v(t) = 3t/(1+t^2)
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) → ∞
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.
3 answers
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
x = integral v dt
= (3/2)(t^2+1)^1 from t = 4 to t = t
= (3/2)(t^2+1) - 3*17/2
as t --> oo, v--> 3t/t^2 = 3/t = 0
as t -->oo, x --> (3/2)t^2 = oo
=[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
x = integral v dt
= (3/2)(t^2+1)^1 from t = 4 to t = t
= (3/2)(t^2+1) - 3*17/2
as t --> oo, v--> 3t/t^2 = 3/t = 0
as t -->oo, x --> (3/2)t^2 = oo
Use Steve's answer for integral !