Question

How would I find the average velocity from t=1 to t=3 using the Mean Value Theorem
given v(t) = t^2 + 3t - 1 ?

Answer 1

would it be v(3) - v(1) all over 3-1?

Answer 2

hmmm

Answer 3

but I have a feeling that that might get me something that is not velocity

Answer 4

no, mvt simply says there exists a point between v(3) and v(1)
that is the same as the mean

Answer 5

why do you think that

Answer 6

m/s /s =/= m/s

Answer 7

what does diving by number hav to do with he units

Answer 8

sec i gotta put stuff downstaira ill help u

Answer 9

http://www.intuitive-calculus.com/images/mean4.gif
http://www.intuitive-calculus.com/mean-value-theorem.html

Answer 10

so should I do
a(3)-a(1)
--------
3-1
to get average velocity?

Answer 11

that according to your link

Answer 12

is the slope

Answer 13

so its the change in velocity

Answer 14

my head's going in circles...

Answer 15

ok wait look at htis part

Answer 16

f'(c)=f(b)-f(a)
-------
b-a

Answer 17

you se that part

Answer 18

im not sure why tha is true, ill have to see the proof but

Answer 19

this means whereever your mean is, it has that as its tangent line

Answer 20

so lets see where this tangent line belongs on your curve

Answer 21

if you keep scrolling in the second link, there's a proof

Answer 22

lets draw something here to see what is going on

Answer 23

\[v(t)=t^2+3t-1\\
v(t)=(t+3/2)^2-1-9/4\\
~~~~~~=(t+3/2)^2-13/4 \]

Answer 24

*thinks harder*
*almost there*
...
so should I find the secant slope
then set it equal to v'(t) and solve for t
which will get me the t value that is when the graph hits the overall velocity average?
no I need average velocity value
so I plug that t into the original given
that should work... I think