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Bladerunner1122

  • 2 years ago

A particle moves along the x-axis with the velocity given by v(t)=3t/(1+t^2) for t >or equal to 0. When t=0, the particle is at the point (4,0). 1. Determine the maximum velocity for the particle. Justify your answer. 2. Determine the position of the particle at any time t. 3. Find the limit of the velocity as t->infinity. 4. Find the limit of the position as t->infinity.

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  1. shubhamsrg
    • 2 years ago
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    dx = vdt use this..this little thing is supposed to do wonders for you ! :D

  2. Bladerunner1122
    • 2 years ago
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    Could you explain how that formula works please?

  3. shubhamsrg
    • 2 years ago
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    have you studied integration ?

  4. Bladerunner1122
    • 2 years ago
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    Just started learning it. It's not my best subject. xD

  5. him1618
    • 2 years ago
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    the definition of velocity is v=dx/dt

  6. Bladerunner1122
    • 2 years ago
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    1. a(t) = 3(1-t^2)/(1+t^2)^2 max v at a=0, or t=1 v(1) = 3/2 2. x(t) = 3/2 log(1+t^2)+C at t=0, x=4, so 4 = 3/2 log(1+0) + C C = 4 x(t) = 3/2 log(1+t^2) + 4 3. v(t) → 0 as t→∞ 4. x(t) → ∞ that's what I've got so far. Does that make sense?

  7. Bladerunner1122
    • 2 years ago
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    @sauravshakya

  8. shubhamsrg
    • 2 years ago
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    you sure differentiated right in the 1st part.. rest all seem perfect! :)

  9. Bladerunner1122
    • 2 years ago
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    At least someone agrees with it so far.

  10. shubhamsrg
    • 2 years ago
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    i missed a question mark there :P i meant you sure differentiated right in the 1st part ???? :D cause i see that wrong..please confirm..

  11. shubhamsrg
    • 2 years ago
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    ohh wait//thats -t^2 // sorry sorry..thats correct..

  12. shubhamsrg
    • 2 years ago
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    my apologies! :P

  13. Bladerunner1122
    • 2 years ago
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    Yeah I just checked. oh well haha

  14. Bladerunner1122
    • 2 years ago
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    Could somebody please check this to make sure it makes sense? 1. dv/dt = [(1+t^2)(3) - 3t(2t)] / (t^2+1)^2 =[3 t^2 + 3 - 6 t^2 ] /(t^2+1)^2 zero when t= 1 (since t always >/=0) so the max v is at t = 1 then at t = 1 v(1) = 3/2 2. x(t) = 3/2 log(1+t^2)+C at t=0, x=4, so 4 = 3/2 log(1+0) + C C = 4 x(t) = 3/2 log(1+t^2) + 4 3. as t --> oo, v--> 3t/t^2 = 3/t = 0 4. as t -->oo, x --> (3/2)t^2 = oo

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