When Were Two Trains Closest Together?

Delber
Messages
19
Reaction score
0

Homework Statement


A train leaves the station at 10:00 and travels due south at a speed of 60 km/h. Another train has been heading due west at 45 km/h and reaches the same station at 11:00. At what time were the two trains closest together?

Homework Equations


c^{2}=a^{2}+b^{2}

The Attempt at a Solution


The trouble I am having is the wording of the question. I think it means the trains left from two different stations and will arrive at the same time together at one station. So the equation for that scenario is:

f(t) = \sqrt{(60-60t)^{2}+(45t)^{2}}

However I get a range outside the limit of one hour. I think I'm just confused about he wording if someone can clarify it for me. Any help is appreciated.
 
Physics news on Phys.org
Delber said:

Homework Statement


A train leaves the station at 10:00 and travels due south at a speed of 60 km/h. Another train has been heading due west at 45 km/h and reaches the same station at 11:00. At what time were the two trains closest together?

Hi Delber! :smile:

It means train 2 arrives at the same station one hour after train 1 left. :smile:
 
Thanks for the clarification.

So the new equation should be:

f(t)=\sqrt{(45-45t)^{2}+(60t)^{2}}?

Edit: Yep, I get the correct answer in the book. Thanks for the help.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top