A community for students.
Here's the question you clicked on:
 0 viewing
virtus
 3 years ago
if v =+/(x1)
however when x=0m, v= 1m/s >0
why do we take v = (x1)
virtus
 3 years ago
if v =+/(x1) however when x=0m, v= 1m/s >0 why do we take v = (x1)

This Question is Closed

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2@virtus do you have a graph along with this?

virtus
 3 years ago
Best ResponseYou've already chosen the best response.0nope i do not have a graph sorry

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2@virtus Is this the full question?

virtus
 3 years ago
Best ResponseYou've already chosen the best response.0a particle moves along a straight line s that its acceleration is given by a =x1 where x is its displacement from the origin. Initially, the particles ias at the origin and has velocity of 1m/s

virtus
 3 years ago
Best ResponseYou've already chosen the best response.0find x as a function of t and describe the motion of the particles

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2OK, so we have \[a=x1\] a=acceleration we are given that at t=0, velocity=1m/s , x=0 Do you agree with this?

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0@ash2326 I dont think x is always 0 when t=0

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2Here it's given that "Initially, the particles ias at the origin and has velocity of 1m/s"

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2We know that acceleration a \[a=\frac{d^2x}{dt^2}\] so we get \[\frac{d^2x}{dt^2}=x1\]

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2Do you understand this?

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2It's better we use differential equations to solve this Do you know differential equations?

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2we have \[a=x1\] we know \[a=\frac {dv}{dt}\] so \[\frac{dv}{dt}=x1\] where v = velocity But here the function is in terms of x, so we'll have to change variables we know that \(v=\frac {dx}{dt}\) \[\frac{dv}{dt}=x1\] multiplying and dividing left side by dx we get \[\frac{dv\times dx}{dt\times dx}=x1\] or \[\frac{dv}{dx}\times\frac {dx}{dt}=x1\] we know that \(v=\frac {dx}{dt}\) so we get \[\frac{dv}{dx}\times v=x1\] do you get these steps @virtus ??

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2it's not over yet, still we have some work. Are you here?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.