# Average Rate of Change (word problem)

by xCanx
Tags: average, rate, word
 P: 45 I'm having trouble understanding this word problem. A construction worker drops a bolt while woring on a high-rise building 320m above the ground. After t seconds, the bolt has fallen a distance of s metres, where s(t) = 320-5t^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)?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,683 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).
P: 45
 Quote by HallsofIvy 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).
Thank you :)

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

 P: 45 Average Rate of Change (word problem) this is what I have so far
 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.
 P: 3,016 $$\frac{s(t+h)-s(t)}{h}$$ is the formula for average velocity. It is a very wordy way of saying $$\frac{\Delta x}{\Delta y}$$, which is the change in y over the change in x. (<--slope) Now in order to see this, you know that $$\Delta y =y_{final}-y_{initial}$$ right? Can you see how that is the same as $$s(t+h)-s(t)$$ ? 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 $$v_{avg}=\frac{s(3+0)-s(o)}{3}$$

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