## 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.

1. shubhamsrg

dx = vdt use this..this little thing is supposed to do wonders for you ! :D

Could you explain how that formula works please?

3. shubhamsrg

have you studied integration ?

Just started learning it. It's not my best subject. xD

5. him1618

the definition of velocity is v=dx/dt

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?

@sauravshakya

8. shubhamsrg

you sure differentiated right in the 1st part.. rest all seem perfect! :)

At least someone agrees with it so far.

10. shubhamsrg

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

ohh wait//thats -t^2 // sorry sorry..thats correct..

12. shubhamsrg

my apologies! :P

Yeah I just checked. oh well haha

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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