Physics help question!

A jetliner, traveling northward, is landing with a speed of 57.0 m/s. Once the jet touches down, it has 846 m of runway in which to reduce its speed to 5.22 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing (take the direction of the plane's motion as positive).

Can someone please explain how to do this! Thanks!!

The average acceleration is -1.90 m/s/s.

The plane never changes direction, and the acceleration always slows it down, so the sign (direction) of the acceleration will be negative. All we need to do is get the magnitude. And for that, we have two equations.

vf = at + vo ==> at = vf - vo where a is acceleration, t is time, vf = 5.22 is final velocity, and vo = 57.0 is initial velocity.

The second equation is

s = 1/2 at^2 + vo t = (1/2 at + vo)t where s = 846 is distance.

Substitute "at" from the first equation into the second to get

s = [1/2 (vf - vo) + vo] t = (vf + vo) (t/2) ==> t = 2s / (vf + vo)

Now solve the first equation for "a" and substitute in the expression for "t" that we just derived:

a = (vf - vo) / t = (vf - vo)(vf + vo) / (2s) = (vf^2 - vo^2) / (2s)

Finally, plug in numbers:

a = (5.22^2 - 57.0^2) / (2 • 846) = -1.90 m/s/s (Answer)