Question

exponential with limits. How do I solve a problem like this: The velocity v(t) of a 137 pound skydiver is approximated by \[v(t) = \frac{ 137 }{ 1.1 }(1 - e^{-24.2t/137})\] where t is in seconds after jumping from a plane. What is the skydiver's terminal velocity? That is, in terms of the constants of the problem \[\lim_{t \rightarrow \infty} v(t) = \] and then How long after jumping does it take the skydiver to reach 52 percent of the terminal velocity?

Answer 1

\[\lim_{t\to\infty}\frac{137}{1.1}(1-e^{-24.2t/137})=\frac{137}{1.1}(1-\lim_{t\to\infty}e^{-24.2t/137})\]what is\[\lim_{t\to\infty}e^{-24.2t/137}\]do you know?

Answer 2

remember that\[e^{-t}=\frac1{e^t}\]

Answer 3

so I plugin infinity to t, think about where it goes. so... that must be going to 0? if the denominator is blowing up by infinity?

Answer 4

yes\[\lim_{t\to\infty}e^{ -24.2t/137}=0\]so what is the terminal velocity?

Answer 5

I guess I don't understand what I'm piecing together here. Terminal velocity is when acceleration ends and a constant maximum speed is reached due to opposing forces. So, we what've gone through so far means that the more amount of time he is falling, his falling speed is approaching 0.

Is that understanding correct so far? Because I'm not sure to get terminal velocity from that point forward.

Answer 6

no, only the part of the formula that depends on t goes to zero\[v_t=\lim_{t\to\infty}\frac{137}{1.1}(1-e^{-24.2t/137})=\frac{137}{1.1}(1-\lim_{t\to\infty}e^{-24.2t/137})=?\]

Answer 7

So I look at is as \[\frac{ 137 }{ 1.1 } (1-0)\] which equals \[\frac{ 137 }{ 1.1 }\] ?

Answer 8

yep

Answer 9

so, \[v(t) = \frac{ 137 }{ 1.1 }\] I'm confused, because there should be another variable there on the right side if that should tell you velocity at a given time.

Answer 10

We are finding terminal velocity which is a constant. there is no need to write this as a function. We have that the terminal velocity is the limit of v(t) as t goes to infinity, at which point it stabilizes and no longer depends upon t.\[v_t=\lim_{t\to\infty}v(t)=\lim\frac{137}{1.1}(1-e^{-24.2t/137})=\frac{137}{1.1}\]that is your terminal velocity; a constant.

Answer 11

oooh ok, when I was seeing "v(t)" and seeing that the next question asks for a time, my mind was on this track of "I'm supposed to be finding an equation that is used for the next part". Also the question gives no unit of measure, so that keeps weirding me out, so I'm going to use meters from now on.
So, we found that when he reaches \[\frac{ 137 }{ 1.1 }\] meters / second, he is at terminal velocity?

Answer 12

correct

Answer 13

Thanks, this is a lot of help. I barely knew how to approach a problem like this.
So, how does knowing the terminal velocity help solve the next part that asks, "How long after jumping does it take him to reach 52% of terminal velocity?" The problem didn't have us figure out an equation that will solve this. And it seems like I need some equation that equates v(t) time and terminal velocity.

Answer 14

just set \[v(t)=\frac{137}{1.1}(1-e^{24.2t/137})=\frac12v_t\]and solve for t

Answer 15

I think part of my constant confusion is that I see v(t) as "velocity at time t", and I read all those previos expressions as \[Velocity At Time T = Some Expression\] but I don't want to solve for velocity given a time, I want to solve for t now that I know terminal velocity. Should I be looking at this as:\[\frac{ 137 }{ 1.1 }(1 - e^{24.2t/137}) = \frac{ 1 }{ 2 }(TerminalVeocity)\]?

Answer 16

actually because it's 52% the terminal velocity we want t for\[\frac{ 137 }{ 1.1 }(1 - e^{24.2t/137}) = 0.52(\text{TerminalVeocity)}\]

Answer 17

wow, having issues with the editor. I have to retype this all again.
The answer I'm getting apparently isn't correct. What did I do wrong?
\[\frac{ 137 }{ 1.1 }(1 - e^{24.2t/137}) = 0.52(\frac{ 137 } { 1.1 })\]
\[1 - e^{24.2t/137} = 0.52(\frac{ 137 } { 1.1 })(\frac{ 1.1 }{ 137 })\]
\[- e^{24.2t/137} = 0.52 - 1\]
\[e^{24.2t/137} = 1 - 0.52\]
\[\ln e^{24.2t/137} = \ln (1 - 0.52)\]
\[\frac{ 24.2t }{ 137 } \ln (e) = \ln (1-0.52)\]
\[24.2t = 137 \ln (1-0.52)\]
\[t = \frac{ 137 \ln (1-0.52) }{ 24.2 }\]

Answer 18

you forgot to make 24.2 negative :P

Answer 19

omg, That was it! I actually did all the algebra correctly this time and my issue was forgetting a minus when rewriting the equation. haha. So, to recap, I can find terminal velocity by taking the limit as time approaches infinity of the velocity equation. And to find time for a certain percentage of terminal velocity, I set the velocity equation equal to the percentage of the terminal velocity and solve for t. Answers were indeed:
\[Terminal Velocity = \frac{ 137 }{ 1.1 }\]
and time after the jump to hit 52% of terminal velocity:
\[-\frac{ 137 \ln (0.48) }{ 24.2 }\]
I'm new to the site. What all can I do to +1 you for all this help other than choosing a best answer? Thank you very much for walking me through this.

Answer 20

Happy to help :)