anonymous
  • anonymous
if v =+/-(x-1) however when x=0m, v= 1m/s >0 why do we take v = -(x-1)
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

ash2326
  • ash2326
@virtus do you have a graph along with this?
anonymous
  • anonymous
nope i do not have a graph sorry
ash2326
  • ash2326
@virtus Is this the full question?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
a particle moves along a straight line s that its acceleration is given by a =x-1 where x is its displacement from the origin. Initially, the particles ias at the origin and has velocity of 1m/s
anonymous
  • anonymous
find x as a function of t and describe the motion of the particles
ash2326
  • ash2326
OK, so we have \[a=x-1\] a=acceleration we are given that at t=0, velocity=1m/s , x=0 Do you agree with this?
anonymous
  • anonymous
@ash2326 I dont think x is always 0 when t=0
ash2326
  • ash2326
Here it's given that "Initially, the particles ias at the origin and has velocity of 1m/s"
anonymous
  • anonymous
oh ya....
ash2326
  • ash2326
anonymous
  • anonymous
ok....
ash2326
  • ash2326
We know that acceleration a \[a=\frac{d^2x}{dt^2}\] so we get \[\frac{d^2x}{dt^2}=x-1\]
ash2326
  • ash2326
Do you understand this?
anonymous
  • anonymous
yeah
ash2326
  • ash2326
It's better we use differential equations to solve this Do you know differential equations?
ash2326
  • ash2326
anonymous
  • anonymous
no
ash2326
  • ash2326
we have \[a=x-1\] we know \[a=\frac {dv}{dt}\] so \[\frac{dv}{dt}=x-1\] where v = velocity But here the function is in terms of x, so we'll have to change variables we know that \(v=\frac {dx}{dt}\) \[\frac{dv}{dt}=x-1\] multiplying and dividing left side by dx we get \[\frac{dv\times dx}{dt\times dx}=x-1\] or \[\frac{dv}{dx}\times\frac {dx}{dt}=x-1\] we know that \(v=\frac {dx}{dt}\) so we get \[\frac{dv}{dx}\times v=x-1\] do you get these steps @virtus ??
anonymous
  • anonymous
thank you i understand
ash2326
  • ash2326
it's not over yet, still we have some work. Are you here?

Looking for something else?

Not the answer you are looking for? Search for more explanations.