Integration (Velocity to Displacement or Position)


by kieran1black2
Tags: displacement, integration, position, velocity
kieran1black2
kieran1black2 is offline
#1
Jun1-07, 06:47 AM
P: 23
please help me with the integration in the word document.
Attached Files
File Type: doc Doc1.doc (16.5 KB, 47 views)
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
G01
G01 is offline
#2
Jun1-07, 10:30 AM
HW Helper
G01's Avatar
P: 2,688
EDIT: I realize you may have work in your document, but I can't yet see it. So, if you have work in the document, ignore my lecture below:)

According to the forum rules, you must show some work to get help. What have you tried? Where are you stuck? Etc.?
dt19
dt19 is offline
#3
Jun1-07, 11:13 AM
P: 47
This is the integration in LaTeX, if anyone else can't see it:

[latex]v = \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30}[/latex]


[latex]s = \int v dt[/latex]


[latex]s = \int \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30} dt[/latex]


Do you know how to integrate exponential functions?


EDIT: I don't know what is up with the LaTeX, but there should be only one expression on each line. So ignore the bit after the second equals sign on the first line. And there shouldn't be an 's' after the fraction on the first line. No idea why it is doing this, there's nothing wrong with the code as I put it in.

G01
G01 is offline
#4
Jun1-07, 02:58 PM
HW Helper
G01's Avatar
P: 2,688

Integration (Velocity to Displacement or Position)


First, I would sugest splitting the integral up into as many smaller integrals as possible. HINT:[tex]e^{a+b}=e^ae^b[/tex]


Using this relationship, plus splitting the integral up, you should end up with two smaller, easier integrals of known forms.
kieran1black2
kieran1black2 is offline
#5
Jun1-07, 03:41 PM
P: 23
i have already tried that... it didnt seem to work... could you show it in latex? so i can see what im doin wrong?
jackiefrost
jackiefrost is offline
#6
Jun1-07, 05:21 PM
P: 137
The 1/30 comes can "come out" of the integrand, right? And the subtraction (e to the blah minus 140000) just yields two integrals: Integral[e to the blah dt] minus Integral[142000 dt]. So that leaves the tricky part: Integral[e to the blah]. That exponent can be broken into two fairly simple expressions by going ahead and doing the division by -100... get it?
kieran1black2
kieran1black2 is offline
#7
Jun2-07, 12:06 AM
P: 23
what i ended up with is

(e^-t/100)/(-30/100) x e^(-1205.525/-100)/t - 142000/30t

that is after integration... is that what you meant?


Register to reply

Related Discussions
integration (displacement or position) Introductory Physics Homework 2
Given an initial position and velocity of a receiver, find the velocity of a ball Introductory Physics Homework 7
Finding Displacement through integration Introductory Physics Homework 3
help on velocity and displacement Introductory Physics Homework 2