Question

A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s^2 for 14.0 s. It runs at constant speed for 70.0 s and slows down at a rate of 3.50 m/s^2 until it stops at the next station. Find the total distance covered?

so far I figured out that I have x=156.8 and Vx = 22.4
but where do i go from there

Answer 1

Your calculations are perfect for the acceleration part.

You should have no problem figuring the distance travelled for 70 seconds at 22.4 m/s.
The slow-down part is the same as the acceleration part, except that the initial velocity u=22.4, final velocity=0 and acceleration is negative (i.e. -3.5 m/s/s).

Take it from here and post us what you get!

Answer 2

So if you could help me a little bit more which equation do i use to figure that out?

Answer 3

For the constant speed part, the formula to use is
Distance = speed * time,
Make sure they are in consistent units.

The deceleration part uses the same formula as the acceleration part, except for the negative value for the acceleration. What formula did you use for the first part where you got correct results?

Answer 4

I used Vx= Vox + axt to find the 22.4
and i used the x=xo + Voxt +1/2 axt^2 to get the 156.8
for the distance part i used 22.4 m/s (70 s) which seconds cancel and i get left with 1568 m

Answer 5

Excellent, the distance for the constant speed portion is correct!

For the deceleration part, use the same formula as for the acceleration part, substitute V_x0=22.4 m/s as you calculated, and Vx_x0t = 0 m/s.
Don't forget that a (in m/s/s) has to be negative because the train slows down, or decelerates.

Tell me what you get.

Answer 6

so then i would use: Vx = Vox + axt
to find the time because we don't have a time so it would be 22.4 = 0 + 3.50(t)
so I got t = 6.4 s

so then i put it in the second equation and got
x= 0 + 22.4(6.4)+ 1/2(-3.50)(6.4)^2
so then that was 143.36-71.68 =
x=71.68 m

Answer 7

Great! You got the right answer.
In fact, you did very well, because the time of slowing down is not given, but you managed to figure it out. Bravo!

That reminds me of another formula you could have used when the starting and ending velocities as well as the acceleration are known:
If
S=distance
u=initial velocity
v=final velocity
a=acceleration (negative if slowing down)

2 a S = v²-u²

From this you can figure out the distance directly, since u, v and a are known. Mind giving it a try?