Terminal Velocity given Time at which velocity is .5Vt

xslc

1. The problem statement, all variables and given/known data

A 9.00 kg object starting from rest falls through a viscous medium and experiences a resistive force R = -bv, where v is the velocity of the object. If the object reaches one-half its terminal speed in 5.93 s,
(a) determine the terminal speed.

2. Relevant equations

v=mg/b(1-e^-bt/m)

3. The attempt at a solution

Tried setting above equation equal to mg/2b and solving. It was wrong.

ehild

Show your work in detail. You were right, the speed after 5.93 s is
mg/(2b) From this condition, you can find b.

ehild

xslc

ok, so:

v = mg/b(1-e^(-bt/m))=mg/(2b)
mg/b's cancel, so 1-e^(-5.93b/m)=.5
.5=e^-5.93b/9
9ln.5=-5.93b
-6.2383=-5.93b
b=1.052
vt = 88.2/b = 88.2/1.052=83.8403

and....i got it.

for some reason i was pluggin in 5.5 not 5.93, thanks for letting me know i was doing it right.

ehild

Trust in yourself and check your calculations

ehild

xslc

one more thing.

i got that, and i got part b, which asked for the time at which the speed is .75vt

i'm now stuck on part c, which says:

(c)How far has the object traveled in the first 5.93 s of motion?

I realize i need to integrate somehow, but i'm not sure exactly what to do...

ehild

You know calculus, don't you?
Think of the definition of velocity: it is the time derivative of displacement. Integrating the velocity with respect to time from zero to a given moment will give the displacement. Try.

ehild

xslc

got it. i was trying to do an indefinite integral and then plug in t, silly me, it was late last night haha.