A jetliner, traveling northward, is landing with a speed of 60 m/s. Once the jet touches down, it has 795 m of runway in which to reduce its speed to 5.6 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing (take the positive direction to be northward).
Vi = 60
5.6 = 60 + a t
a t = - 54.4
so
t = -54.4/a
795 = 60 t + (1/2) a t^2
795 = 60 (-54.4/a) + .5 (2960/a)
795 a = -3264 + 1480
a = -2.24 m/s^2
To find the average acceleration of the plane during landing, we can use the formula:
average acceleration = (final velocity - initial velocity) / time
In this case, the final velocity of the plane is 5.6 m/s, and the initial velocity is 60 m/s. We need to find the time it takes for the plane to reduce its speed from 60 m/s to 5.6 m/s.
To find the time, we can use the equation of motion:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance
Plugging in the values, we have:
(5.6 m/s)^2 = (60 m/s)^2 + 2 * acceleration * 795 m
31.36 m^2/s^2 = 3600 m^2/s^2 + 1590 m * acceleration
Rearranging the equation, we get:
1590 m * acceleration = -3600 m^2/s^2 + 31.36 m^2/s^2
1590 m * acceleration = -3568.64 m^2/s^2
Dividing both sides by 1590 m, we get:
acceleration = -2.247 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which is northward.
Therefore, the average acceleration of the plane during landing is 2.247 m/s^2 southward.