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March 29, 2017

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An airplane lands and starts down the runway at a southwest velocity of 43 m/s. What constant acceleration allows it to come to a stop in 1.2 km?Answer is in ___ m/s2.

  • Physics - ,

    Vf^2=Vi^2+2a*distance

  • Physics - ,

    V = sqrt (2 a X)
    X = 1200 m
    V = 43 m/s

    Solve for a. Add a minus sign because there must be deceleration. Or make the direction of the acceleration 180 degrees opposite from the landing direction.

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