What is the time to reach 98% of terminal velocity?

[rockytriton]

Ok, there's a differential equation that is:

dv/dt = 9.8 - (v/5), v(0) = 0

to represent a falling object. So the solution ends up being

v = 49(1 - e^t/5)

and the equilibrium solution (terminal velocity) is v = 49.

Now I have a problem that says "find the time that must elapse for the object to reach 98% of its limiting velocity"

To do this, I am doing the following:

terminal velocity * 98% = the equation...
49 * 0.98 = 49(1 - e^t/5)

div by 49...
0.98 = 1 - e^t/5

move 1 over and multiply both sides by -1...
1 - 0.98 = e^t/5

use ln...
ln(1 - 0.98) = t/5

mult by 5...

5 * ln(1 - 0.98) = t

so:
t = -19.56


now... in the book's solution, it states:
T = 5 * ln(50) ~= 19.56

So, where is this 50 coming from? And my answer looks to be right except for the sign, where did I go wrong?

 

[StatusX]

It should be v = 49(1 - e^(-t/5)). Your solution has the velocity exponentially growing.

 

[rockytriton]

thanks! I knew it would be something stupid that I missed. I still don't get where they got the "5 * ln(50)" from though...

 

[HallsofIvy]

0.98 = 1 - e^-t/5

e^-t/5= 1-0.98= 0.02= 2/100= 1/50
That's where the "50" is from.

Now -t/5= ln(1/50)= -ln(50) so t= 5 ln(50).