# Can anyone help me finding the ans?plz explain!

## Mar 27, 2018

(2) decreasing speed

#### Explanation:

Let the height of the weight be $y$. Then it is easy to see that
${y}^{2} + {\left(\frac{x}{2}\right)}^{2} = {l}^{2}$
where $l = 1 \text{ m}$ is the length of each rod.

Differentiating, we get

$2 y \frac{\mathrm{dy}}{\mathrm{dt}} + \frac{1}{4} 2 x \frac{\mathrm{dx}}{\mathrm{dt}} = 0$

Thus

$\frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{1}{4} \frac{x}{y} \frac{\mathrm{dx}}{\mathrm{dt}}$

In the problem, if we choose the stationary pivot as the origin and the $X$ axis towards the left, then $x > 0$ and $\frac{\mathrm{dx}}{\mathrm{dt}} < 0$, So

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1}{4} \frac{x}{y} | \frac{\mathrm{dx}}{\mathrm{dt}} |$

It is stated that $\frac{\mathrm{dx}}{\mathrm{dt}}$ is a constant. On top of this, $x$ is decreasing while $y$ is increasing. So$\frac{x}{y}$ is decreasing with times - and thus the weight is moving up with decreasing speed.

When $y$ is 0.4 m , we can see that

${\left(\frac{x}{2}\right)}^{2} = {l}^{2} - {y}^{2} = \left({1}^{2} - {0.4}^{2}\right) {\text{ m"^2 = 0.84" m}}^{2}$

Thus $x = 0.92$ m, and these values say that

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1}{4} \frac{0.92}{0.4} | \frac{\mathrm{dx}}{\mathrm{dt}} | \approx 0.57 | \frac{\mathrm{dx}}{\mathrm{dt}} |$