Question

A jetliner, traveling northward, is landing with a speed of 57.4 m/s. Once the jet touches down, it has 799 m of runway in which to reduce its speed to 6.92 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing (take the direction of the plane's motion as positive).

Answer 1

Let a be the acceleration of the jet, then using

v2 - u2 = 2as, where, v = 6.92 m/s, u = 57.4 m/s,

s = 799 m, we have,

6.922 - 57.42 = 2*a*799

Or, 47.8864 - 3294.76 = 1598a

Or, - 3246.8736 = 1598a

Or, a = -(3246.8736/1598) = -2.031

Thus, magnitude of acceleration = 2.03 m/s2

Direction of acceleration is opposite to the direction of motion of the jet. (It is obvious from the negative sign appering in the expression of acceleration or from the fact that speed of the jet is decreasing).