Math / Algebra

QuestionSlope As a Rate of Change Slope is often referred to as a rate of change because it compares the change in one variable (the rise) to the change in another variable (the run). Chris runs each day as part of his daily exercise. The graph shows his distance from home as he runs his route.
Calculate his rate of change (speed) for each segment of the graph and describe what is happening in each segment. Don't forget to include units in your calculations! \begin{tabular}{|c|c|c|} \hline Segment & \begin{tabular}{c} Slope (Rate of \\ Change) \end{tabular} & \\ \hline AB & & \\ \hline BC & & \\ \hline CD & & \\ \hline DE & & \\ \hline \end{tabular} 17

Studdy Solution

STEP 1

1. The x-axis represents time in minutes.
2. The y-axis represents distance in meters.
3. Slope is calculated as the change in distance divided by the change in time.
4. Each segment represents a different part of Chris's run.
5. We need to calculate the slope for each segment and describe what is happening.

STEP 2

1. Calculate the slope for Segment AB.
2. Calculate the slope for Segment BC.
3. Calculate the slope for Segment CD.
4. Calculate the slope for Segment DE.
5. Calculate the slope for Segment EF.
6. Describe what is happening in each segment.

STEP 3

Calculate the slope for Segment AB.
The coordinates for Segment AB are (0, 0) and (5, 200).
The formula for slope (m) is: $m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$
For Segment AB: $m_{AB} = \frac{200 - 0}{5 - 0} = \frac{200}{5} = 40$
The slope for Segment AB is $40 \, \text{m/min}$ .

STEP 4

Calculate the slope for Segment BC.
The coordinates for Segment BC are (5, 200) and (10, 200).
For Segment BC: $m_{BC} = \frac{200 - 200}{10 - 5} = \frac{0}{5} = 0$
The slope for Segment BC is $0 \, \text{m/min}$ .

STEP 5

Calculate the slope for Segment CD.
The coordinates for Segment CD are (10, 200) and (15, 400).
For Segment CD: $m_{CD} = \frac{400 - 200}{15 - 10} = \frac{200}{5} = 40$
The slope for Segment CD is $40 \, \text{m/min}$ .

STEP 6

Calculate the slope for Segment DE.
The coordinates for Segment DE are (15, 400) and (20, 400).
For Segment DE: $m_{DE} = \frac{400 - 400}{20 - 15} = \frac{0}{5} = 0$
The slope for Segment DE is $0 \, \text{m/min}$ .

STEP 7

Calculate the slope for Segment EF.
The coordinates for Segment EF are (20, 400) and (25, 0).
For Segment EF: $m_{EF} = \frac{0 - 400}{25 - 20} = \frac{-400}{5} = -80$
The slope for Segment EF is $-80 \, \text{m/min}$ .

STEP 8

Describe what is happening in each segment.
- Segment AB: Chris is running away from home at a speed of $40 \, \text{m/min}$ . - Segment BC: Chris is stationary, as the distance does not change, with a speed of $0 \, \text{m/min}$ . - Segment CD: Chris resumes running away from home at a speed of $40 \, \text{m/min}$ . - Segment DE: Chris is stationary again, with a speed of $0 \, \text{m/min}$ . - Segment EF: Chris is running back towards home at a speed of $80 \, \text{m/min}$ .

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Studdy Solution

STEP 1

STEP 2

1. Calculate the slope for Segment AB.
2. Calculate the slope for Segment BC.
3. Calculate the slope for Segment CD.
4. Calculate the slope for Segment DE.
5. Calculate the slope for Segment EF.
6. Describe what is happening in each segment.

STEP 3

STEP 4

Calculate the slope for Segment BC.
The coordinates for Segment BC are (5, 200) and (10, 200).
For Segment BC: $m_{BC} = \frac{200 - 200}{10 - 5} = \frac{0}{5} = 0$
The slope for Segment BC is $0 \, \text{m/min}$ .

STEP 5

Calculate the slope for Segment CD.
The coordinates for Segment CD are (10, 200) and (15, 400).
For Segment CD: $m_{CD} = \frac{400 - 200}{15 - 10} = \frac{200}{5} = 40$
The slope for Segment CD is $40 \, \text{m/min}$ .

STEP 6

Calculate the slope for Segment DE.
The coordinates for Segment DE are (15, 400) and (20, 400).
For Segment DE: $m_{DE} = \frac{400 - 400}{20 - 15} = \frac{0}{5} = 0$
The slope for Segment DE is $0 \, \text{m/min}$ .

STEP 7

Calculate the slope for Segment EF.
The coordinates for Segment EF are (20, 400) and (25, 0).
For Segment EF: $m_{EF} = \frac{0 - 400}{25 - 20} = \frac{-400}{5} = -80$
The slope for Segment EF is $-80 \, \text{m/min}$ .

STEP 8

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Studdy Solution

STEP 1

STEP 2

1. Calculate the slope for Segment AB.2. Calculate the slope for Segment BC.3. Calculate the slope for Segment CD.4. Calculate the slope for Segment DE.5. Calculate the slope for Segment EF.6. Describe what is happening in each segment.

STEP 3

STEP 4

Calculate the slope for Segment BC.The coordinates for Segment BC are (5, 200) and (10, 200).For Segment BC: mBC=200−20010−5=05=0 m_{BC} = \frac{200 - 200}{10 - 5} = \frac{0}{5} = 0 mBC​=10−5200−200​=50​=0The slope for Segment BC is 0 m/min 0 \, \text{m/min} 0m/min.

STEP 5

Calculate the slope for Segment CD.The coordinates for Segment CD are (10, 200) and (15, 400).For Segment CD: mCD=400−20015−10=2005=40 m_{CD} = \frac{400 - 200}{15 - 10} = \frac{200}{5} = 40 mCD​=15−10400−200​=5200​=40The slope for Segment CD is 40 m/min 40 \, \text{m/min} 40m/min.

STEP 6

Calculate the slope for Segment DE.The coordinates for Segment DE are (15, 400) and (20, 400).For Segment DE: mDE=400−40020−15=05=0 m_{DE} = \frac{400 - 400}{20 - 15} = \frac{0}{5} = 0 mDE​=20−15400−400​=50​=0The slope for Segment DE is 0 m/min 0 \, \text{m/min} 0m/min.

STEP 7

STEP 8

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1. Calculate the slope for Segment AB.
2. Calculate the slope for Segment BC.
3. Calculate the slope for Segment CD.
4. Calculate the slope for Segment DE.
5. Calculate the slope for Segment EF.
6. Describe what is happening in each segment.

Calculate the slope for Segment BC.
The coordinates for Segment BC are (5, 200) and (10, 200).
For Segment BC: $m_{BC} = \frac{200 - 200}{10 - 5} = \frac{0}{5} = 0$
The slope for Segment BC is $0 \, \text{m/min}$ .

Calculate the slope for Segment CD.
The coordinates for Segment CD are (10, 200) and (15, 400).
For Segment CD: $m_{CD} = \frac{400 - 200}{15 - 10} = \frac{200}{5} = 40$
The slope for Segment CD is $40 \, \text{m/min}$ .

Calculate the slope for Segment DE.
The coordinates for Segment DE are (15, 400) and (20, 400).
For Segment DE: $m_{DE} = \frac{400 - 400}{20 - 15} = \frac{0}{5} = 0$
The slope for Segment DE is $0 \, \text{m/min}$ .