Final answer:
To find the deceleration, use the kinematic equation v^2 = u^2 + 2as, yielding a deceleration of 486 m/s^2 when a pilot goes from 54 m/s to a stop over 3 meters.
Step-by-step explanation:
The question is asking for the deceleration of a pilot who landed from a free-fall using tree branches and snow to come to a stop. Deceleration is the rate at which an object slows down, and it is a vector quantity, meaning it has both magnitude and direction. In this case, the deceleration will be in the opposite direction to the pilot's initial velocity (upwards).
To calculate the deceleration, we can use the kinematic equation that relates velocity, acceleration, and displacement:
v2 = u2 + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration (deceleration in this case), and s is the displacement.
The pilot's final velocity when he comes to a stop is 0 m/s, and we're given his initial velocity as 54 m/s. The displacement, s, over which the deceleration occurs is 3.0 meters. Plugging in the values and rearranging the equation to solve for a, we get:
0 = 542 + 2a(3)
a = -542 / (2 × 3)
a = -486
This means the pilot's deceleration was 486 m/s2, a tremendous deceleration force.