Math / Calculus

QuestionEurrent Attempt in Progress
A village wishes to measure the quantity of water that is piped to a factory during a typical morning. A gauge on the water line gives the flow rate (in cubic meters per hour) at any instant. The flow rate is about $140 \mathrm{~m}^{3} / \mathrm{hr}$ at 6 am and increases steadily to about 280 $\mathrm{m}^{3} / \mathrm{hr}$ at 9 am . Using only this information, give your best estimate of the total volume of water used by the factory between 6 am and 9 am.
Best estimate = i $\square$ $\mathrm{m}^{3}$

Studdy Solution

STEP 1

What is this asking? How much water, in cubic meters, flowed through the pipe in those three hours, if it started at a rate of 140 and went up to 280 at a constant rate? Watch out! The flow rate changes!
Don't just multiply one of the rates by the time.

STEP 2

1. Find the average flow rate.
2. Calculate the total water usage.

STEP 3

Since the flow rate increases steadily, we can use the average flow rate as if it were constant for those three hours.
It's like if you drove a car and steadily increased your speed--you could find your average speed and use that to figure out how far you went!

STEP 4

To find the average, we add the starting and ending rates and divide by 2! $\text{Average flow rate} = \frac{\text{Starting rate} + \text{Ending rate}}{2}$ $\text{Average flow rate} = \frac{140 \frac{\text{m}^3}{\text{hr}} + 280 \frac{\text{m}^3}{\text{hr}}}{2}$ $\text{Average flow rate} = \frac{420 \frac{\text{m}^3}{\text{hr}}}{2}$ $\text{Average flow rate} = 210 \frac{\text{m}^3}{\text{hr}}$ So, the average flow rate is 210 $\frac{\text{m}^3}{\text{hr}}$ !

STEP 5

Now we can multiply the average flow rate by the time to get the total volume.
The time is 3 hours (from 6 am to 9 am).
Remember, $\frac{\text{m}^3}{\text{hr}} \cdot \text{hr} = \text{m}^3$ , which is what we want!

STEP 6

$\text{Total volume} = \text{Average flow rate} \cdot \text{Time}$ $\text{Total volume} = 210 \frac{\text{m}^3}{\text{hr}} \cdot 3 \text{ hr}$ $\text{Total volume} = 630 \text{ m}^3$ So, our best estimate is 630 $\text{m}^3$ !

STEP 7

Best estimate = 630 $\text{m}^3$

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Studdy Solution

STEP 1

What is this asking? How much water, in cubic meters, flowed through the pipe in those three hours, if it started at a rate of 140 and went up to 280 at a constant rate? Watch out! The flow rate changes!
Don't just multiply one of the rates by the time.

STEP 2

1. Find the average flow rate.
2. Calculate the total water usage.

STEP 3

Since the flow rate increases steadily, we can use the average flow rate as if it were constant for those three hours.
It's like if you drove a car and steadily increased your speed--you could find your average speed and use that to figure out how far you went!

STEP 4

STEP 5

Now we can multiply the average flow rate by the time to get the total volume.
The time is 3 hours (from 6 am to 9 am).
Remember, $\frac{\text{m}^3}{\text{hr}} \cdot \text{hr} = \text{m}^3$ , which is what we want!

STEP 6

$\text{Total volume} = \text{Average flow rate} \cdot \text{Time}$ $\text{Total volume} = 210 \frac{\text{m}^3}{\text{hr}} \cdot 3 \text{ hr}$ $\text{Total volume} = 630 \text{ m}^3$ So, our best estimate is 630 $\text{m}^3$ !

STEP 7

Best estimate = 630 $\text{m}^3$

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Studdy Solution

STEP 1

What is this asking? How much water, in cubic meters, flowed through the pipe in those three hours, if it started at a rate of 140 and went up to 280 at a constant rate? Watch out! The flow rate *changes*!Don't just multiply one of the rates by the time.

STEP 2

1. Find the average flow rate.2. Calculate the total water usage.

STEP 3

Since the flow rate increases *steadily*, we can use the average flow rate as if it were constant for those three hours.It's like if you drove a car and steadily increased your speed--you could find your average speed and use that to figure out how far you went!

STEP 4

STEP 5

Now we can multiply the average flow rate by the time to get the total volume.The time is 3 hours (from 6 am to 9 am).Remember, m3hr⋅hr=m3\frac{\text{m}^3}{\text{hr}} \cdot \text{hr} = \text{m}^3 hrm3​⋅hr=m3, which is what we want!

STEP 6

STEP 7

Best estimate = 630 m3\text{m}^3m3

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? How much water, in cubic meters, flowed through the pipe in those three hours, if it started at a rate of 140 and went up to 280 at a constant rate? Watch out! The flow rate changes!
Don't just multiply one of the rates by the time.

1. Find the average flow rate.
2. Calculate the total water usage.

Since the flow rate increases steadily, we can use the average flow rate as if it were constant for those three hours.
It's like if you drove a car and steadily increased your speed--you could find your average speed and use that to figure out how far you went!

Now we can multiply the average flow rate by the time to get the total volume.
The time is 3 hours (from 6 am to 9 am).
Remember, $\frac{\text{m}^3}{\text{hr}} \cdot \text{hr} = \text{m}^3$ , which is what we want!

Best estimate = 630 $\text{m}^3$