Calculating Average Water Flow Rate with f(t) Function

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To calculate the average water flow rate given by the function f(t) over the interval 0<t<4, the correct formula is the integral method, specifically \(\frac{1}{4}\int f(t) dt\) evaluated from 0 to 4. Using the formula f(4)-f(0) / (4-0) can lead to incorrect results, especially if f(t) is constant. If f(t) is constant, the average flow rate simplifies to that constant value. The integral approach accurately reflects the total water flow over the time period divided by the duration. This method is applicable for any function f(t) representing the rate of water flow.
fiziksfun
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the rate at which water is sprayed is given by the function f(t)

during the time interval 0<t<4, what is the average rate of water flow??

I'm confused whether to use the formula

f(4)-f(0) / (4-0)

OR

\frac{1}{4}\intf(t) evaluated from 0 to 4.

help!
 
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Hi fiziksfun ! :smile:

Hint: Suppose f(t) is a constant, C. Then f(4) = f(0) = C.

So which formula is right? :smile:
 
i'm kind of slow, so i still don't understand :[
can you help me more??
 
Sure! :smile:

If f(t) = C, a constant, then obviously the average of f(t) is C.

But f(4)-f(0) / (4-0) = (C - C)/4 = 0, which obviously is wrong. :frown:

And ∫f(t)/4 = ∫C/4 (evaluated from t = 0 to 4) = C, which equally obviously is right! :smile:

This works for any f(t), because f(t) is the rate of water, so ∫f(t) is the total water.

And so the average rate of water = total/time = ∫f(t)/4. :smile:
 
ah, i think i understand, thank you!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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