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Answer :
Explanation:
It is given that,
Distance covered by the airplane, [tex]d=8.72\ km=8.72\times 10^3\ m[/tex]
Time taken, t = 35.9 s
Acceleration of the airplane, [tex]a=3.03\ m/s^2[/tex]
(a) Let u is the initial speed of the airplane at the beginning of the 35.9 seconds. It can be calculated using the second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
[tex]u=\dfrac{d-(\dfrac{1}{2}at^2)}{t}[/tex]
[tex]u=\dfrac{8.72\times 10^3-(\dfrac{1}{2}\times 3.03\times (35.9)^2)}{35.9}[/tex]
u = 188.50 m/s
(b) Let v is the speed of the airplane at the end of the 35.9 seconds. It can be calculated using the first equation of motion as :
[tex]v=u+at[/tex]
[tex]v=188.50+3.03\times 35.9[/tex]
v = 297.27 m/s
Hence, this is the required solution.
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