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Posted by on Saturday, February 6, 2010 at 11:12am.

A train is traveling down a straight track at 22 m/s when the engineer applies the brakes, resulting in an acceleration of -1.0 m/s2 as long as the train is in motion. How far does the train move during a 40 s time interval starting at the instant the brakes are applied?

Distance = velocity(initial) * time + 1/2 * acceleration * time^2

D = 22 * 40 + 1/2 * (-1) * 1600

D = 880 + (-800)

D = 80
What's wrong with my approach?
Thanks.

  • Physics - , Saturday, February 6, 2010 at 11:27am

    You used the equation for displacement, not distance.

    Here is a clue: what is final velocity at t=0?
    vf=22 -1*40=22-40= -18/s, so it is moving back toward the origin.
    What was average velocity (22+(-18))/2=
    = 2m/s
    What displacement does it travel in 40 sec?
    d= avg velocity*time=80m
    Now, what distance did it travel?
    distance= 22+18/2=20m/s x 40 sec=800m

  • Physics - , Saturday, February 6, 2010 at 11:38am

    I see my mistakes but the homework site doesn't accept 800m.

  • Physics - , Saturday, February 6, 2010 at 12:50pm

    Ana: I reread the dang problem. Key phrase: " as long a the train in motion" For some of the time of 40 seconds, the train is stopped.

    Vf^2=Vi^2+2ad
    0=22^2-2d
    d= 22^2/2 m

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