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virtus

  • 2 years ago

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

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  1. ash2326
    • 2 years ago
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    @virtus do you have a graph along with this?

  2. virtus
    • 2 years ago
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    nope i do not have a graph sorry

  3. ash2326
    • 2 years ago
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    @virtus Is this the full question?

  4. virtus
    • 2 years ago
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    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

  5. virtus
    • 2 years ago
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    find x as a function of t and describe the motion of the particles

  6. ash2326
    • 2 years ago
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    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?

  7. sauravshakya
    • 2 years ago
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    @ash2326 I dont think x is always 0 when t=0

  8. ash2326
    • 2 years ago
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    Here it's given that "Initially, the particles ias at the origin and has velocity of 1m/s"

  9. sauravshakya
    • 2 years ago
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    oh ya....

  10. ash2326
    • 2 years ago
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    @virtus ??

  11. virtus
    • 2 years ago
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    ok....

  12. ash2326
    • 2 years ago
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    We know that acceleration a \[a=\frac{d^2x}{dt^2}\] so we get \[\frac{d^2x}{dt^2}=x-1\]

  13. ash2326
    • 2 years ago
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    Do you understand this?

  14. virtus
    • 2 years ago
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    yeah

  15. ash2326
    • 2 years ago
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    It's better we use differential equations to solve this Do you know differential equations?

  16. ash2326
    • 2 years ago
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    @virtus ??

  17. virtus
    • 2 years ago
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    no

  18. ash2326
    • 2 years ago
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    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 ??

  19. virtus
    • 2 years ago
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    thank you i understand

  20. ash2326
    • 2 years ago
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    it's not over yet, still we have some work. Are you here?

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