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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

User Afron
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1 Answer

5 votes

Answer:

x = 100.64 [m]

Step-by-step explanation:

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


v_(f) = v_(i) + (a*t)\\


v_(f) ^(2) = v_(i) ^(2) + (2*a*x)\\

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]

User Dwarfy
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