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anonymous
 3 years ago
if v =+/(x1)
however when x=0m, v= 1m/s >0
why do we take v = (x1)
anonymous
 3 years ago
if v =+/(x1) however when x=0m, v= 1m/s >0 why do we take v = (x1)

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

anonymous
 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?

anonymous
 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

anonymous
 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?

anonymous
 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 ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you i understand

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?
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