Question

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.

Answer 1

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 <b>displacement</b> 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

Answer 2

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

Answer 3

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