Answer:
The plane needs 1,56 seconds to clear the intersection.
Explanation:
This is a case of uniformly accelerated rectilinear motion.
[tex]V_0^{2} = V_f^{2} - 2ad[/tex]
[tex]V_0=\sqrt{V_0^{2} } = ?[/tex]
Vf=50 m/s
[tex]V_f^{2} = (50 m/s)^{2} = 2500 m^{2}/s^{2}[/tex]
a = -5.4 [tex]m/s^{2}[/tex] (Negative because is decelerating)
d = displacement needed to clear the intersection. It should be the width of the intersection plus the lenght of the plane.
d= 59,7m + 25 m = 84.7 m
Calculating [tex]V_0[/tex]:
[tex]V_0^{2} = V_f^{2} - 2ad[/tex]
[tex]V_0^{2}= 2500 \frac{m^{2} }{s^{2} } - 2(-5.4\frac{m}{s^{2} })(84.7 m)[/tex]
[tex]V_0^{2}= 3,414.76 \frac{m^{2} }{s^{2} }[/tex]
[tex]V_0= \sqrt{3,414.76} = 58.44 \frac{m}{s}[/tex]
Otherwise:
[tex]t = \frac{V_f-V_0}{a} =\frac{50\frac{m}{s} - 58.44\frac{m}{s} }{-5.4 \frac{m}{s^{2} } } = 1.56 s[/tex]
