Vectors & 3D Geometry
Average velocity of midpoint of sliding ladder
MJMT_Full_Test_02
Grade 12
Question:
A ladder of 3m length leans against a wall. The ladder forms a vertical angle of 30° with the wall. The top slides down at 20 cm/s. The bottom slides away at 20 cm/s at time $t$. The average velocity of a person halfway up the ladder for the first $t$ seconds is
$\frac{\sqrt{3}+\sqrt{2}}{5}\left[\left(1-\frac{1}{\sqrt{2}}\right)\hat{i}+\left(1-\frac{3}{\sqrt{2}}\right)\hat{j}\right]$
$\frac{\sqrt{3}+\sqrt{2}}{10}\left[\left(1-\frac{1}{\sqrt{2}}\right)\hat{i}+\left(1-\frac{\sqrt{3}}{2}\right)\hat{j}\right]$
$\frac{\sqrt{3}+\sqrt{2}}{5}\left[(\sqrt{2}-1)\hat{i}+(\sqrt{2}-3)\hat{j}\right]$
$\frac{\sqrt{3}+\sqrt{2}}{10}\left[(\sqrt{2}-1)\hat{i}+(\sqrt{2}-3)\hat{j}\right]$
Step-by-Step Solution
Key Concept: At $t=0$: $x_0=3/2$, $y_0=3\sqrt3/2$. When $|dx/dt|=|dy/dt|=0.2$ m/s, $x=y=3/\sqrt2$. Midpoint is at $(x/2,y/2)$. Average velocity = $\Delta(x/2,y/2)/t$.
Average velocity $=\frac{\sqrt3+\sqrt2}{10}[(\sqrt2-1)\hat i+(\sqrt2-3)\hat j]$.
Correct Answer: 4