alanab 0 #1 November 13, 2003 trying to help a roommate out with her calc homework... can anyone here please explain this problem?? A skydiver jumps from a plane when it is 1000 meters above earth. letting y be her distance above earth and t the elapsed time of her jump, her hight above the earth for all t up to the time her chute opens is y=f(t)=1000-4.9t2 (that's squared) where lenghts are measured in meters and time in seconds. (in this model we are ignoring aair resistance and any sidewise movement. calculate her average velocity v(2,t2 (thats sub 2)) for t2 (thats sub 2)=2.1, 2.01, 2.001, 2.0001. from these results, estimate her velocity at t=2. compare your estimate with her velocity at t=2. calculate her speed at t=2. Quote Share this post Link to post Share on other sites
pilotdave 0 #2 November 13, 2003 That's a math problem disguised as a skydiving post. It burns my eyes! Dave Quote Share this post Link to post Share on other sites
Guest #3 November 13, 2003 Acceleration in the absence of air resistance is constant at 32 ft/sec². mh"The mouse does not know life until it is in the mouth of the cat." Quote Share this post Link to post Share on other sites
pilotdave 0 #4 November 13, 2003 The problem is in meters. I'm thinking about it but i cant see what the average velocity over the first 2 seconds has to do with the velocity at the end of the first 2 seconds. The only difference between the speed and velocity, as far as I can see, will be a negative sign in front of the velocity. To calculate velocty from positon, you just need to differentiate it. v(t) = -9.8t average velocity is the velocity at the end minus the velocity at the start, divided by 2, right? So for t=2.1, v(avg) = -10.29 t=2.01, v(avg) = -9.849 t=2.001, v(avg) = -9.8049 t=2.0001, v(avg) = -9.80049 The actual velocity would be 9.8(2) = -19.6, speed=19.6 But that's all I know. Dave Quote Share this post Link to post Share on other sites
DShiznit 0 #5 November 13, 2003 Quotetrying to help a roommate out with her calc homework... can anyone here please explain this problem?? A skydiver jumps from a plane when it is 1000 meters above earth. letting y be her distance above earth and t the elapsed time of her jump, her hight above the earth for all t up to the time her chute opens is y=f(t)=1000-4.9t2 (that's squared) where lenghts are measured in meters and time in seconds. (in this model we are ignoring aair resistance and any sidewise movement. calculate her average velocity v(2,t2 (thats sub 2)) for t2 (thats sub 2)=2.1, 2.01, 2.001, 2.0001. from these results, estimate her velocity at t=2. compare your estimate with her velocity at t=2. calculate her speed at t=2. Google it, you'll find the answer. Quote Share this post Link to post Share on other sites
happythoughts 0 #6 November 13, 2003 So... the way that you are caculating the time, it appears that you are going to be in freefall for the entire 1000 ft. Ruh-roh! Quote Share this post Link to post Share on other sites
captainbb7 0 #7 November 13, 2003 The way I read that is to calculate the average velocity on the following intervals... (2,2.1) (2,2.01) (2,2.001), etc... The open interval is shrinking closer to the point {2}. Then they want you to compare the avg velocity with the instantaneous velocity at t=2. Its analogous to the secant line approximation to the slope of the tangent line. Does that help at all? So you prob want to calculate the average velocity by: v_{avg} = change in distance / change in time for those intervals and compare that to the instantaneous vel you get by using the time deriv of the position. Does *that* help? Quote Share this post Link to post Share on other sites
flyhi 24 #8 November 13, 2003 Quotecalculate her average velocity v(2,t2 (thats sub 2)) for t2 (thats sub 2)=2.1, 2.01, 2.001, 2.0001. Average velocity is a tricky term here. I used an interval equal to the distance from 2. I agree the velocity equation is v(t) = -9.8 * t . Using an Excel spread sheet, I found V avg (2.1) = -(19.27+20.21)/2 = -19.74m/sec V avg (2.01) = -(18.847+18.941)/2 = -18.894 m/sec V avg(2.001) = -(18.8047+18.8141)/2 = -18.8094 m/sec V avg (2.001) = -(18.80047+18.80141) = -18.80094 m/sec I would estimate from that, that her velocity at v(2) would be -18.8 m/sec as we insert more and more zeros into the equation. That is also the velocity we calculate. Her speed would be 18.8 m/sec. Velocity requires position. Speed does not.Shit happens. And it usually happens because of physics. Quote Share this post Link to post Share on other sites
captainbb7 0 #9 November 13, 2003 That looks good... but dont you mean velocity requires direction also, since it is a vector quantity... and that makes speed the magnitude of that vector. Math is cool.... and I cant think straight today... Quote Share this post Link to post Share on other sites
narcimund 0 #10 November 13, 2003 QuoteAcceleration in the absence of air resistance is constant at 32 ft/sec2. In the absence of air resistance, wingloading becomes moot. First Class Citizen Twice Over Quote Share this post Link to post Share on other sites
captainbb7 0 #11 November 13, 2003 Actually... I dont agree...I really cant think today... To calculate the average velocity, calculate the distance at each time using y(t) = -9.8/2*t^2+1000. Then the avg_vel is change in distance / change in time. This results in: -20.0900 for (2,2.1) -19.6490 for (2,2.01) -19.6049 for (2,2.001) -19.6005 for (2,2.0001) the avg velocity approaches -19.6. The instantaneous velocity at t=2 is given by dy/dt = -9.8*t which evaluated at t=2 yields -19.6. So in the limit as the change in time goes to zero, the avg velocity approaches the instantaneous velocity. Quote Share this post Link to post Share on other sites
flyhi 24 #12 November 13, 2003 Oops! I used 9.4 instead of 9.8m/sec. I concur.Shit happens. And it usually happens because of physics. Quote Share this post Link to post Share on other sites
Squeak 17 #13 November 14, 2003 QuoteAcceleration in the absence of air resistance is constant at 32 ft/sec². mh 9.1 metres per sec per secYou are not now, nor will you ever be, good enough to not die in this sport (Sparky) My Life ROCKS! How's yours doing? Quote Share this post Link to post Share on other sites
prepheckt 0 #14 November 14, 2003 Quote9.1 metres per sec per sec If I remember my physics class, I believe it is 9.8m/sec/sec"Dancing Argentine Tango is like doing calculus with your feet." -9 toes Quote Share this post Link to post Share on other sites
kallend 1,644 #15 November 14, 2003 Fascinating. Just fascinating.... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
TB99 0 #16 November 14, 2003 9.81 m/s/s ... looks like the work's already done, tho ... Trailer 11/12 was the best. Thanks for the memories ... you guys rocked! Quote Share this post Link to post Share on other sites
happythoughts 0 #17 November 14, 2003 So if someone fell from 22K and had 60 seconds of freefall... it would be 9.81 * 60 * 60 m/s? Ouch. Let's not forget terminal. Quote Share this post Link to post Share on other sites
masher 1 #18 November 14, 2003 QuoteAcceleration in the absence of air resistance is constant at 32 ft/sec². We're using _real_ units here, metres and seconds and things.... g=9.81 ms^-2 . Her average velocity is equal to the integral of v with repect to t from 0 to t, divided by t. v=-at=-9.8*t /t v(ave)= | v dt /0 ------------ t /t = | -at dt /0 ------------ t t = [-0.5*9.8*t^2] ----------------0 t But because we started from zero velocity, then v(ave) is given by v(ave)=v/2 In this case, t=2.1, 2.01, 2.001 ... Her average velocities are: t=2.1 v(ave)= -10.290 ms^-1 t=2.01 v(ave)= -9.849 ms^-1 t=2.001 v(ave)= -9.805 ms^-1 t=2.0001 v(ave)=-9.800 ms^-1 From this we can say that her average velocity is tending to -9.8 ms^-1. -ve, because she's going down. It's an arbitrary choice of reference frame, but I'll stick with it. From the above equation, v(ave)=v/2. So, v=-19.6 ms^-1 Speed at t=2 is 19.6 ms^-1.-- Arching is overrated - Marlies Quote Share this post Link to post Share on other sites
masher 1 #19 November 14, 2003 QuoteSo if someone fell from 22K and had 60 seconds of freefall... it would be 9.81 * 60 * 60 m/s? Ouch. Let's not forget terminal. We're doing what all physicists love to do. We're ignoring air resistance. There is no terminal velocity.-- Arching is overrated - Marlies Quote Share this post Link to post Share on other sites
kelpdiver 2 #20 November 14, 2003 QuoteTo calculate the average velocity, calculate the distance at each time using y(t) = -9.8/2*t^2+1000. Then the avg_vel is change in distance / change in time. This results in: -20.0900 for (2,2.1) -19.6490 for (2,2.01) -19.6049 for (2,2.001) -19.6005 for (2,2.0001) the avg velocity approaches -19.6. The instantaneous velocity at t=2 is given by dy/dt = -9.8*t which evaluated at t=2 yields -19.6. So in the limit as the change in time goes to zero, the avg velocity approaches the instantaneous velocity. You guys are making this a lot harder than it needs to be. The velocity (we seem to be ignoring the forward component) increases at a constant rate up until what, about 10 seconds for terminal? But no, the average velocity cannot equal the instantaneous velocity in this semi real world scenario. Since the person started at rest and constantly accelerated to the exit velocity at time=2s (or any under 10), the average velocity will be half the terminal speed. Quote Share this post Link to post Share on other sites
kallend 1,644 #21 November 14, 2003 QuoteQuoteAcceleration in the absence of air resistance is constant at 32 ft/sec². We're using _real_ units here, metres and seconds and things.... g=9.81 ms^-2 . Her average velocity is equal to the integral of v with repect to t from 0 to t, divided by t. v=-at=-9.8*t /t v(ave)= | v dt /0 ------------ t /t = | -at dt /0 ------------ t t = [-0.5*9.8*t^2] ----------------0 t But because we started from zero velocity, then v(ave) is given by v(ave)=v/2 In this case, t=2.1, 2.01, 2.001 ... Her average velocities are: t=2.1 v(ave)= -10.290 ms^-1 t=2.01 v(ave)= -9.849 ms^-1 t=2.001 v(ave)= -9.805 ms^-1 t=2.0001 v(ave)=-9.800 ms^-1 From this we can say that her average velocity is tending to -9.8 ms^-1. -ve, because she's going down. It's an arbitrary choice of reference frame, but I'll stick with it. From the above equation, v(ave)=v/2. So, v=-19.6 ms^-1 Speed at t=2 is 19.6 ms^-1. Very clever I'm sure. It's a pity you mis-read the question so your answers are incorrect.... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
kallend 1,644 #22 November 14, 2003 QuoteQuoteSo if someone fell from 22K and had 60 seconds of freefall... it would be 9.81 * 60 * 60 m/s? Ouch. Let's not forget terminal. We're doing what all physicists love to do. We're ignoring air resistance. There is no terminal velocity. Not true. REAL physicists include air resistance and the variation of air density with altitude. See http://www.iit.edu/~kallend/skydive/ for an example... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
kallend 1,644 #23 November 14, 2003 QuoteQuoteTo calculate the average velocity, calculate the distance at each time using y(t) = -9.8/2*t^2+1000. Then the avg_vel is change in distance / change in time. This results in: -20.0900 for (2,2.1) -19.6490 for (2,2.01) -19.6049 for (2,2.001) -19.6005 for (2,2.0001) the avg velocity approaches -19.6. The instantaneous velocity at t=2 is given by dy/dt = -9.8*t which evaluated at t=2 yields -19.6. So in the limit as the change in time goes to zero, the avg velocity approaches the instantaneous velocity. You guys are making this a lot harder than it needs to be. The velocity (we seem to be ignoring the forward component) increases at a constant rate up until what, about 10 seconds for terminal? But no, the average velocity cannot equal the instantaneous velocity in this semi real world scenario. Since the person started at rest and constantly accelerated to the exit velocity at time=2s (or any under 10), the average velocity will be half the terminal speed. The average over the entire fall is what you state, but that is not the question asked. You mis-read the question. Captainbb is correct. No calculus needed.... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
rehmwa 2 #24 November 14, 2003 QuoteNo calculus needed. Not true. No calculus needed for the "estimate". Just delta y/delta t work. But the last part does require a simple derivative. The question also says compare your estimate to the velocity. So the derivative of the y(t) equation is needed to get the actual velocity. The point is to show the student that taking the derivative is quick and easy. Much easier that doing the 'estimate' technique of the limit of delta t as t goes to zero. By doing both techniques and comparing, the point is made if the student is paying attention. The funny parts of this post is the discussions about what is the gravitational constant. It's not needed as the the y(t) equation is already provided in the question. So no one needs to do the integrals. (It's a math class, not a physics class). ... Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants Quote Share this post Link to post Share on other sites
rehmwa 2 #25 November 14, 2003 Unless they just want them to take the derivative once and just plug and chug all those number into it, then I'd say college calc being dumbed down since I was a freshman (a very long time ago)...... ... Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants Quote Share this post Link to post Share on other sites
