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A Cesna-182 Skylan starts at rest at one end of a runway undergoes a uniform acceleration of 9.73 m/s2 for 15.2s before takeoff. What is its velocity at takeoff?

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ackahedem582

Answer:

147.896m/s.

Explanation:

initial velocity(u) = 0, since it started from rest.

acceleration (a) = 9.73m/s2

time taken (t) = 15.2s

Using the first equation of motion:

v = u + at

v = 0 + 9.73(15.2)

v = 147.896m/s

hence the velocity at the takeoff is 147.896m/s.

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