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Question
A jetliner, traveling northward, is landing with a speed of 59.7 m/s. Once the jet touches down, it has 723 m of runway in which to reduce its speed to 7.05 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing (take the direction of the plane's motion as positive).

Virtually ALL kinematics problems of this type can be solved in the exact same way.
The two applicable kinematics equations are:
Df=1/2*a*t^2+Vo*t+Do and Vf=a*t+Vo
Since these two equation contain 6 variables and since you have two equations any time you can identify 4 of the 6 variables the other 2 can always be found!
In this case the variables are: Do=0m (assume this unless the question indicates otherwise), Df=723m (given), Vo=59.7m/s (given), Vf=0m/s (the plane comes to rest), a=? (Unknown) and t=? (Unknown).
Since we know 4 of the variables the other 2 can always be found. Using the displacement equation:
Df=1/2*a*t^2+Vo*t+Do  becomes  732=1/2*a*t^2+59.7*t+0
And the velocity equation becomes:
Vf=a*t+Vo  0=a*t+59.7
Solving the velocity equation for the time t:
t=-59.7/a
And then substitute this time into the displacement equation:
Df=1/2*a*t^2+Vo*t+Do  then   732=1/2*a*t^2+59.7*t+0=1/2*a*(-59.7/a)^2+59.7*(-59.7/a)
a=-59.7^2/(2*732)=-2.43m/s^2

Physics