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anonymous
 3 years ago
A particle moves along the xaxis 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.
anonymous
 3 years ago
A particle moves along the xaxis 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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shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1dx = vdt use this..this little thing is supposed to do wonders for you ! :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Could you explain how that formula works please?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1have you studied integration ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just started learning it. It's not my best subject. xD

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the definition of velocity is v=dx/dt

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01. a(t) = 3(1t^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?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1you sure differentiated right in the 1st part.. rest all seem perfect! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0At least someone agrees with it so far.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1i missed a question mark there :P i meant you sure differentiated right in the 1st part ???? :D cause i see that wrong..please confirm..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1ohh wait//thats t^2 // sorry sorry..thats correct..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah I just checked. oh well haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Could 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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