Question

Help pls...very complicated and i've got a deadline

Answer 1

what is the question

Answer 2

its attached

Answer 3

not sure but I think you just place the [3,4] into the equation and solve

Answer 4

each indidividually

Answer 5

then I think you divide

Answer 6

i know ut has something to d with derivatives

Answer 7

Sorry I am just not to familiar with these

Answer 8

its no problem...thanx so much for trying to help though

Answer 9

someone so be along hopefully to help just be patient

Answer 10

Familiar w/ Calculus or no?

Answer 11

the average velocity is delta s/ delta t=\[ \frac{ \Delta s }{ \Delta t}\]

Answer 12

Then the velocity is the first derivative of displacement. So differentiate that.

Answer 13

I would love to differentiate it but I only learnt differentiation today and am not very familiar with it. I just realised that the homework is due in a couple of minutes

Answer 14

Sounds like you waited far too long.
The differentiation is 2t-9

Answer 15

yeah..that was what i got too

Answer 16

(x+h)^2-9(x+h)+17-[x^2-9x+17]
average= -------------------------------
h
solve that

Answer 17

Excellent, now plug in the times to give you the changes in velocity.
For the last part, you can directly substitute 4 in to the velocity equation.

With the post above, I'd HIGHLY recommend you use the differentiated equation as it's linear, and will be far easier to deal with.

Answer 18

at [3,4], t=1, so i had 2(1)-9 and it gave me -7

Answer 19

however, upon entering -7, i was told my answer was wrong, my instataneous velocity was correct though after plugging it into 2t-9

Answer 20

For part 1, look at the change in velocity over the change in time.
2(4) - 9 = -1
2(3) - 9 = -3

[-1 - (-3)]/(4-3) = 2 / 1 = 2.

Answer 21

Your dv is 2, the dt is 1. So your velocity is 2m/s

Answer 22

for instantaneous velocity you have to solve the limit x->0 of deltas/delta t

Answer 23

No, if you differentiate the equation, you can plug 4 right in for t. That IS your instantaneous velocity.

Answer 24

it still says am wrong though

Answer 25

Which part?

Answer 26

for all of the first part

Answer 27

i got the instanteneous velocity right though

Answer 28

@qpHalcy0n

Answer 29

For an average you're going to want to add the instantaneous velocities and divide by 2.

Answer 30

So
(-1 + -3) / 2 = -2m/s

Answer 31

i could hug you from afar right now..@qpHalcy0n and every other person that helped...thanx you all i beat my deadline....i wont procrastinate again...smile...cheers!!!!

Answer 32

Yes you will.
But you're welcome anyways ;]

Answer 33

\[
\text{i. }\frac{s(4)-s(3)}{1}={4^2-9(4)+17-3^2+9(3)-17}\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =16-36-9+27\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\\
\text{ii. }\frac{s(4)-s(3.5)}{0.5}=2[4^2-9(4)+17-3.5^2+9(3.5)-17]\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2[16-36-12.25+31.5]\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1.25\\
\text{iii. }\frac{s(5)-s(4)}{1}={5^2-9(5)+17-4^2+9(4)-17}\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =25-45-16+36\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0\\
\text{iv. }\frac{s(4.5)-s(4)}{0.5}=2[4.5^2-9(4.5)+17-4^2+9(4)-17]\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2[20.25-40.5-16+36]\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-0.5\\
\text{B. First, differentiate }s\text{ with respect to t.}\\
\ \ \ \ \ s'(t)=\lim_{\delta\to0}[\frac{s(t+\delta)-s(t)}\delta]\\
\ \ \ \ \ \ \ \ \ \ \ =\lim_{\delta\to0}[\frac{(t+\delta)^2-9(t+\delta)+17-t^2+9t-17}\delta]\\
\ \ \ \ \ \ \ \ \ \ \ =\lim_{\delta\to0}[\frac{t^2+2t\delta+\delta^2-9t-9\delta-t^2+9t-17}\delta]\\
\ \ \ \ \ \ \ \ \ \ \ =\lim_{\delta\to0}[\frac{2t\delta+\delta^2-9\delta}\delta]\\
\ \ \ \ \ \ \ \ \ \ \ =\lim_{\delta\to0}[2t+\delta-9]\\
\ \ \ \ \ \ \ \ \ \ \ =2t-9\\
\text{Next, find the derivative where }t=4.\\
\ \ \ \ \ s'(4)=2(4)-9=8-9=-1
\]

Answer 34

remember that given a position function x(t), derivative of x [x'(t)] gives you velocity and the second derivative of x(t) [x''(t)] gives you the acceleration..
you have the formulas, play around with it

Answer 35

did you get what oldrin was saying here? thjat was i am trying to explain awhile age.lol..have fun n gud luck now

Answer 36

Merci Beaucoup!!!