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if v =+/-(x-1) however when x=0m, v= 1m/s >0 why do we take v = -(x-1)

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

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Other answers:

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
find x as a function of t and describe the motion of the particles
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?
@ash2326 I dont think x is always 0 when t=0
Here it's given that "Initially, the particles ias at the origin and has velocity of 1m/s"
oh ya....
ok....
We know that acceleration a \[a=\frac{d^2x}{dt^2}\] so we get \[\frac{d^2x}{dt^2}=x-1\]
Do you understand this?
yeah
It's better we use differential equations to solve this Do you know differential equations?
no
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 ??
thank you i understand
it's not over yet, still we have some work. Are you here?

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