Solve 10km Train Acceleration Problem in 6 Mins

[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

Homework Equations

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 separate 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 traveled 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.

 

[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 into 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 into seconds which is 360

Is this correct

 

[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

 

[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 traveled 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 traveling 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).

If we add this time onto the 6 minutes required to accelerate, you'll get your answer.

 

[speedy46]

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

does the 6 mintues have to be converted into seconds

 

[astrorob]

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

does the 6 mintues have to be converted into 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]

No problem!

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

Memorising those five equations of motion (listed under classic version) will enable you to solve any further questions like this, I strongly advise it!