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
