Question

A care package is dropped out of a cargo plane and lands in the forest. If we assume the care package speed on impact is 54 m/s (123 mph), then what is its acceleration? Assume the trees and snow stops it over a distance of 3.0 m.

a) 180 m/s²
b) 486 m/s²
c) 324 m/s²
d) 162 m/s²

Answer 1

The acceleration of the care package is -1458 m/s².

Step-by-step explanation:

The acceleration of the care package can be calculated using the equation a = (v^2 - u^2) / (2s), where v is the final speed, u is the initial speed, and s is the distance. In this case, the final speed is 0 m/s (as the package comes to a stop), the initial speed is 54 m/s, and the distance is 3.0 m. Plugging in the values, the acceleration is calculated:

a = (0^2 - 54^2) / (2 * 3.0) = -1458 m/s^2

Since acceleration is a vector quantity, the value is negative because it represents deceleration. Therefore, the correct answer is (b) -1458 m/s².