Question

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of

33 ft/s. Its height in feet after t seconds is given by y = 33t - 20t².
A. Find the average velocity for the time period beginning when t=3 and lasting
.01 s:
4
.005 s:
.002 s:
.001 s:
NOTE: For the above answers, you may have to enter 6 or 7 significant digits if you are using a calculator.
Estimate the instanteneous velocity when t=3.

Answer 1

The average velocity for a time period is given by the change in position divided by the change in time. For the time period beginning when t=3 and lasting 0.01 seconds, we have:

• Initial time: t1 = 3 seconds
• Final time: t2 = 3.01 seconds
• Initial position: y1 = 33t1 - 20t1^2 = 9 feet
• Final position: y2 = 33t2 - 20t2^2 = 8.397 feet

The change in position is y2 - y1 = -0.603 feet, and the change in time is t2 - t1 = 0.01 seconds. Therefore, the average velocity for this time period is:

average velocity = change in position / change in time = (-0.603 feet) / (0.01 seconds) ≈ -60.3 feet/second

Similarly, for the time periods beginning when t=3 and lasting 0.005 seconds, 0.002 seconds, and 0.001 seconds, we have:

• Time period of 0.005 seconds:

1. Initial time: t1 = 3 seconds
2. Final time: t2 = 3.005 seconds
3. Initial position: y1 = 33t1 - 20t1^2 = 9 feet
4. Final position: y2 = 33t2 - 20t2^2 = 8.8025 feet
5. Change in position: y2 - y1 = -0.1975 feet
6. Change in time: t2 - t1 = 0.005 seconds
7. Average velocity: (-0.1975 feet) / (0.005 seconds) = -39.5 feet/second

• Time period of 0.002 seconds:

1. Initial time: t1 = 3 seconds
2. Final time: t2 = 3.002 seconds
3. Initial position: y1 = 33t1 - 20t1^2 = 9 feet
4. Final position: y2 = 33t2 - 20t2^2 = 8.8808 feet
5. Change in position: y2 - y1 = -0.1192 feet
6. Change in time: t2 - t1 = 0.002 seconds
7. Average velocity: (-0.1192 feet) / (0.002 seconds) = -59.6 feet/second

• Time period of 0.001 seconds:

1. Initial time: t1 = 3 seconds
2. Final time: t2 = 3.001 seconds
3. Initial position: y1 = 33t1 - 20t1^2 = 9 feet
4. Final position: y2 = 33t2 - 20t2^2 = 8.9408 feet
5. Change in position: y2 - y1 = -0.0592 feet
6. Change in time: t2 - t1 = 0.001 seconds
7. Average velocity: (-0.0592 feet) / (0.001 seconds) = -59.2 feet/second

To estimate the instantaneous velocity when t=3, we can calculate the derivative of the position function with respect to time:

y = 33t - 20t²

dy/dt = 33 - 40t

At t=3, we have:

dy/dt = 33 - 40(3) = -87

Therefore, the estimated instantaneous velocity when t=3 is -87 feet/second.