Motion In A Plane Ques 40
40. A ship A is moving Westwards with a speed of $10 km h^{-1}$ and a ship B $100 km$ South of A, is moving Northwards with a speed of $10 km h^{-1}$. The time after which the distance between them becomes shortest, is :
[2015]
(a) $5 h$
(b) $5 \sqrt{2} h$
(c) $10 \sqrt{2} h$
(d) $0 h$
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Answer:
Correct Answer: 40.(a)
Solution:
- (a) Let $t$ be the time passed and $x$ be the distance covered by the two ships after which the distance between them becomes shortest.
distance(s) between the ships is given by,
$ S=\sqrt{(100-x)^{2}+x^{2}} $
For $S$ to be minimum, $\frac{d s}{d t}=0$
$ \begin{alignedat} & \Rightarrow \quad \frac{d s}{d t}=\frac{1}{9[(100-x)^{2}+x^{2}]^{-1 / 2}} \\ & \Rightarrow \quad 4-2(100-x)+2x=0 \\ & \quad 4 x-200=0 \Rightarrow x=50 \text{ m} \end{aligned} $
So, after both the ships have covered $50 m$, distance between them becomes shortest. Time taken for it, will be
$ t=\frac{x}{10}=\frac{50}{10}=5 hr . $
BETA
