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A stealth bomber is moving at 799km/h west and travels to a top velocity of 1,1325km/h west in 4.8 minutes.
What is the stealth bomber's acceleration in m/s??

Respuesta :

pstnonsonjoku

Answer:

10.1 ms^-2

Explanation:

From;

v = u + at

v = final velocity

u = initial velocity

a= acceleration

t = time taken

To convert the velocity from Km/hr to m/s we us;

x * 1000/3600

Where x is the velocity in Km/hr

So;

799 * 1000/3600 = 221.9 m/s

11325 * 1000/3600 = 3145.8 m/s

4.8 minutes = 4.8 * 60 = 288 s

Applying the formula;

a= v - u/t

a = 3145.8 - 221.9/288

a = 10.1 ms^-2

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