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 :)
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
Sure! I'd be happy to help you understand how to solve these problems.
a) To find the average velocity of the jumper from 3.00 seconds to the time when he is at the lowest point, we need to calculate the displacement and divide it by the time interval.
Looking at the data provided, we can see that the lowest point occurs at t = 3.40 seconds when the height is 4.10 feet. The initial height at t = 3.00 seconds is 7.20 feet.
The displacement (change in height) is calculated as final height - initial height:
Displacement = 4.10 ft - 7.20 ft = -3.10 ft
(Here, the negative sign indicates that the jumper is moving downward.)
The time interval is 3.40 seconds - 3.00 seconds = 0.40 seconds.
Average velocity is calculated as displacement / time interval:
Average velocity = -3.10 ft / 0.40 s ≈ -7.75 ft/s
So, the average velocity of the jumper from 3.00 seconds to the lowest point is -7.75 ft/s (or you can just write it as 7.75 ft/s downward).
It seems that your initial answer was incorrect because you might have mistakenly taken the displacement as positive instead of negative.
b) To estimate the velocity of the jumper at t = 3.75 seconds using linear approximation, we can use the concept of average velocity. Let's look at the two closest points in the data: (t1, x1) = (3.70 s, 5.20 ft) and (t2, x2) = (3.80 s, 7.15 ft).
The time interval is 3.80 s - 3.70 s = 0.10 s.
The displacement is calculated as the final position minus the initial position:
Displacement = x2 - x1 = 7.15 ft - 5.20 ft = 1.95 ft
Average velocity is calculated as displacement / time interval:
Average velocity = 1.95 ft / 0.10 s = 19.5 ft/s
So, the estimated velocity of the jumper at t = 3.75 seconds is approximately 19.5 ft/s.
It appears that your initial answer was incorrect because you might have used different data points or made a calculation mistake.
c) Unfortunately, with just the information provided in part (b) (the estimate of the velocity at t = 3.75 seconds), we cannot directly estimate the acceleration at t = 3.70 seconds. We would need additional information, such as the rate of change of velocity, to estimate the acceleration at that specific time.
I hope this explanation helps you understand how to solve these problems. If you have any further questions, please feel free to ask!