1) A jetliner, traveling northward, is landing with a speed of 69.3 m/s. Once the jet touches down, it has 787 m of runway in which to reduce its speed to 13.6 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing (take the direction of the plane's motion as positive).
I know that the formula for average acceleration is change in velocity / elasped time.
To get the change in velocity I did 69.3-13.6. I am not sure what the elasped time is.
a ={v(final)² - v(initial)²}/2•s =
=(13.6² - 69.3²)/2•787= - 2.93 m/s²
To find the average acceleration of the plane during landing, you correctly identified that the formula for average acceleration is change in velocity divided by elapsed time.
In this case, the change in velocity is the difference between the initial velocity (69.3 m/s) and the final velocity (13.6 m/s). Therefore, the change in velocity is: 69.3 m/s - 13.6 m/s = 55.7 m/s.
To determine the elapsed time, you need to consider the distance travelled and the average velocity. The plane starts from rest (initial velocity is 0 m/s), and the distance it needs to cover is 787 m.
Since the average velocity is the total distance divided by the elapsed time, you can rearrange the equation to solve for the elapsed time: elapsed time = distance / average velocity.
Therefore, the elapsed time is: 787 m / ((69.3 m/s + 13.6 m/s) / 2) = 787 m / (82.9 m/s / 2) = 787 m / 41.45 m/s = 18.99 seconds (rounded to two decimal places).
Now that you have the change in velocity (55.7 m/s) and the elapsed time (18.99 s), you can calculate the average acceleration by dividing the change in velocity by the elapsed time: average acceleration = change in velocity / elapsed time = 55.7 m/s / 18.99 s.
Finally, calculating the value gives us: average acceleration = 2.93 m/s^2 (rounded to two decimal places).
So, the average acceleration of the plane during landing is approximately 2.93 m/s^2 in the positive direction (since the direction of the plane's motion is considered positive in this case).