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?
Physics - bobpursley, 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=
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 - Ana, Saturday, February 6, 2010 at 11:38am
I see my mistakes but the homework site doesn't accept 800m.
Physics - bobpursley, 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.
d= 22^2/2 m