An airplane covers a straight-line distance of 8.72km in 35.9 s, during which time it has a constant forward acceleration of 3.03 m/s^2. 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?

Explanation:

It is given that,

Distance covered by the airplane,

Time taken, t = 35.9 s

Acceleration of the airplane,

(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 :

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 :

v = 297.27 m/s

Hence, this is the required solution.

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