Two Trains: 500 km Apart in 2.549 Hours

[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

 

[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

 

[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]

 

[Tom McCurdy]

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

 

[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 dimensions of the original equations when putting it under the integral

 

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

 

[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 $500 \mbox{km}$?

 

[Doc Al]

What i did was

Trying to solve

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

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.

 

[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 $s$
~

 

[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 $$\sqrt{86}-6$$