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Average Rate of Change (word problem) 
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#1
Sep1107, 03:44 PM

P: 45

I'm having trouble understanding this word problem.
A construction worker drops a bolt while woring on a highrise building 320m above the ground. After t seconds, the bolt has fallen a distance of s metres, where s(t) = 3205t^2, o < t < 8 a. Find the average velocity during the first, third and eighth seconds. I have to use f(x + h)  f(x)/h I got 5(t +h) The answer is 5m/s for the first one. How do I get the rest of them (third and eighth second)? 


#2
Sep1107, 03:53 PM

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the "first second" is from t= 0 to t=1, of course: average speed during the first second is s(1) s(0) since here h= 1. The "third second" is between t= 2 and t= 3 so average speed during the third second is s(3) s(2) and the "eighth" second is between t= 7 and t= 8 so average speed for the eighth second is s(9) s(8).



#3
Sep1107, 03:58 PM

P: 45

so I plug in h into 5(t+h)^2???? 


#4
Sep1107, 04:41 PM

P: 45

Average Rate of Change (word problem)
this is what I have so far 


#5
Sep1207, 06:27 AM

P: 14

You are making the problem too difficult. Velocity is the change of position over time. The formula you list is the formula for position.



#6
Sep1207, 11:16 AM

P: 3,015

[tex]\frac{s(t+h)s(t)}{h}[/tex] is the formula for average velocity.
It is a very wordy way of saying [tex]\frac{\Delta x}{\Delta y}[/tex], which is the change in y over the change in x. (<slope) Now in order to see this, you know that [tex]\Delta y =y_{final}y_{initial}[/tex] right? Can you see how that is the same as [tex]s(t+h)s(t)[/tex] ? It is saying: take the position of some initial time (t) PLUS some amount h and subtract the original initial time from it. So in your second problem after 3 seconds, since your initial time is 0 and your final time is 3 then t+h means 0+3 which is of course 3. You see h is the change in time from 0 to 3. So now back to the problem; you should evaluate it as [tex]v_{avg}=\frac{s(3+0)s(o)}{3}[/tex] 


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