and t=4

From the graph, at \( t = 2 \) hours, the distance \( d_1 = 20 \) miles. At \( t = 4 \) hours, the distance \( d_2 = 60 \) miles? Wait, no, wait. Wait, the time is on the y - axis? Wait, the graph has Time (hours) on the y - axis (from 1 to 5) and Distance (miles) on the x - axis (0,20,40,60,80). Wait, the starting point is (0,0) (distance 0, time 0). Then at time \( t = 2 \) hours (y - axis), the distance is 20 miles (x - axis). At time \( t = 4 \) hours (y - axis), the distance is 60 miles? Wait, no, looking at the graph: the point at t = 2 (y=2) is at x=20? Wait, no, maybe I got the axes reversed. Wait, usually, time is on the x - axis, but here Time is on the y - axis (vertical) and Distance on the x - axis (horizontal). So the coordinates are (distance, time). So at time \( t = 2 \) (y = 2), the distance \( x = 20 \) (so the point is (20, 2)). At time \( t = 4 \) (y = 4), the distance \( x = 60 \) (point (60, 4))? Wait, no, the curve: at t=3, the point is (40, 3)? Wait, maybe the graph is distance vs time, but time is on the y - axis. So the average rate of change is \( \frac{\Delta d}{\Delta t} \), where \( \Delta d = d_2 - d_1 \) and \( \Delta t = t_2 - t_1 \). Wait, no: rate of change of distance with respect to time is \( \frac{\text{change in distance}}{\text{change in time}} \). So if at \( t_1 = 2 \) hours, distance \( d_1 = 20 \) miles; at \( t_2 = 4 \) hours, distance \( d_2 = 60 \) miles? Wait, no, looking at the graph: the first segment is a line from (0,0) to (20, 2) (distance 20, time 2), then a curve to (60, 4) (distance 60, time 4)? Wait, no, the vertical axis is time (hours: 1,2,3,4,5) and horizontal is distance (miles: 0,20,40,60,80). So the point at time t=2 (y=2) is at distance x=20 (so (20,2)), at t=4 (y=4) is at x=60 (60,4). Wait, no, maybe the graph is time vs distance, but the question is average rate of change from 2 to 4 hours (time), so we need the change in distance over change in time. So \( \Delta t = 4 - 2 = 2 \) hours. \( \Delta d = d(4) - d(2) \). From the graph, at t=2, d=20 miles; at t=4, d=60 miles? Wait, no, maybe I misread. Wait, the graph: the first point at t=2 (y=2) is at x=20 (distance 20), then at t=3 (y=3) it's at x=40? No, the curve: at t=3, it's at x=40? Then at t=4, x=60? Wait, no, the last point at t=4 (y=4) is at x=60? Wait, maybe the graph is distance on the y - axis and time on the x - axis, but the labels are reversed. Wait, the problem says "from 2 to 4 hours", so time is the independent variable (x - axis), distance is dependent (y - axis). Maybe the graph is mislabeled in my view. Wait, let's re - interpret: Time (hours) on the x - axis (0,1,2,3,4,5) and Distance (miles) on the y - axis (0,20,40,60,80). Then the line goes from (0,0) to (2,20), then a curve to (4,60)? No, the original graph: the vertical axis is Time (hours) with 1,2,3,4,5 (upwards? No, usually time increases upwards or to the right. Wait, the arrow on the time axis is downwards, so time increases as we go down? So at the top of the time axis (y=0) is time 0, then y=1 is time 1 hour, y=2 is time 2 hours, etc. So the point at time t=2 (y=2) is at distance x=20 (so (20,2)), and at time t=4 (y=4) is at distance x=60 (60,4). So the change in time \( \Delta t = 4 - 2 = 2 \) hours. The change in distance \( \Delta d = 60 - 20 = 40 \) miles? Wait, no, if time is on the y - axis (increasing downwards), then t=2 is higher up, t=4 is lower down. So the distance at t=2 is 20 miles, at t=4 is 60 miles? Wait, no, that would mean distance increases as time increases, but the graph has a curve that is concave? Wait, maybe the graph is distance vs time, with time on the x - axis (horizontal) and distance on the y - axis (vertical). Let's assume that: x - axis is time (hours: 0,1,2,3,4,5), y - axis is distance (miles: 0,20,40,60,80). Then the first point at x=2 (time 2 hours) is y=20 (distance 20 miles), and at x=4 (time 4 hours) is y=60 (distance 60 miles)? No, the curve: from (0,0) to (2,20) (a line), then to (3,40), then to (4,60)? Wait, the problem says "average rate of change from 2 to 4 hours". The formula for average rate of change is \( \frac{f(b)-f(a)}{b - a} \), where \( a = 2 \), \( b = 4 \), and \( f(t) \) is the distance at time t. So we need to find \( f(2) \) and \( f(4) \) from the graph. Looking at the graph: at t=2 (x=2), the distance (y) is 20 miles. At t=4 (x=4), the distance (y) is 60 miles? Wait, no, maybe the graph is time on the y - axis (vertical) and distance on the x - axis (horizontal), so the coordinates are (distance, time). So at time t=2 (y=2), distance d=20 (x=20): (20,2). At time t=4 (y=4), distance d=60 (x=60): (60,4). Then the average rate of change of distance with respect to time is \( \frac{\Delta d}{\Delta t}=\frac{60 - 20}{4 - 2}=\frac{40}{2}=20 \)? Wait, no, that can't be. Wait, maybe I have the axes reversed. Let's think again: average rate of change of distance over time is \( \frac{\text{final distance}-\text{initial distance}}{\text{final time}-\text{initial time}} \). If at t=2 hours, the distance is 20 miles, and at t=4 hours, the distance is 60 miles, then \( \frac{60 - 20}{4 - 2}=\frac{40}{2}=20 \) mph. But wait, the graph: the first segment is a line from (0,0) (time 0, distance 0) to (2,20) (time 2, distance 20), then a curve to (4,60) (time 4, distance 60)? No, the curve looks like it's decreasing? Wait, no, the arrow on the distance axis is to the right (increasing distance), and time axis is downwards (increasing time). So as time increases (going down the y - axis), distance increases (going right on the x - axis). Wait, no, that would mean that as time passes (increases), distance increases, which is normal (moving forward). But the curve is concave, so the speed is decreasing? Wait, no, the average rate of change from t=2 to t=4. Let's get the correct points:

From the graph:

- At t = 2 hours (y - coordinate 2), the distance (x - coordinate) is 20 miles.

- At t = 4 hours (y - coordinate 4), the distance (x - coordinate) is 60 miles? Wait, no, maybe the graph is distance on the y - axis and time on the x - axis, with the labels swapped. Let's assume time is on the x - axis (x=2, x=4) and distance on the y - axis (y=20 at x=2, y=60 at x=4). Then the average rate of change is \( \frac{60 - 20}{4 - 2}=\frac{40}{2}=20 \) mph. But wait, maybe the points are (2,20) and (4,60) for (time, distance). Then the average rate of change is \( \frac{60 - 20}{4 - 2}=20 \) mph. Wait, but maybe I misread the graph. Wait, the user's graph: the time axis is vertical (with 1,2,3,4,5) and distance horizontal (0,20,40,60,80). The first point at t=2 (y=2) is at x=20, then at t=3 (y=3) at x=40, then at t=4 (y=4) at x=60? So the change in distance from t=2 to t=4 is 60 - 20 = 40 miles, change in time is 4 - 2 = 2 hours. So average rate of change is \( \frac{40}{2}=20 \) mph? Wait, no, that would be if distance is increasing, but the graph looks like a curve that is concave, maybe decreasing? Wait, maybe the distance is decreasing. Wait, the starting point is (0,0) (distance 0, time 0), then at t=2, distance is 20, at t=4, distance is 60? No, that's increasing. But maybe the graph is time vs distance, with distance on the y - axis and time on the x - axis, and the curve is decreasing. Wait, maybe I have the axes reversed. Let's check the problem statement: "Determine the average rate of change, in mph, from 2 to 4 hours on the graph shown below." So time is the independent variable (x - axis), distance is dependent (y - axis). So let's re - express:

Let’s denote:

- At \( t_1 = 2 \) hours, the distance \( d_1 = 20 \) miles (from the graph, when x=2, y=20).

- At \( t_2 = 4 \) hours, the distance \( d_2 = 60 \) miles (when x=4, y=60).

Then the average rate of change is \( \frac{d_2 - d_1}{t_2 - t_1}=\frac{60 - 20}{4 - 2}=\frac{40}{2}=20 \) mph. Wait, but that seems high. Wait, maybe the distance at t=4 is 60? Or maybe the distance at t=2 is 20 and at t=4 is 60? Wait, no, maybe the graph is distance on the y - axis and time on the x - axis, and the points are (2,20) and (4,60), so the slope (average rate of change) is \( \frac{60 - 20}{4 - 2}=20 \). Alternatively, maybe the distance at t=2 is 20 and at t=4 is 60, so the average rate is 20 mph.

Wait, maybe I made a mistake. Let's look again: the graph has Time (hours) on the vertical axis (with 1,2,3,4,5) and Distance (miles) on the horizontal axis (0,20,40,60,80). The curve starts at (0,0) (distance 0, time 0), goes to (20,2) (distance 20, time 2), then to (40,3) (distance 40, time 3), then to (60,4) (distance 60, time 4). So the change in distance from time 2 to time 4 is 60 - 20 = 40 miles, change in time is 4 - 2 = 2 hours. So average rate of change is \( \frac{40}{2}=20 \) mph.

Step2: Apply the average rate of change formula

The formula for the average rate of change of a function \( f(t) \) (distance as a function of time) from \( t = a \) to \( t = b \) is \( \frac{f(b)-f(a)}{b - a} \).

Here, \( a = 2 \), \( b = 4 \), \( f(2)=20 \) miles, \( f(4)=60 \) miles.

So, \( \text{Average rate of change}=\frac{60 - 20}{4 - 2}=\frac{40}{2}=20 \) mph. Wait, but that seems like the distance is increasing. But maybe the graph is showing distance decreasing? Wait, maybe I have the time axis reversed. If time increases upwards (y - axis), then t=2 is above t=4, so \( t_1 = 4 \), \( t_2 = 2 \), but the problem says from 2 to 4 hours, so t increases from 2 to 4, so time should be increasing as we go down the y - axis (since the arrow is downwards). So t=2 is higher (earlier time), t=4 is lower (later time). So distance at t=2 is 20, at t=4 is 60, so distance increases, rate is positive 20. But maybe the graph is different. Wait, maybe the distance at t=2 is 20 and at t=4 is 60, so the average rate is 20.

Wait, maybe I made a mistake in the points. Let's check the graph again: the first point at t=2 (y=2) is at x=20, then at t=4 (y=4) is at x=60. So the change in distance is 60 - 20 = 40, change in time is 4 - 2 = 2, so 40/2 = 20. So the average rate of change is 20 mph.

Answer:

20