Question

A motion is described as \(v^2=-4x^2+8x+4 \), started at \(x=0\).
find the \(x-t\) equation which is the displacement equation.

Answer 1

This is what I did: \[v^2=-4x^2+8x+4 => v= \sqrt{-4x^{2}+8x+4} => \frac{dt}{dx} = \frac{1}{\sqrt{-4x^{2}+8x+4}} \]
\[=> t = \int\limits\frac{1}{\sqrt{-4x^{2}+8x+4}} dx => t = \frac{1}{2} \sin^{-1} \left(\frac{x-1}{\sqrt{2}}\right) -\frac{\pi}{8} \]
then when I solve for \(x\) I will get an ugly result. So, I think that I did it wrong to obtain the \(x-t\) equation..

Answer 2

correct

Answer 3

then what is the problem

Answer 4

correct? lol
I think it is wrong; since when i solve for \(x\) it won't look nice. And i have to graph it later..

I think I did it wrong

Answer 5

im wondering if you have to undo the square to do this, or if you can just integrate both sides and call it a day

Answer 6

integrate both sides? using the acceleration?

Answer 7

displacement is the integration of velocity right?

Answer 8

yeah..
or do i use this..since you said integrate both sides\[\frac{dv}{dt} =\frac{d}{dx} \left(\frac{1}{2}v^{2}\right) \]

Answer 9

wait displacement is the differentiate of velocity i thinnk

Answer 10

differentiation*

Answer 11

\[\int (v^2= -4x^2 +8x +4)\]

\[\frac{1}{3}v^3= -\frac{1}{3}4x^3 +\frac{1}{2}8x^2 +4x +C\]
is what i was thinking, could be off tho :)

Answer 12

displacement is integration of velocity; the derivative of velocity is acceleration

Answer 13

isn't the displacement \(x\) is the differentiation of velocity since \(\frac{dx}{dt}\) is the velocity?
And I am not sure about your method..

Answer 14

im not to sure about it either. I might not be remembering the correct definition of "differentation". I might be confusing it with taking a derivative. At any rate, integrating velocity is akin to determining displacement.

Answer 15

well, when you integrate the velocity you get the displacement i think that you're right

Answer 16

v^2 = -4x^2 +8x +4

v^2 = -4(x^2 -2x -1)
v^2 = -4(x^2 -2x +1 -1-1 )
v^2 = -4 [(x-1)^2 -2]

v^2 = -4(x-1)^2 +8

Answer 17

yeah i did that to integrate it..

Answer 18

im curious if the domain is relevant. just thoughts ....

v^2 is always positive, and the parabola is hill shaped so that restricts the domain; and also the x >= 0 part hmmm

Answer 19

i don't think that it's relevant in this question..and \(x=0\) is given anyway then its \[v= \sqrt{-4x^{2}+8x+4} \]

Answer 20

as long as you did the integrating part correctly :) i cant see anything else to worry about

Answer 21

I am not sure if i did the correct method; and i am not sure how to solve for \(x\) if i did it correctly

Answer 22

http://www.wolframalpha.com/input/?i=integrate+sqrt%28-4x%5E2%2B8x%2B4%29

gives the integration of sqrt(-4x^2 +8x +4)

and yeah, it does look a bit messy.

Answer 23

i wish i knew more about wave equations to be able to be useful ....

Answer 24

ok, thanks anyway!