Probably really simple, I'm just not seeing it
 

If a ball is thrown into the air with a velocity of 40ft/s, its height i feet is after T seconds is given by y=40T-16T^2. Find the average velocity for the time period beginning when T = 2 and lasting

.5 sec
.1 sec
.05sec
.01 sec.

I got .5 sec at -32. Which is the correct answer. But when I try the same way for the other numbers I get the wrong answer. For the -32 I just put in .5 for T. Thanks.

i had a question just like that on my physics midterm i just took

just plug in the numbers into the equation

To find the average velocity you find the area under the curve from [a,b] divided by (b-a). Since the antiderivative of velocity is the position function. Just plug in (y(b)-y(a))/(b-a). If you case, the first b would be 2.5 and a will be 2.

You're given the formula of the position as a function of time. People seem to forget that the derivative gives you the slope of the tangent, which is the instantaneous velocity. Because this is often what we want, we often simply call it velocity.

However, sometimes we want the average velocity, which is simply the slope of the secant of the function (during some interval). It can be proven, using calculus, that the slope of the secant is equal to the integral of the function divided by the length of the interval.

Anyway, back to your original problem. You have the position as a function of time. So, simply find the position at your first point in time and then find your position at the second point in time. Divide the difference in position with the length of time and that will be your average velocity.

I hope this helps. If you don't understand any part of this, please say so and I will either explain it in more detail or explain it more simply, depending on what you don't understand.
Good luck!

KnifeMissle

You wouldn't happen to go to U of W do you?

This is probably more appropriate as a personal message if for no other reason than to not bore other people with our personal and trivial chatter.

I did go the U(W) but I haven't been in school for some time. I did recently (last week or so) go to DC to attend Prof. Aho's presentation on modern compilers.

What gave me away? The fact that I'm a geeky geek from Waterloo?

The answers in the book are as follows
-32
-25.6
-24.8
-24.16

What I ended up doing was working backwards from these answers. 1 minus each of the numbers provided at the top of this topic gave what t is. I'm not sure why this is.

I'm sorry, was I not clear earlier? Did my post not make any sense? Did you not read it?

Let me explain this again. The average velocity is the slope of the secant of the function during the interval in question. It is analogous to how the instantaneous velocity is the slope of the tangent at a point. Basically, the secant is a tangent during an interval instead of at a point.

So, let's examine the slope of the secant line during these intervals.

The slope of the secant is the rise divided by the run. So, if we let t be the delta time (rather than simply the variable), the formula for delta y looks like this...
Since the slope of the secant is simply delta y - delta x = (-16t^2 - 24t)/t, we can just plug your numbers into this formula and it will all work out.

This is simple, and it does not need calculus.

Simply plug in the time period over which you want to know the average velocity into the distance equation, and then divide that answer by the time duration.

I.e for t=.5 then y=40*.5-16*.5^2=16.
then divide that answer by the time to get average velocity: avgV=y(t)/t=16/.5=32


Don't overthing the problems.

The average velocity will determine the position of the ball at time t, so take the position and divide it by that time.

By this logic, the answer to the second value, 0.1, is avgV=y(t)/t=3.84/0.1 = 38.4

I don't think you can ever overthink a problem but you can definitely underthink it!

This is the incorrect answer. Can you figure out what you missed?
At least you are correct in saying that this is simple and not a question of calculus...