Question
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.
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.
Answers
Damon
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)
= (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)