Two Trains: 500 km Apart in 2.549 Hours

  • Thread starter Tom McCurdy
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In summary, two trains, A and B, leave from the same station in opposite directions. Train A has a velocity of 120 km/hr and train B has an acceleration of 20 km/hr/hr. The question is when will they be 500 km apart? The conversation discusses different approaches to solving this problem, including using integrals and kinematics equations. Ultimately, they come to the conclusion of solving a differential equation with given initial conditions and plugging in the desired distance to find the time. The final answer is approximately 2.549 hours.
  • #1
Tom McCurdy
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I am trying to help a friend through this problem

train problem - 2 trains going oposite direction leaving from same station - train A going 120 km/hr (velocity)
- train B going 20 km / hr / hr (acceler) Time when are they 500 km apar

What i did was

Trying to solve

[tex] 500= \int_{0}^{t} (20x^2+120x) [/tex]

however when I put that into the solver I am getting negitive answers
with guess and check with the integral I cam out to around 2.549
 
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  • #2
So then I guess you could do

[tex] 500= \frac{20t^3}{3}+60t^2}-(\frac{20x^3}{3}+60x^2}) [/tex]

[tex] 500=\frac{20t^3}{3}+60t^2} [/tex]

but hmm when you solve that you still get negitive answers
What am I doing wrong
 
  • #3
lol I just inserted the
[tex] \frac{20t^3}{3}+60t^2 [/tex] into y1
then 500 into y2
find where they intersect
it should be 2.54841

[edit--- this is wrong]
 
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  • #4
Can someone please show me why I can't go back and do basic math??
 
  • #5
I know you can solve it with .5*20*x^2+120x=500 but the thing I don't get is why you reduce the dimensions of the original equations when putting it under the integral
 
  • #6
Set up a differential equation

[tex] x^{\prime \prime} (t) = 20 \frac{\mbox{km}}{\mbox{h}^2}[/tex]

with ICs [tex] x^\prime (0) = 120 \frac{\mbox{km}}{\mbox{h}}, \ x(0) = 0[/tex] and solve for [itex]x(t)[/tex]. Once you've done that, just plug in [itex] x(t_0) = 500 \mbox{km}[/itex] and solve for [itex]t_0[/itex].
 
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  • #7
The reason that your integral didn't work is that it is set up wrong. Why would you integrate distance (although your integrand isn't quite distance either, but it's close enough for me! :smile:) to get distance?

If you want it illustrated more clearly that your method doesn't work, then just tell me:What are the units of your [tex]\left(20 \frac{\mbox{km}}{\mbox{h}^2}\right)t^3[/tex] term? Do they match the units of [itex]500 \mbox{km}[/itex]?
 
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  • #8
Tom McCurdy said:
What i did was

Trying to solve

[tex] 500= \int_{0}^{t} (20x^2+120x) [/tex]
Don't know why you are integrating (instead of just using kinematics equations), but this is the integral you want:
[tex] 500= \int_{0}^{t} (20t +120) dt [/tex]

Time is in hours, of course.
 
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  • #9
Time to nitpick:

To clean up notation, you should actually use

[tex] \int_0^t (20s + 120) ds[/tex]

or some other random variable instead of [itex]s[/itex]
~
 
  • #10
Thanks... yeah I got that now... I had added an extra dimension for no reason. And I agree the kinematic equations work much easyier but he wasn't allowed to do it. We got what we think is the right answer now of [tex] \sqrt{86}-6[/tex]
 

Related to Two Trains: 500 km Apart in 2.549 Hours

What is the distance between the two trains?

The distance between the two trains is 500 km.

What is the speed of each train?

The speed of each train is 500 km / 2.549 hours = 196.4 km/h.

How long does it take for the trains to meet?

The trains will meet in 2.549 hours.

What is the formula used to calculate the distance between two objects?

The formula used to calculate the distance between two objects is distance = rate * time.

How would the outcome change if the trains were traveling in opposite directions at different speeds?

If the trains were traveling in opposite directions at different speeds, the outcome would change as the distance between the two trains would be decreasing at a faster rate. The time it takes for the trains to meet would also change depending on the relative speeds of the trains.

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