speedy46

## Homework Statement

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

S=u×t
V=u+at

## 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

speedy46
Thank you for the help this has sent me on me way

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:
speedy46
where does the 0.5 come from

astrorob
where does the 0.5 come from

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:

$$s = 0.5at^2$$

speedy46
Thank you

astrorob
No problem.

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

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

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:
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.

speedy46
do i divide the 360 / 16.67 = 21.595

astrorob
so to work out the acceleration I will use the formula

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

do i divide the 360 / 16.67 = 21.595

Compare these two.

It should be 16.67 / 360.

speedy46
which is 0.0463

V=u+at

17.8819 + 0.0463 * 360 = 34.5499

Last edited:
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.

speedy46
s= 0.5 * 0.0463 * 360 2 = 3000.24

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).

speedy46
time = distance / time = 7000 / 16.67 = 419.91 + 6 = 425.91

does the 6 mintues have to be converted in to seconds

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

does the 6 mintues have to be converted in to seconds

Yes,

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

speedy46
419.91 + 360 = 779.91 seconds

astrorob
Correct!

It may be nicer to convert it into minutes (but this is not necessary!)

that is, 780/60 = 13 minutes.

Either way is right so long as you state the units.

speedy46
Thank you for helping me much appreciated

astrorob