Two objects A and B are connected by a rigid rod that has a length L. The objects slide along perpendicular guide rails. If A slides to the left with a constant speed v, what is the velocity of B when a=60$^\circ$?
I know the answer is .566 v, but I am completely unsure why.
I have tried saying that $$ y=L*sin(a), dy/da=d/da L*sina,dy/da=L cosa,dy/da=Lcos60=L/2 $$ but that obviously is the wrong aproach, since I need $dy/dt$, not $dy/da$.
You can write:
$x^2+y^2 = z^2$
where $x$ is the location of A with respect to where the rails meet, and $y$ is the location of B with respect to the same spot. And $z^2$ is just L, the length of the rod, which is constant.
Differentiate this wrt to $t$, and you get
$2x \frac{dx}{dt}+ 2y \frac{dy}{dt} = 0$
We know the angle is 60 degrees, so we can determine what $x$ and $y$ should be using $y=Lsin(\theta)$ and $x=Lcos(\theta)$. We also know $\frac{dx}{dt}=v$.
Plug in, solve for $\frac{dy}{dt}$