Question

1. oscar participates in a charity walk. the graph shows his distance in miles from the water stop as a function of time. how many miles did oscar walk? explain your answer. see example 3

for the graph shown, find the rate of change over the interval. see example 4

1. ( 3 leq t leq 6 )
2. ( 7 leq t leq 10 )

Explanation:

Problem 26: Oscar's Charity Walk Distance

Step 1: Analyze the graph's segments

The graph has two main segments: from start (0 min, 1.5 miles? Wait, looking at the y-axis, start is at (0, 1.5?) Wait, no, the start point: when time \( t = 0 \), distance \( y \) is 1.5? Wait, the first segment is from \( t = 0 \) to \( t = 30 \) min, going from \( y = 1.5 \) (start) to \( y = 0 \) (water stop). The second segment is from \( t = 30 \) min to \( t = 60 \) min, going from \( y = 0 \) to \( y = 1.5 \) (finish). Wait, actually, the distance from the water stop: so when he starts, he's 1.5 miles from water stop, then walks to water stop (distance 1.5 miles), then walks back to finish, which is 1.5 miles from water stop. So total distance is the sum of the two segments.

Step 2: Calculate each segment's distance

First segment: from start (distance 1.5 miles from water stop) to water stop (0 miles from water stop). So distance walked here is \( 1.5 - 0 = 1.5 \) miles. Second segment: from water stop (0 miles) to finish (1.5 miles from water stop). So distance walked here is \( 1.5 - 0 = 1.5 \) miles.

Step 3: Sum the distances

Total distance = \( 1.5 + 1.5 = 3 \) miles. Wait, but let's check the graph's y-axis. Wait, the start point: when \( t = 0 \), \( y \) is 1.5? Wait, the graph's y-axis: the first point is at (0, 1.5) (start), then at (30, 0) (water stop), then at (60, 1.5) (finish). So the distance from water stop: so when moving from start to water stop, he's decreasing distance by 1.5 miles (walking towards water stop), then increasing by 1.5 miles (walking away from water stop to finish). So total distance is \( 1.5 + 1.5 = 3 \) miles.

Step 1: Identify points on the graph

For \( t = 3 \) and \( t = 6 \), find \( d(t) \). From the graph, when \( t = 3 \), \( d(t) = 4 \) (since at \( t = 3 \), height is 4? Wait, the graph: at \( t = 0 \), \( d(t) = 0 \); \( t = 6 \), \( d(t) = 8 \); wait, no, let's check the grid. The x-axis (time) has grid lines, and y-axis (height) too. Wait, for \( t = 3 \): looking at the graph, at \( t = 3 \), what's \( d(t) \)? Let's see, the line from \( t = 0 \) to \( t = 6 \) is increasing. At \( t = 6 \), \( d(t) = 8 \). At \( t = 3 \), since it's linear, the midpoint: from \( t = 0 \) (0) to \( t = 6 \) (8), so at \( t = 3 \), \( d(t) = 4 \).

Step 2: Apply rate of change formula

Rate of change (slope) is \( \frac{\Delta d(t)}{\Delta t} = \frac{d(6) - d(3)}{6 - 3} \).

Step 3: Calculate values

\( d(6) = 8 \), \( d(3) = 4 \), \( \Delta t = 6 - 3 = 3 \). So rate of change is \( \frac{8 - 4}{3} = \frac{4}{3} \approx 1.33 \) ft/s. Wait, no, wait: wait, the y-axis is height (ft), x-axis time (s). Wait, at \( t = 6 \), height is 8 ft; at \( t = 3 \), height is 4 ft. So \( \Delta d = 8 - 4 = 4 \), \( \Delta t = 6 - 3 = 3 \). So rate is \( \frac{4}{3} \) ft/s? Wait, no, maybe I misread the graph. Wait, the graph: from \( t = 0 \) to \( t = 6 \), it goes from (0,0) to (6,8). So slope is \( \frac{8 - 0}{6 - 0} = \frac{8}{6} = \frac{4}{3} \) ft/s. But for the interval \( 3 \leq t \leq 6 \), since it's a straight line, the slope is the same. So \( d(3) = 4 \), \( d(6) = 8 \), so \( \frac{8 - 4}{6 - 3} = \frac{4}{3} \) ft/s.

Step 1: Identify points on the graph

For \( t = 7 \) and \( t = 10 \), find \( d(t) \). The graph from \( t = 6 \) to \( t = 12 \) is decreasing. At \( t = 6 \), \( d(t) = 8 \); at \( t = 12 \), \( d(t) = 0 \). Let's find \( d(7) \) and \( d(10) \). The slope from \( t = 6 \) to \( t = 12 \) is \( \frac{0 - 8}{12 - 6} = \frac{-8}{6} = -\frac{4}{3} \) ft/s. For \( t = 7 \): \( t = 7 \) is 1 second after \( t = 6 \), so \( d(7) = 8 + (-\frac{4}{3})(1) = \frac{24}{3} - \frac{4}{3} = \frac{20}{3} \) ft. For \( t = 10 \): \( t = 10 \) is 4 seconds after \( t = 6 \), so \( d(10) = 8 + (-\frac{4}{3})(4) = 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} \) ft.

Step 2: Apply rate of change formula

Rate of change is \( \frac{\Delta d(t)}{\Delta t} = \frac{d(10) - d(7)}{10 - 7} \).

Step 3: Calculate values

\( d(10) - d(7) = \frac{8}{3} - \frac{20}{3} = \frac{-12}{3} = -4 \). \( \Delta t = 10 - 7 = 3 \). So rate of change is \( \frac{-4}{3} \) ft/s (or approximately -1.33 ft/s).

Answer:

Oscar walked 3 miles. He walked 1.5 miles to the water stop (from 1.5 miles away to 0 miles) and then 1.5 miles from the water stop to the finish, so \( 1.5 + 1.5 = 3 \) miles.

Problem 27: Rate of Change for \( 3 \leq t \leq 6 \)