a 747 jet, traveling at a velocity of 70. meters per second north, touches down on a runway. the jet slows to a rest at the rate of 2.0 meters per second^2. calculate the total distance traveled on the runway as it is brought to rest.

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Applying Newton's laws of motion

V^2 = U^2+2aS, where V = 0, final velocity, U = 70 m/s, initial velocity and acceleration, a = -2 m/s^2, decelerating, S is distance

0 = U^2 - 2as; 2aS = U^2

S = sqrt (U^2/2a) = sqrt ( 70^2/2*2) = 1225 m

Distance = 1225 m

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