Final answer:
The deceleration of a pilot who landed at 54 m/s and was stopped over 3.0 m is -81 m/s², calculated using the kinematic equation for uniform accelerated motion.
Step-by-step explanation:
To calculate the deceleration of a pilot who landed at 54 m/s and was stopped by trees and snow over a distance of 3.0 m, we can use the formula for deceleration (a), which is derived from the kinematic equation: v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (or deceleration, in this case), and s is the stopping distance.
Since the pilot comes to a stop, the final velocity (v) is 0 m/s, the initial velocity (u) is 54 m/s, and the stopping distance (s) is 3.0 m. Rearranging the equation to solve for a, we have: a = -(u2) / (2s). Plugging in the values, we get: a = -(54 m/s) 2 / (2 × 3.0 m), which gives us a = -486 m/s2 / 6 m = -81 m/s2. The negative sign indicates deceleration.