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A train has a mass of 5.22 multiplied by 106 kg and is moving at 82.0 km/h. The engineer applies the brakes, resulting in a net backward force of 1.87 multiplied by 106 N on the train. The brakes are held on for 21.0 s.

  • Physics -

    You forgot to ask a question.

    It is possible to use your data to compute a deceleration rate, and the velocity change after 21 s of braking.

    a = F/m = 1.86*10^6/5.22*10^6
    = 0.3563 m/s^2.
    That is the deceleration rate.

    After 21.0 s, the velocity will be lowered by a*t = 7.48 m/s

    The original speed was 82.0 km/h = 22.78 m/s.

    The final speed is 15.30 m/s.
    That can be converted back to km/h if you wish.

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