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Physics 9 Online
OpenStudy (anonymous):

A train travels at a speed of 24 m/s. Then it slows down uniformly at 0.065m/s^2 until it stops. What distance does the train travel while slowing down?

OpenStudy (anonymous):

24 m/s / .065 m/s^2 = 369.231 s D = vit + 1/2at^2 D = 24 m/s * 369.231s + 1/2 (.065 m/s^2) * (369.231s)^2 D = 8861.544m + .0325 m/s^2 * 136331.5314s^2 D = 8861.544m + 4430.775m D = 13292.319m ^_^ I g2g to sleep, g'luck!

OpenStudy (anonymous):

thank you and night :D

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