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Sagot :
Given data
*The given initial speed of an airplane is u = 0 m/s
*The engine exerts a constant force is F = 78.0 kN = 78.0 × 10^3 N
*The given mass is m = 9.20 × 10^4 kg
*The given final speed of the plane is v = 75.0 m/s
The acceleration of the airplane is calculated by using the relation as
[tex]\begin{gathered} F=ma \\ a=\frac{F}{m} \\ =\frac{78.0\times10^3}{9.20\times10^4} \\ =0.848m/s^2 \end{gathered}[/tex]The formula for the distance of the runway is given by the kinematic equation of motion as
[tex]\begin{gathered} v^2=u^2+2as \\ s=\frac{v^2-u^2}{2a} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} s=\frac{(75.0)^2-(0)^2}{2(0.848)} \\ =3316.62\text{ m} \end{gathered}[/tex]Hence, the distance of the runway is s = 3316.62 m
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