Ajet airplane touches down on an airport runway with a speed of 207 km/hr. it slows at a constant rate to a stop over a time period of 7.43 s. how far does the jet travel along the runway before stopping?

So distance travel by airplane before stopping is 213.637 m

Explanation:

We have given initial speed of of the airplane u = 207 km /hr

Time t = 7.43 sec

207 km/hr =

As the plane finally stops its final velocity v = 0 m/sec

According to first equation of motion v = u+at

So 0=57.5+a×7.43

a= -7.738

Now according to third equation of motion

s = 213.637 m

answer: electrons

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explanation: