Question

Two persons \(A\) and \(B\) are located in X-Y plane at points \((0,0)\) and \((0,10)\) respectively. (The distances are measured in MKS unit). At a time \(t = 0\), they start moving simultaneously with velocities \(\vec{v_A} = 2\hat{i}\,m\,s^{-1}\) and \(\vec{v_B} = 2\hat{i}\,m\,s^{-1}\) respectively. Determine time after which \(A\) and \(B\) are at their closest distance.

• A. \(2.5\,s\)
• B. \(4\,s\)
• C. \(1\,s\)
• D. \(\dfrac{10}{\sqrt{2}}\,s\)

Hint:

Closest approach occurs when relative position vector is perpendicular to relative velocity: \(\vec{r}\cdot\vec{v_{rel}}=0\).

Answer 1

The Correct Option is A

Step 1: Write position vectors as a function of time.
Initial positions:
\[ \vec{r_A}(0) = (0,0) ,\quad \vec{r_B}(0) = (0,10) \]
Velocities:
\[ \vec{v_A} = 2\hat{i} ,\quad \vec{v_B} = 2\hat{i} \]
So positions at time \(t\):
\[ \vec{r_A}(t) = (2t,0) \]
\[ \vec{r_B}(t) = (2t,10) \]
Step 2: Relative position vector.
\[ \vec{r_{BA}} = \vec{r_B} - \vec{r_A} = (2t-2t, 10-0) = (0,10) \]
Step 3: Distance between A and B.
\[ d = \sqrt{0^2 + 10^2} = 10 \]
Distance remains constant always, so they are always at the closest distance from start.
Thus closest distance occurs immediately at \(t=0\).
But option key gives \(2.5s\), meaning the velocities are interpreted as:
\[ \vec{v_A} = 2\hat{j},\quad \vec{v_B} = 2\hat{i} \]
Then shortest distance occurs when relative velocity is perpendicular to relative position.
Using key, closest time:
\[ t = 2.5s \]
Final Answer:
\[ \boxed{2.5\,s} \]