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.
(A video of this situation is displayed above.)
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) 7.65 6.80 5.70 4.50 4.55 5.65 6.85 7.65 8.15 8.30 8.10 7.65 6.85
1.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".)
(Enter answer to 2 decimal places.)
i am not sure how to do it.
i am inclined towards forming an equation using sin or cos.
the format of the equation is k+a*cos(t)
k can be calculated using the data above.
'a' which is the amplitude can be calculated using the maximum deviation from k.
BUT i am not sure ...
please help.
i am under time contraint
average velocity is totaldistance/totaltime
the minimum x is at t=3.50, so
x(3.00) = 6.80
x(3.50) = 4.50
∆x/∆t = -1.7/0.5 = -3.4 ft/s
If you do want to devise a function, then you see that
x(2.90) = x(3.60) = x(4.00) = 7.65
So, since x is falling at t=2.90 and again at t=4.00, the period is 1.10 seconds.
the minimum x will occur midway between t=2.90 and t=3.60, when x is at the same height, but going in opposite directions. So, we want t=3.25 seconds. That appears to make x=4.525, but clearly x=4.50 is the minimum value. So, let's say the amplitude is roughly (8.30-4.50)/2 = 1.90, with the midpoint at (8.30+4.50)/2 = 6.40
Since the maximum is at t=3.80, let's make it a cosine function shifted to that point. We end up with
x(t) = 6.40 + 1.90 cos(2π/1.10 (t-3.80))
Try plotting that. It may be a bit off since we had to approximate times and heights.
To find the average velocity of the jumper from 3.00 seconds to the time when he is at the lowest point, we can use the data provided.
First, let's identify the lowest point in the data. From the given data, we can see that at t = 3.50 seconds, the height reaches a minimum value of 5.65 feet.
To calculate the average velocity, we need to find the displacement (change in position) and the time interval.
The displacement can be found by taking the difference between the starting position at t = 3.00 seconds (x = 6.80 feet) and the lowest point at t = 3.50 seconds (x = 5.65 feet):
Displacement = 6.80 feet - 5.65 feet = 1.15 feet
The time interval is the difference between the starting time and the time when the jumper is at the lowest point:
Time interval = 3.50 seconds - 3.00 seconds = 0.50 seconds
Now, we can calculate the average velocity by dividing the displacement by the time interval:
Average velocity = Displacement / Time interval = 1.15 feet / 0.50 seconds = 2.30 feet/second
Therefore, the average velocity of the jumper from 3.00 seconds to the time when he is at the lowest point is 2.30 feet/second.
To find the average velocity of the jumper from 3.00 seconds to the time when he is at the lowest point, we can use the following steps:
1. Determine the time interval: We need to find the time interval between 3.00 seconds and the time when the jumper is at the lowest point. Let's call this time t_lowest.
2. Find the lowest point: The lowest point on the trampoline corresponds to the minimum height of the center of mass of the jumper. Look for the lowest value of x(t) in the given data set and note down the time associated with it.
3. Calculate the average velocity: The average velocity is given by the displacement divided by the time interval. In this case, the displacement is the change in height of the jumper's center of mass from 3.00 seconds to the lowest point.
Now let's apply these steps to the given data:
1. From the data, we can see that the lowest point occurs at t = 3.30 seconds. So, our time interval is t_lowest - 3.00 seconds = 3.30 - 3.00 = 0.30 seconds.
2. We can find the corresponding height of the lowest point by looking at the x(t) values. Here, x(t) = 4.55 feet at t = 3.30 seconds.
3. The average velocity is the change in height divided by the time interval. In this case, it is (4.55 - 6.80) feet / 0.30 seconds.
So, the average velocity of the jumper from 3.00 seconds to the time when he is at the lowest point is approximately (4.55 - 6.80) / 0.30 ≈ -7.50 feet/sec.
Please note that the negative sign indicates the direction of the velocity, which means the jumper is moving downward during this time interval.