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

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

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

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

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

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
@virtus ??
 one year 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 \[\frac{d^2x}{dt^2}=x1\]
 one year ago

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

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
@virtus ??
 one year 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 ??
 one year ago

virtus Group TitleBest ResponseYou've already chosen the best response.0
thank you i understand
 one year 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?
 one year ago
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