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Bladerunner1122
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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.
 one year ago
 one year ago
Bladerunner1122 Group Title
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.
 one year ago
 one year ago

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shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
dx = vdt use this..this little thing is supposed to do wonders for you ! :D
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.1
Could you explain how that formula works please?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
have you studied integration ?
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.1
Just started learning it. It's not my best subject. xD
 one year ago

him1618 Group TitleBest ResponseYou've already chosen the best response.0
the definition of velocity is v=dx/dt
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.1
1. 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?
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.1
@sauravshakya
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
you sure differentiated right in the 1st part.. rest all seem perfect! :)
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.1
At least someone agrees with it so far.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
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..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
ohh wait//thats t^2 // sorry sorry..thats correct..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
my apologies! :P
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.1
Yeah I just checked. oh well haha
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago
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