# Average Rate of Change (word problem)

1. Sep 11, 2007

### xCanx

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)?

2. Sep 11, 2007

### 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).

3. Sep 11, 2007

### xCanx

Thank you :)

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

4. Sep 11, 2007

### xCanx

this is what I have so far

5. Sep 12, 2007

### kev1829

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. Sep 12, 2007

$$\frac{s(t+h)-s(t)}{h}$$ 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
$$v_{avg}=\frac{s(3+0)-s(o)}{3}$$

Last edited: Sep 12, 2007