Question

An object is dropped straight down from a helicopter. The object falls faster and faster but its acceleration (rate of change of its velocity) decreases over time because of air resistance. The acceleration is measured in ft/sec and recorded every second after the drop for 4 sec, as shown. Find the following estimates.

t 0 1 2 3 4
a 32.00 19.41 11.77 7.14 4.33

Find an upper estimate for the speed when t = 4. ft/sec

Answer 1

The upper estimate for the speed of the object at t = 4 seconds, accounting for decreasing acceleration due to air resistance, is calculated as 82.75 ft/s.

Step-by-step explanation:

To estimate the upper speed of the object at t = 4 seconds after being dropped from the helicopter, we can approximate the speed using the given acceleration values at each second. Since air resistance decreases the acceleration over time, we will assume a constant acceleration within each one-second interval between recorded values. This is not exact but gives an upper bound estimate.

For the first second (from t = 0 to t = 1), we use the initial acceleration of 32.00 ft/s².

Multiply this by 1 second to get a speed of 32.00 ft/s at t = 1. For each subsequent second, multiply the average of the accelerations at the beginning and end of the interval by the time interval (1 second) and add it to the previous speed:

• Between t = 1 and t = 2, we use the average acceleration (32.00 + 19.41) / 2 = 25.705 ft/s². After 1 second this adds 25.705 ft/s to the speed, giving an estimate of 32.00 + 25.705 = 57.705 ft/s at t = 2.
• Between t = 2 and t = 3, we use (19.41 + 11.77) / 2 = 15.59 ft/s². This adds 15.59 ft/s, resulting in 57.705 + 15.59 = 73.295 ft/s at t = 3.
• Finally, between t = 3 and t = 4, we use (11.77 + 7.14) / 2 = 9.455 ft/s². After 1 second, this adds 9.455 ft/s, leading to 73.295 + 9.455 = 82.75 ft/s at t = 4.

Therefore, an upper estimate for the speed of the object at t = 4 seconds is 82.75 ft/sec.