A jetliner, travelling northward, is landing with a speed of 79 m/s. Once the jet touches down, it has 710 m of runway in which to reduce its speed to 7.4 m/s. Compute the the average accelaration (magnitude and direction) of the plane during landing.
v^2=u^2+2fs or f=(v^2-u^2)/2s
=(7.4^2-79^2)/(2*710)
=(54.76-6241)/1420=-4.36m/s^2
Negative acceleration (i.e. against the direction of motion)of 4.36m/s^2
To compute the average acceleration of the plane during landing, we can use the following formula:
average acceleration = (final velocity - initial velocity) / time
In this case, the final velocity is the speed at which the plane comes to a stop (7.4 m/s), and the initial velocity is the speed at the beginning of the landing (79 m/s).
First, we need to find the time it takes for the plane to come to a stop. We can use the equation of motion:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance
Rearranging the equation, we get:
acceleration = (final velocity^2 - initial velocity^2) / (2 * distance)
Plugging in the given values:
- final velocity = 7.4 m/s
- initial velocity = 79 m/s
- distance = 710 m
acceleration = (7.4^2 - 79^2) / (2 * 710)
Now we can calculate the acceleration:
acceleration = (-585.56) / (1420)
Therefore, the average acceleration of the plane during landing is approximately -0.411 m/s^2, in the opposite direction of the initial velocity (northward).