find the average velocity given the function s(t) = -4.9t^2 + 27t + 25

[mrnerd]

The position of an object moving along a line is given by the function s(t) = 4.9t^2 + 27t + 25
find the average velocity over the interval [0, 1 + h]

s(0) = 25

s(1+h) = -4.9(1+h)^2 + 27(1 + h) + 25
(-4.9h - 4.9)(h + 1) + 27(h + 1) + 25
-4.9h^2 - 9.8h - 22.1 + 27h + 25

(s(1 + h) - s(0))/(1 + h - 0) = (-4.9h^2 + 17.2h + 22.1)/(1 + h)

this doesn't fit evenly, so I then used the quadratic formula to try and get to the next step, and I got:
(-17.2 + 27)/-9.8 and (-17.2 - 27)/-9.8
which leads to (h + 1)(h + 4.510204082) as answers and
h + 4.510204082 as my final answer
needless to say this is the wrong answer, so where have I gone wrong?

thanks for your help

 

[stapel]

The position of an object moving along a line is given by the function s(t) = 4.9t^2 + 27t + 25
find the average velocity over the interval [0, 1 + h]

s(0) = 25

s(1+h) = -4.9(1+h)^2 + 27(1 + h) + 25
(-4.9h - 4.9)(h + 1) + 27(h + 1) + 25
-4.9h^2 - 9.8h - 22.1 + 27h + 25

(s(1 + h) - s(0))/(1 + h - 0) = (-4.9h^2 + 17.2h + 22.1)/(1 + h)

this doesn't fit evenly...

What do you mean when you say that "this" (what?) "doesn't fit evenly" (into what?)?

...so I then used the quadratic formula to try and get to the next step...

What did you set equal to zero? What was the logical basis for this step?

...and I got:
(-17.2 + 27)/-9.8 and (-17.2 - 27)/-9.8
which leads to (h + 1)(h + 4.510204082) as answers....

How did you get to this point? What were all of your steps?

Note: It may be more helpful to leave things in factored form:

. . . . .s(1 + h) = 4.9(1 + h)² + 27(1 + h) + 25

. . . . .s(1 + h) - s(0) = [4.9(1 + h)² + 27(1 + h) + 25] - [25] = 4.9(1 + h)² + 27(1 + h)

. . . . .[s(1 + h) - s(0)] / [1 + h] = [4.9(1 + h)² + 27(1 + h)] / [1 + h] = 4.9(1 + h) + 27

...and so forth.

 

[Ishuda]

The position of an object moving along a line is given by the function s(t) = 4.9t^2 + 27t + 25
find the average velocity over the interval [0, 1 + h]

s(0) = 25

s(1+h) = -4.9(1+h)^2 + 27(1 + h) + 25
(-4.9h - 4.9)(h + 1) + 27(h + 1) + 25
-4.9h^2 - 9.8h - 22.1 + 27h + 25

(s(1 + h) - s(0))/(1 + h - 0) = (-4.9h^2 + 17.2h + 22.1)/(1 + h)

this doesn't fit evenly, so I then used the quadratic formula to try and get to the next step, and I got:
(-17.2 + 27)/-9.8 and (-17.2 - 27)/-9.8
which leads to (h + 1)(h + 4.510204082) as answers and
h + 4.510204082 as my final answer
needless to say this is the wrong answer, so where have I gone wrong?

thanks for your help

I believe you are approaching this the wrong way [unless, of course, your instructor told you to average the two end points]. The way I would do it, starting with a function f(x), would be to think of it this way:
First let
dx = (b-a)/n
where the average of the function is between a and b and n is some arbitrarily large number. Now let
x_j = a + j dx, j = 0, 1, 2, ..., n-1.
Then the average (A) is about
A ~ [ f(x₀) + f(x₁) + f(x₂) + ... + f(x_n-1) ] / n
or
A ~ \(\displaystyle \dfrac{1}{b-a}\) \(\displaystyle \Sigma_{j=0}^{n-1}\,\, f(x_j)\,\, dx\)
or letting n become very large (go to infinity)
A = \(\displaystyle \dfrac{1}{b-a}\) \(\displaystyle \int_{a}^{b}\,\, f(x)\,\, dx\)

EDIT: Since you have a position function