Final answer:
The deceleration experienced by a pilot falling from an aircraft and stopping over 3.0 m is calculated using the equation v² = u² + 2as. The pilot's deceleration would be -486 m/s², indicating a rapid decrease in velocity over a short distance.
Step-by-step explanation:
To calculate the deceleration undergone by a pilot falling from a plane and stopping over a certain distance, we must utilize the formula for deceleration. The final velocity (v) of the pilot upon impact was 54 m/s, and he was brought to a stop (final velocity of 0 m/s) over a distance of 3.0 m. To find deceleration (a), we can use the equation v² = u² + 2as, where u is the initial velocity (0 m/s in this case), v is the final velocity, a is the deceleration, and s is the distance over which deceleration occurs.
By rearranging this equation to solve for a, we get a = (v² - u²) / (2s). Substituting the given values, the deceleration is (54² - 0²) / (2 × 3.0) = 2916 / 6 = 486 m/s². Due to the negative sign, which indicates a reduction in speed, the deceleration is -486 m/s². This value is significant as it shows the intense forces that a body can undergo in such events, highlighting the miracle that some pilots survived such free falls during World War II.