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A jet 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?

Answer :

Answer:

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 = [tex]=\frac{207\times 1000m}{3600sec}=57.5m/sec[/tex]

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[tex]m/sec^2[/tex]

Now according to third equation of motion [tex]v^2=u^2-2as[/tex]

[tex]0^2=57.5^2-2\times 7.738\times s[/tex]

s = 213.637 m

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