Problem number 1 10 pts a parachutist jumps at an altitude...

Name: problem number 1 10 pts a parachutist jumps at an altitude of 3000 meters if the parachute does not open she is expected to approach a terminal speed of 30 msec g 10 ms2 a 4 points assuming 58988
Uploaded: 2021-01-15T20:11:02-08:00
Duration: 12 min 58 s
Channel: Bobby Barnes
Description: problem number 1 10 pts a parachutist jumps at an altitude of 3000 meters if the parachute does not open she is expected to approach a terminal speed of 30 msec g 10 ms2 a 4 points assuming 58988

Problem Number 1 (10 pts) A parachutist jumps at an altitude of 3000 meters. If the parachute does not open, she is expected to approach a terminal speed of 30 m/sec. (g = 10 m/s2 ) (a) [4 points] Assuming that air resistance is proportional to speed, about how long does it take her to reach 90% of the terminal speed? (b) [2 points] How far has she traveled in reaching 90% of the terminal speed? After her parachute opens, her speed is slowed to 3m/sec. As she hits the ground, she flexes her knees to absorb the shock. (c) [4 points] How far must she bend her knees in order to experience deceleration no greater than 10 g? Assume that her knees are like a spring with a resisting force proportional to displacement (mechanical energy is conserved).

Transcript

00:02 So the first thing we want to do with this is to determine the velocity for this.

00:10 Well, if this is our position function, t, then to get the velocity, we just take the derivative of this with respect to time.

00:20 So we'll take the derivative of this.

00:24 So let's go ahead and plug s in.

00:26 And the first thing i'm going to do is pull that 160 out just due to the constant rule.

00:31 So this would be 160 d by d t of 1 fourth t minus 1 plus e to the negative t to the fourth now we can use the sum and difference rule to distribute this across so 160 and then i'll also use the constant rule as well to pull that one -fourth out so it'll be one -fourth d by d t of t minus d by d t of one plus d by d t of e to the negative t to the fourth.

01:05 And we can go ahead and take each of these derivatives by themselves.

01:09 So first the derivative of t, well, that's just one.

01:12 The derivative of any constant to include one is going to be zero.

01:16 And now to take the derivative of e to the negative t fourth, we're going to need to use chain rule.

01:25 So first, let's write out what we have over here.

01:27 So one -fourth times one gives us one -fourth.

01:30 Now, chain rule remember says we take the derivative of the outside function, which is going to be e to the t.

01:37 And so that just gives us e to the t, and then we plug back in the negative t fourth.

01:42 And then we take the derivative of what we have on the inside, which is t to the negative fourth.

01:49 And now that there, again, we would just pull out the negative one fork, take the derivative of t.

01:56 And so then this becomes negative one fourth.

01:59 So let's go ahead and write all that up.

02:01 So we have 160, 1 4th, plus, or not plus, minus, because it's negative, negative, 1 4th, 160, e to the negative t 4th.

02:16 I'm just to surrebeat that 160 just to make it look a little bit prettier because i don't like to have to look at fraction.

02:21 This 160 is not there.

02:23 I was getting ahead of myself.

02:25 Yeah, i don't like to look at fractions if i don't have to.

02:28 So distributing that gives 40 minus 40e to the negative t4.

02:33 So this is going to be our velocity function based off of time.

02:41 Now the next thing they want us to do is to show, so b, they want us to show that our acceleration is going to be 10 minus 1 fourth of our velocity.

03:00 So first, let's just go ahead and plug it in over here to see what we.

03:04 Get just so we can kind of like check that this is true.

03:08 So it will be 10 minus 1 fourth and then we're taking this and plugging it in for v.

03:20 So 40 minus 40 e to the negative t over 4.

03:26 And then we distribute the 1 fourth so that gives us 10 minus 10 plus 10 e to the negative t to the fourth, which those 10 cancel out with each other, and we're going to just be left with 10e to the negative t to the 4th.

03:45 So this right here is what we want to show.

03:48 So we haven't actually shown this yet.

03:50 This is just what we're trying to get to.

03:53 All right.

03:54 So to actually get the acceleration, the proper way, though.

03:58 So acceleration is supposed to be the derivative with respect to time of our velocity function.

04:03 So our velocity function, we found in the last step.

04:07 So this will be d by d t of 40 minus 40 e to the negative t to the fourth.

04:15 And now we can go ahead and distribute this.

04:20 So that gives d by d t of 40 minus.

04:24 So we can use the sum and scalar rule for this over here on the right side, or difference in scalar.

04:33 So it would be 40 d by d t of e to the negative t fourth.

04:39 And again, the derivative of 40 is just 0.

04:42 And then we already took the derivative e to the negative t fourth, but let's just go ahead and repeat those steps again.

04:49 So this is going to be 0 minus 40.

04:52 So we use chain rule.

04:54 So first, we just have e to the negative t fourth by itself because derivative x is just e to the x.

05:01 Then we take the derivative of what was on the inside, which was that negative t fourth.

05:07 And again, this is negative 1 fourth.

05:12 And if we multiply everything together, so negative 40 times negative 1 fourth, is going to be 10e to the negative t fourth.

05:20 So you can see these two are the same.

05:27 It checks out...

Question

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