5) Placido pulls a rope attached to a wagon through a pulley at a rate of $q \mathrm{~m} / \mathrm{s}$. With dimension as in Figure 5: (a) Find a formula for the speed of the wagon in terms of $q$ and the variable $x$ in the figure. (b) Find the speed of the wagon when $x=0.6$ if $q=0.5$ $\mathrm{m} / \mathrm{s}$. FIGURE 5

Step 1 ：Let's denote the horizontal distance the wagon moves as \(y\), the height of the pulley as \(h\), and the length of the rope as \(r\). Then we have \(r^2 = y^2 + h^2\).

Step 2 ：Differentiating both sides with respect to time, we get \(2r \frac{dr}{dt} = 2y \frac{dy}{dt} + 2h \frac{dh}{dt}\).

Step 3 ：Since the height of the pulley doesn't change, \(\frac{dh}{dt} = 0\). So the equation simplifies to \(r \frac{dr}{dt} = y \frac{dy}{dt}\).

Step 4 ：The rate at which the length of the rope changes is the speed at which Placido pulls the rope, which is given as \(q\). So we can substitute \(\frac{dr}{dt} = q\) into the equation to get \(r q = y \frac{dy}{dt}\).

Step 5 ：The speed of the wagon is \(\frac{dy}{dt}\), which is what we're trying to find. We can solve the equation for \(\frac{dy}{dt}\) to get \(\frac{dy}{dt} = \frac{r q}{y}\).

Step 6 ：From the Pythagorean theorem, we know that \(r = \sqrt{y^2 + h^2}\). Substituting this into the equation gives \(\frac{dy}{dt} = \frac{\sqrt{y^2 + h^2} q}{y}\).

Step 7 ：Finally, from the figure, we can see that \(y = h x\). Substituting this into the equation gives \(\frac{dy}{dt} = \frac{\sqrt{(h x)^2 + h^2} q}{h x} = \frac{\sqrt{x^2 + 1} q}{x}\).

Step 8 ：This is the formula for the speed of the wagon in terms of \(q\) and \(x\).

Step 9 ：Final Answer: \(\boxed{\frac{dy}{dt} = \frac{\sqrt{x^2 + 1} q}{x}}\)