An airplane covers a straight-line distance of 8.72 km in 35.9 s, during which time it has a constant forward acceleration of 3.03 m/s².

a) What is the speed of the airplane, in m/s, at the beginning of the 35.9 s?

b) What is the speed of the airplane, in m/s, at the end of the 35.9 s?

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Answer :

Muscardinus Muscardinus · Answer 1 · 2024-07-20T16:09:23-04:00

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.