Difference between revisions of "Talk:135: Substitute"
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1. It takes the raptor 25m/s / 4m/s^2 = 6.25s to reach it's top speed, during which I can run 6.25s * 6m/s = 37.5m. Add on my 40m head start, and I can reach a spot 77.5m away from the raptor before he gets me. In the same time, the raptor can run 4m/s^2 * (6.25s)^2 / 2 = 78.125m. I'm eaten before he's fully up to speed. | 1. It takes the raptor 25m/s / 4m/s^2 = 6.25s to reach it's top speed, during which I can run 6.25s * 6m/s = 37.5m. Add on my 40m head start, and I can reach a spot 77.5m away from the raptor before he gets me. In the same time, the raptor can run 4m/s^2 * (6.25s)^2 / 2 = 78.125m. I'm eaten before he's fully up to speed. | ||
Therefore, I have to solve for when the raptors location, r(t) = 4m/s^2 * t^2 /2 - 40, and my location, m(t) = 6m/s*t, are equal. Dropping units, we get 2t^2 -40 = 6t, or 2t^2 - 6t - 40 = 0. Dividing by 2 I get t^2 - 3t - 20=0. Using the quadratic equation, I get (3 +/- sqrt(89))/2, roughly equal to 6.217s and -3.217s. Plugging that back into m(t), I get 37.302m for my terminal run. [[User:Blaisepascal|Blaisepascal]] ([[User talk:Blaisepascal|talk]]) 22:18, 14 September 2012 (UTC) | Therefore, I have to solve for when the raptors location, r(t) = 4m/s^2 * t^2 /2 - 40, and my location, m(t) = 6m/s*t, are equal. Dropping units, we get 2t^2 -40 = 6t, or 2t^2 - 6t - 40 = 0. Dividing by 2 I get t^2 - 3t - 20=0. Using the quadratic equation, I get (3 +/- sqrt(89))/2, roughly equal to 6.217s and -3.217s. Plugging that back into m(t), I get 37.302m for my terminal run. [[User:Blaisepascal|Blaisepascal]] ([[User talk:Blaisepascal|talk]]) 22:18, 14 September 2012 (UTC) | ||
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| + | For 1 and 2 the solution depends on whether the raptors can accelerate at 2m/s, or they actually increase their speed at this rate. If they just accelerate, It should be possible to do tight circles, and even wind yourself slowly towards another location. I believe this is possible even treating yourself and the raptors as point masses. [[Special:Contributions/2.102.215.18|2.102.215.18]] 13:19, 17 July 2013 (UTC) | ||
Revision as of 13:19, 17 July 2013
Rikthoff (talk) The issue date is off, as i can't find a create date for the image. Can anyone fix?
1. It takes the raptor 25m/s / 4m/s^2 = 6.25s to reach it's top speed, during which I can run 6.25s * 6m/s = 37.5m. Add on my 40m head start, and I can reach a spot 77.5m away from the raptor before he gets me. In the same time, the raptor can run 4m/s^2 * (6.25s)^2 / 2 = 78.125m. I'm eaten before he's fully up to speed. Therefore, I have to solve for when the raptors location, r(t) = 4m/s^2 * t^2 /2 - 40, and my location, m(t) = 6m/s*t, are equal. Dropping units, we get 2t^2 -40 = 6t, or 2t^2 - 6t - 40 = 0. Dividing by 2 I get t^2 - 3t - 20=0. Using the quadratic equation, I get (3 +/- sqrt(89))/2, roughly equal to 6.217s and -3.217s. Plugging that back into m(t), I get 37.302m for my terminal run. Blaisepascal (talk) 22:18, 14 September 2012 (UTC)
For 1 and 2 the solution depends on whether the raptors can accelerate at 2m/s, or they actually increase their speed at this rate. If they just accelerate, It should be possible to do tight circles, and even wind yourself slowly towards another location. I believe this is possible even treating yourself and the raptors as point masses. 2.102.215.18 13:19, 17 July 2013 (UTC)
