Posted by **Tiffany** on Sunday, March 25, 2012 at 1:28am.

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)

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.

- calculus -
**Damon**, Sunday, March 25, 2012 at 8:53am
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)

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