# Help Laws of Motion and question help

1. May 2, 2013

### physicsguru97

1. The problem statement, all variables and given/known data
A train moves off from a station and accelerates uniformly for 30.0 s over a distance of 225 m. It continues with the speed aquired for another 135 s, then the driver applies the brakes and the train comes to rest with uniform retardation at the next station in a further 10.0 s. Calculate the distance between the two stations

2. Relevant equations

s= ut + 1/2at^2

3. The attempt at a solution
i am still stuck please someone help me go through the steps

2. May 2, 2013

### Clever_name

Can you find the acceleration over the 225m distance given you have time?

Last edited: May 2, 2013
3. May 2, 2013

### physicsguru97

No I don't understand which time to use

4. May 2, 2013

### Clever_name

well the question states that the train travels 225m in 30s - try using these.

5. May 2, 2013

### physicsguru97

So I divide those 2 then times it by 135 to get the acceleration?

6. May 2, 2013

### Clever_name

use the equation you stated, s= ut + 1/2at^2

7. May 2, 2013

### physicsguru97

But I don't know how to get acceleration would u be 0 because it started from rest and is the time 30 if I put in equation?

8. May 2, 2013

### Clever_name

that is correct.

9. May 2, 2013

### physicsguru97

Yes but how do you get acceleration?

10. May 2, 2013

### Clever_name

as i said before use this equation..... s= ut + 1/2at^2
you know what s, u, and t are can you not solve the equation for a?

11. May 2, 2013

### physicsguru97

a = 2(s - ut)/t^2 would I rearrange to this??

12. May 2, 2013

### Clever_name

Yes that is correct however you know that u=0 right? because as you said earlier the train starts from rest, and u represents the initial velocity of the train.

13. May 2, 2013

### physicsguru97

Yes I know that so after I've calculated the acceleration what do I do next?

14. May 2, 2013

### Clever_name

Well what you're trying to do is find the total distance to the second train station.... so you're told the train travels at this acceleration you've found for another 135s, you need to find the change in distance with this acceleration.