1. May 20, 2008

### speedy46

1. The problem statement, all variables and given/known data
A train accelerates uniformly from rest to a velocity of 60km/h in 6 minutes after which the velocity is kept constant calculate the time taken to travel 10km

2. Relevant equations
S=u×t
V=u+at

3. The attempt at a solution
I have worked out the 60km/h to meters per seconds I get 16.67m/s and do not know how to contintue

2. May 20, 2008

### speedy46

Thank you for the help this has sent me on me way

3. May 20, 2008

### astrorob

Sorry, let me repost after making a complete hash of the last attempt to answer you!

Still seperate the journey into two legs;

The accelerating leg.
The constant speed leg.

Start by working out the acceleration of the train in the first leg of the journey:

$$v=u+at$$

then using this you can find the distance travelled in the first leg using the equation:

$$s = 0.5at^2$$

so now we're onto the second part of the journey, we know the speed and the total distance we want to travel, but we have already done some that distance in accelerating, so the distance we still need to travel is total distance - distance done in first leg.

Now simply use $$s = ut$$ and solve for t.

Adding the 6 minutes we are told it takes for the train to accelerate you will get your answer.

Last edited: May 20, 2008
4. May 20, 2008

### speedy46

where does the 0.5 come from

5. May 20, 2008

### astrorob

It's another one of the equations of motion formulas for constant acceleration.

http://en.wikipedia.org/wiki/Equations_of_motion

Written in its entirety it's:

$$s = ut+0.5at^2$$

but in your case, the train accelerates uniformly from rest so that u = 0, thus:

$$s = 0.5at^2$$

6. May 20, 2008

### speedy46

Thank you

7. May 20, 2008

### astrorob

No problem.

8. May 22, 2008

### speedy46

to work out the initial velocity in the first equation

V=u+at

do I convet 60 km/h in to 17.8819 m/s

9. May 22, 2008

### astrorob

Yes, it's always wise to convert units you're given into SI units.

10. May 22, 2008

### speedy46

so to work out the acceleration I will use the formula

accerleration (m/s2) = change in speed m/s / time taken for change

for my case it will be

accerleration (m/s2) = 17.8819 m/s divided by 360 = 20.132
changing the 6 minutes in to seconds which is 360

Is this correct

Last edited: May 22, 2008
11. May 22, 2008

### astrorob

60km/h = 60'000/3600m/s ~= 16.67m/s - you actually stated this in your initial post :)

Then divide through by the 360 to get the acceleration.

12. May 22, 2008

### speedy46

do i divide the 360 / 16.67 = 21.595

13. May 22, 2008

### astrorob

Compare these two.

It should be 16.67 / 360.

14. May 22, 2008

### speedy46

which is 0.0463

V=u+at

17.8819 + 0.0463 * 360 = 34.5499

Last edited: May 22, 2008
15. May 22, 2008

### astrorob

Correct, now use this in the equation:

$$s = 0.5at^2$$

You now know a (0.046) and t (360).

This will give you the distance travelled in the accelerating leg of the journey.

16. May 22, 2008

### speedy46

s= 0.5 * 0.0463 * 360 2 = 3000.24

17. May 22, 2008

### astrorob

correct, now the question asks for how long it takes for the train to travel 10km (10000m)

We know it takes 6 minutes to accelerate travelling 3000.24m in the process.

So all that's left is to travel ~7000m at a constant speed.

We use:

distance / speed = time

we know the distance (7000) and the speed (16.67m/s) so we can work out the time required to do this leg of the journey (in seconds).

18. May 22, 2008

### speedy46

time = distance / time = 7000 / 16.67 = 419.91 + 6 = 425.91

does the 6 mintues have to be converted in to seconds

19. May 22, 2008

### astrorob

Yes,

The 419.91 figure is in seconds, and the 6 is in minutes. These need to be the same unit.

20. May 22, 2008

### speedy46

419.91 + 360 = 779.91 seconds