A ship $$A$$ is moving westward with a speed of $$10$$ kmph and a ship $$B$$, $$100 ~\text{km}$$ South of $$A$$, is moving northward with a speed of $$10$$ $$\text{kmph}$$. The time after which the distance between them becomes the shortest is:
1. $$0$$ h
2. $$5$$ h
3. $$5\sqrt{2}$$ h
4. $$10\sqrt{2}$$ h

Subtopic: Â Relative Motion |
Â 50%
From NCERT
NEET - 2015
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Two particles $$\mathrm{A}$$ and $$\mathrm{B}$$, move with constant velocities $$\overrightarrow{{v}_1}$$ and $$\overrightarrow{{v}_2}$$ respectively. At the initial moment, their position vectors are $$\overrightarrow{{r}_1}$$ and $$\overrightarrow{{r}_2}$$ respectively. The condition for particles $$\mathrm{A}$$ and $$\mathrm{B}$$ for their collision will be:
1.$$\dfrac{\vec{r_1}-\vec{r_2}}{\left|\vec{r_1}-\vec{r_2}\right|}=\dfrac{\vec{v_2}-\vec{v_1}}{\left|\vec{v_2}-\vec{v_1}\right|}$$
2. $$\vec{r_1} \cdot \vec{v_1}=\vec{r_2} \cdot \vec{v_2}$$   ${\mathrm{}}_{}$
3. $$\vec{r_1} \times \vec{v_1}=\vec{r_2} \times \vec{v_2}$$
4. $$\vec{r_1}-\vec{r_2}=\vec{v_1}-\vec{v_2}$$

Subtopic: Â Relative Motion |
Â 70%
From NCERT
NEET - 2015
To view explanation, please take trial in the course.
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