# 1-D Kinematics question, pls help

1. Apr 25, 2008

### 9giddjl

1-D Kinematics question, pls help:)

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

vav= s/t?

3. The attempt at a solution

In the first part of the equation 'A train moves from a station...over a distance of 225 m'. It states that accelerates uniformily for 30. s over distance of 225 m and then continues that speed for another 135 s, so i did 135 s/30 s and then multiplied the answer which is 4.5 by 225 m= 1012.5, therefore it travelled 1012.5 m in the first 165 sec. The part of the equation i dnt understand is the last part 'then the driver applies the brakes.....' and i do not understand what uniform retardation is. Could someone please help me?
I would really appreciate it, thankyou!:)

2. Apr 25, 2008

### t-bone-steak

there seems to be some values missing in the question, you need a value of the acceleration or the maximum speed, but it doesnt give you one...

3. Apr 25, 2008

### astrorob

Separate the journey into three legs;

The first leg is the departure with which we can use the equation of motion:

$$s = ut + \frac{at^{2}}2$$

to find the acceleration. Then use this to find the max speed obtained after acceleration.

Alternatively, you can just use;

$$s = \frac{(u + v)}2$$

and solve for v, the final speed after acceleration.

Both of these equations will yield the same result.

Are you familiar with these equations of motion?

Last edited: Apr 25, 2008
4. Apr 25, 2008

### Khushveen

Hmm, first line you have u( initial speed ) =0 ,s= 225m, so using s= ut + 1/2at^2 u get a=o.5m/s^2.

Using v= u + at we get final speed (v) = 15m/s

For second line, its linear speed, so to calculate distance covered u use v=d/t, so d = 15*135 = 2025m

And for last part, to calculate retardation u use v= u + at, where v=0, u=15 and t=10...a= -1.5m/s^2

Using s= ut + 0.5at^2, we get distance during retardation, and finally add all distance

5. Apr 25, 2008

### astrorob

You really shouldn't just give the answer..

6. Apr 25, 2008

### Khushveen

Sorry, am new here, just learning how to post ^^

7. Apr 26, 2008

### 9giddjl

Thankyou for that, i understand the first part. Could someone pls tell me what uniform retardation means, is that decceleration? and also why do u make the initial speed 15 m/s in the last part, that was the final speed for the first part but why is it the initial speed in the last? i understand why the final speed is 0 but not why u is 15.
thankyou veri much for helping tho!!

8. Apr 27, 2008

### astrorob

Well,

Uniform retardation means constant deceleration.

As for your second question, think of the journey in three parts.

The train accelerates to 15m/s, then travels a distance at THIS constant speed, then uniformly decelerates to 0m/s.

0m/s > acceleration > 15m/s > constant speed > deceleration > 0m/s

In the last leg of our journey, we're dealing with the deceleration, so the initial speed before deceleration is 15m/s.

9. Apr 28, 2008

### 9giddjl

Oh I get it now! Thankyou very much!

10. Apr 28, 2008

### astrorob

You're welcome