4. A particle starts at the point (5, 0) at t = 0 and moves along the x-axis in such a way that at time t > 0 the velocity is given by v(t)=t/(1+t^2)

Question

a. Determine the maximum velocity attained by the particle. Justify your answer.
b. Determine the position of the particle at t = 6.
c. Find the limiting value of the velocity as t increases without bound.
d. Does the particle ever pass the point (500,0)? Explain.

Answer 1

dv/dt = [ (1+t^2) -t(2t) ] / (1+t^2)^2
= (1-t^2) /(1+t^2)^2
zero when
t = + or - 1
when t = +1
v = 1/2
when t = -1 that was before we started
so v max is 1/2 at t = 1

integrate dx = v dt to get x(6), remember to add 5
(1/2)ln(t^2+1) from t = 0 to 6
it is 0 when t = 0
so
x = 5 + .5 ln 37 when x = 6

looks like 1/t for large t, so approaches 0

sure it reaches all +x
because as t ---> oo
x = 5 + ln (oo)