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.
(If you want to see the data, look at the video 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
x(t) (feet) 7.20 6.80 6.10 5.05 4.00 3.80 4.60 5.80 6.90 7.60 8.00 7.90
Find the average velocity (in feet/sec) of the jumper from 3.00 seconds to the time when he is at the lowest point.
(Enter answer to 2 decimal places.)
- sin responder
Find the linear approximation to the velocity (in feet/sec) of the jumper at t=3.75.
(Enter answer to 2 decimal places.)
- sin responder
Using the information from your previous answer, what would be the best estimate for the acceleration (in feet/sec2) at t=3.7?
(Enter answer to 2 decimal places.)
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please til 18 ag
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 change in displacement and divide it by the change in time.
1. Identify the displacement at 3.00 seconds and at the time when the jumper is at the lowest point:
- At 3.00 seconds, x(t) = 6.80 feet.
- At the lowest point, x(t) = 3.80 feet (as seen in the data).
2. Calculate the change in displacement:
Δx = x(final) - x(initial) = 3.80 feet - 6.80 feet = -3 feet
3. Calculate the change in time:
Δt = t(final) - t(initial) = (time at the lowest point) - 3.00 seconds
Using the given information, we can determine the time when the jumper is at the lowest point. From the data, we can see that the height begins to decrease at 3.40 seconds and reaches its minimum at 3.50 seconds.
Therefore, Δt = 3.50 seconds - 3.00 seconds = 0.50 seconds.
4. Calculate the average velocity:
Average Velocity = Δx / Δt
Average Velocity = -3 feet / 0.50 seconds
Now, let's proceed to find the average velocity.
Average Velocity = -6 feet/second (rounded to 2 decimal places).
To find the linear approximation to the velocity of the jumper at t = 3.75 seconds, we can use the concept of instantaneous velocity. The linear approximation assumes that the change in position is relatively small over a small interval of time.
1. Observe the data provided and find the position at 3.70 seconds and 3.80 seconds:
- At 3.70 seconds, x(t) = 6.90 feet
- At 3.80 seconds, x(t) = 7.60 feet
2. Calculate the change in displacement within the small interval:
Δx = x(final) - x(initial) = 7.60 feet - 6.90 feet = 0.70 feet
3. Calculate the change in time:
Δt = t(final) - t(initial) = 3.80 seconds - 3.70 seconds = 0.10 seconds
4. Calculate the instantaneous velocity or the linear approximation to the velocity:
Velocity = Δx / Δt
Velocity = 0.70 feet / 0.10 seconds
Therefore, the linear approximation to the velocity at t = 3.75 seconds is 7 feet/second (rounded to 2 decimal places).
Finally, to estimate the acceleration at t = 3.7 seconds, we can use the concept of average acceleration within a small interval of time.
1. Recall the linear approximation to the velocity at t = 3.75 seconds, which was found to be 7 feet/second.
2. Subtract the velocity at t = 3.75 seconds from the velocity at t = 3.70 seconds to find the change in velocity:
Δv = 7 feet/second - (previous velocity at t = 3.70 seconds)
Since the previous velocity at t = 3.70 seconds is not provided in the data, we cannot directly compute the average acceleration. Therefore, we are unable to estimate the acceleration at t = 3.70 seconds.