The average velocity for 3.00 <= t <= 3.35 is total_distance / total time = (4.10-7.20)/0.35 = -8.857 ft/s
The average velocity for 3.70 <= t <= 3.80 is (7.60-7.15)/0.10 = 4.5 ft/s
for 3.60 <= t <= 3.70 v = 8.0
for 3.70 <= t <= 3.80 v = 4.5
for 3.80 <= t <= 3.90 v = 2.0
So one approximation would be (2.0-8.0)/0.3 = -2.0 ft/s^2
another would be (2.0-4.5)/0.2 = -1.25
A person is jumping straight up and down on a trampoline. The height of the center of mass of the person is measured every tenth of a second. It takes just over one second to complete one full bounce.
𝑡(seconds)2.90|3.00|3.10|3.20|3.30|3.40|3.50|3.60|3.70|3.80|3.90|4.00|4.10
𝑥(𝑡) (feet) 7.60|7.20|6.35|5.25|4.10|4.10|5.20|6.35|7.15|7.60|7.80|7.55|7.15
a) Find the average velocity (in feet/sec) of the jumper from 3.00 seconds to the time when he is at the lowest point. (If there is more than one lowest point in your data, compute the average velocity to the first "lowest point".)
(my incorrect answer: 10.33ft/sec)
b) Using a linear approximation, estimate the velocity (in feet/sec) of the jumper at 𝑡=3.75 .
(my incorrect answer): 4.6ft/sec
c) Using the information from your previous answer, what would be the best estimate for the acceleration (in feet/ sec^2 ) at 𝑡=3.70 ?
I've just completed the hardest test I've ever taken and did quite well, except for this problem, which would've gotten me an A. I'm curious to how these can be worked out. Would anyone like to enlighten me? Thank you :)
1 answer
3 answers