Particle Approaching a Fixed Point
Problem Analysis
A particle moves with constant velocity vector $\vec{V}$.
Magnitude $|\vec{V}| = v = 5$ m/s.
At $t=0$, its velocity component towards the fixed point P is $u = 3$ m/s.
After $\Delta t = 6$ s, it reaches the position closest to P.
Geometry of Approach
Let the particle be at distance $r$ from P at $t=0$.
The velocity component towards P is the projection of $\vec{V}$ onto the position vector $\vec{r}$.
$$ v \cos \theta = u $$
where $\theta$ is the angle between the velocity vector and the position vector.
From the diagram, $\cos \theta = \frac{x}{r}$, where $x$ is the distance along the path from the current position to the closest point (where $\vec{r} \perp \vec{v}$).
So, $\frac{v x}{r} = u \implies \frac{x}{r} = \frac{u}{v} = \frac{3}{5}$.
Distance to Closest Approach
The particle reaches the closest point after $\Delta t = 6$ s.
Since velocity is constant, the distance traveled along the path is:
$$ x = v \cdot \Delta t = 5 \times 6 = 30 \text{ m} $$
Calculate Minimum Distance
Using the ratio $\frac{x}{r} = \frac{3}{5}$:
$$ r = x \times \frac{5}{3} = 30 \times \frac{5}{3} = 50 \text{ m} $$
The closest distance $d_{\min}$ is the perpendicular distance from P to the line of motion.
Using Pythagoras theorem in the triangle formed by $r$, $x$, and $d_{\min}$:
$$ d_{\min} = \sqrt{r^2 – x^2} $$
$$ d_{\min} = \sqrt{50^2 – 30^2} = \sqrt{2500 – 900} = \sqrt{1600} = 40 \text{ m} $$
Answer: 40 m