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A Lockheed F-117 Nighthawk stealth bomber starts at rest at the south end of a runway and undergoes a uniform acceleration of 17.98 m/s2 to the north. it takes the plane 7s to reach a velocity of 60.16 m/s to the north. How far does the plane travel along the runway

Respuesta :

rafaleo84

Answer:

x = 100.64 [m]

Explanation:

To solve this problem we must use the following kinematics equation.

[tex]v_{f} = v_{i} + (a*t)\\[/tex]

[tex]v_{f} ^{2} = v_{i} ^{2} + (2*a*x)\\[/tex]

where:

Vi = initial velocity [m/s]

Vf = final velocity = 60.16 [m/s]

a = acceleration = 17.98 [m/s^2]

t = time = 7 [s]

Now replacing the values, we have:

(60.16)^2 = 0 + (2*17.98*x)

x = 100.64 [m]