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virtus
Group Title
if v =+/(x1)
however when x=0m, v= 1m/s >0
why do we take v = (x1)
 2 years ago
 2 years ago
virtus Group Title
if v =+/(x1) however when x=0m, v= 1m/s >0 why do we take v = (x1)
 2 years ago
 2 years ago

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

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

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

virtus Group TitleBest ResponseYou've already chosen the best response.0
a 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
 2 years ago

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

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

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

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
oh ya....
 2 years ago

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

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

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

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
we 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 ??
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.0
thank you i understand
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
it's not over yet, still we have some work. Are you here?
 2 years ago
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