A jetliner, traveling northward, is landing with a speed of 65 m/s. Once the jet touches down, it has 785 m of runway in which to reduce its speed to 5.0 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing (take the positive direction to be northward).
Vf^2=Vi^2+2*a*distance. compute a, if it turns out negative, it must be in the direction opposite the displacement vector.
To compute the average acceleration of the plane during landing, we can use the equation:
average acceleration = (final velocity - initial velocity) / time
In this case, the final velocity is 5.0 m/s, the initial velocity is 65 m/s, and we need to find the time it takes for the plane to reduce its speed to the final velocity over a distance of 785 m.
To find the time, we can use the equation of motion:
final velocity² = initial velocity² + 2 * acceleration * distance
In this case, the initial velocity is 65 m/s, the final velocity is 5.0 m/s, the distance is 785 m, and we need to solve for the acceleration. Rearranging the equation, we get:
acceleration = (final velocity² - initial velocity²) / (2 * distance)
Now we can substitute the values into the equation to find the acceleration:
acceleration = (5.0 m/s)² - (65 m/s)² / (2 * 785 m)
Calculating this, we get:
acceleration = (-4225 m²/s² - 4225 m²/s²) / (1570 m)
Simplifying, we have:
acceleration = -8450 m²/s² / 1570 m
The magnitude of the average acceleration is the absolute value of the result, which is:
magnitude of acceleration = 5.38 m/s²
Since the positive direction is defined as northward, and the acceleration is negative, the direction of the acceleration is southward.
Therefore, the average acceleration of the plane during landing is approximately 5.38 m/s² southward.