# Helping friend

1. Mar 21, 2005

### Tom McCurdy

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

$$500= \int_{0}^{t} (20x^2+120x)$$

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

2. Mar 21, 2005

### Tom McCurdy

So then I guess you could do

$$500= \frac{20t^3}{3}+60t^2}-(\frac{20x^3}{3}+60x^2})$$

$$500=\frac{20t^3}{3}+60t^2}$$

but hmm when you solve that you still get negitive answers
What am I doing wrong

3. Mar 21, 2005

### Tom McCurdy

lol I just inserted the
$$\frac{20t^3}{3}+60t^2$$ into y1
then 500 into y2
find where they intersect
it should be 2.54841

[edit--- this is wrong]

Last edited: Mar 21, 2005
4. Mar 21, 2005

### Tom McCurdy

Can someone please show me why I can't go back and do basic math??

5. Mar 21, 2005

### Tom McCurdy

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 dimesnions of the original equations when putting it under the integral

6. Mar 21, 2005

### Data

Set up a differential equation

$$x^{\prime \prime} (t) = 20 \frac{\mbox{km}}{\mbox{h}^2}$$

with ICs $$x^\prime (0) = 120 \frac{\mbox{km}}{\mbox{h}}, \ x(0) = 0$$ and solve for $x(t)[/tex]. Once you've done that, just plug in [itex] x(t_0) = 500 \mbox{km}$ and solve for $t_0$.

Last edited: Mar 21, 2005
7. Mar 21, 2005

### Data

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! ) 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 $$\left(20 \frac{\mbox{km}}{\mbox{h}^2}\right)t^3$$ term? Do they match the units of $500 \mbox{km}$?

Last edited: Mar 21, 2005
8. Mar 21, 2005

### Staff: Mentor

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

Time is in hours, of course.

Last edited: Mar 21, 2005
9. Mar 21, 2005

### Data

Time to nitpick:

To clean up notation, you should actually use

$$\int_0^t (20s + 120) ds$$

or some other random variable instead of $s$
~

10. Mar 21, 2005

### Tom McCurdy

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 $$\sqrt{86}-6$$