Hi,
I understand now what my problem was.
I can also 'see' why it's true through the algebra, but I think the fact the independent variable is a derivatie of time, makes it a bit un-intuitiable for me.
I tried plotting it but it came as something ##\delta## like, so it didn't helped so much...
Yes, you're right. It dosen't! my professor accidently wrote it as:
\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)}}
and that's how I tried to solve it...
thank you!
If you mean
\left(\frac{\dot{q}}{const}\right)^{2}-\dot{q}-1=0\Rightarrow?
It's by dividing two sides by const and multiplying by
\sqrt{1+\left(\dot{q}\right)^{2}}
and than squaring both sides.
If I'll continue I'll need to solve quadratic equation which it's results will depend on the...
Homework Statement
\frac{d}{dt}\left(\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}\right)=0\Rightarrow\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A\Rightarrow q=At+B
Homework Equations
Why it ok to say that...
Of course it does :smile:
But speaking in energetic terms - is the work (Force*Length [Joule]) will be the same,
no matter what's the configuration of the engine comparing to the wagons?
If so, why did the 'Push-Pull' train was invented?
Couldn't they just leave the engine in the rear part...
Hi,
I have a question about where's the most efficient place for a train engine -
Let's say a train has to do a some route back and forth, the route has curves and positive and negative slopes.
In real life - What's the best position for the enginge?
My intuition says it will be most efficient...