# How far apart are the stations?

1. Dec 7, 2016

### Starrrrr

1.
A train leaves a station from rest with a constant acceleration of 0.3m/s2. It reaches a maximum speed after 3 minutes and maintains this speed for a further 5 minutes, when it slows down with a deceleration −2 m/s2 until it comes to rest at the next station. How far apart are the stations?

2.
s=1/2(a)(t)^2
s2-s1 or s1-s2 not sure.

3. The attempt at a solution: a1=0.3ms^-2 t1=180s a2=-2ms^-2 t2=300s
Then I used the s=1/2at^2 formula and found that s1=4860m and s2=-90000m
This is the part I'm stuck on is it s2-s1 or s1-s2 to find how far apart the stations are?

When I use the s2-s1 it gives me a negative value and I think there is no such thing as a negative distance.

2. Dec 7, 2016

### jbriggs444

There are three portions of the trip, not just two.

The equation $s=\frac{1}{2}at^2$ only applies if certain conditions are met. What are some of those conditions?

Edit: I may have misunderstood your difficulty. The approach that I have in mind is adding up the distance for the three segments of the journey, not subtracting.

3. Dec 7, 2016

### Starrrrr

I have been trying to find the third trip but I still dont understand.

4. Dec 8, 2016

### TomHart

There are 3 segments of the trip that you need to calculate individually. The first segment is where the train is accelerating during the whole segment. The second segment is where the train is moving at constant velocity during the whole segment. The third segment is where the train is decelerating during the whole segment. The distance between the two stations is the sum of those three distances: s1 + s2 + s3

5. Dec 8, 2016

### Starrrrr

ok thanks ill try and solve it, I have been doing this question for 3 hours, the funny thing is I find the hardest qs easy and easy qs hard :D

6. Dec 8, 2016

7. Dec 8, 2016

### TomHart

I got the same result.

8. Dec 8, 2016

### Starrrrr

Ok thank you sooo much. ;)