91Ó°ÊÓ

• Initial velocity, U =\({\bf{54}}\;{\rm{m/s}}\).
• Final velocity, V =0.
• Height, D, at which the trees and the snow stopped the pilot = \({\bf{3}}.{\bf{0}}\;{\bf{m}}\).

The velocity decreases; hence it is a case of deceleration.

When there is a change in the velocity, whether it is due to change in magnitude or direction, we can calculate the acceleration from the equation of motion as

\({V^2} - {U^2} = 2ad\)

Here V is the final velocity, U is the initial velocity, a is the acceleration, and d is the distance that was covered to achieve the final velocity.

Substituting the values in the above expression, we get

\(\begin{array}{c}{(0)^2} - {(54)^2} = 2 \times (3) \times a\\ - 2916 = 2 \times (3) \times a\\a = \frac{{ - 2916}}{6}\\a = - 486\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

Hence the deceleration is\( - {\bf{486}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).