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Ship A is moving west at a speed of  10\ km\ h^{-1}, and ship B is moving north at a speed of 100\ km south of A. The time at which the distance between them becomes shortest is:

Option: 1

0\ h

Option: 2

5\ h

Option: 3

5\sqrt2\ h\

Option: 4

10\sqrt2\ h\

Answers (1)


Let the shortest distance between Ship A and B is PQ.

Therefore, sin\ 45^{\circ}=\frac{PQ}{OQ }                 

Or,     PQ=100\times\frac{1}{\sqrt2}=50\sqrt2\ m

Also |v_{AB}|=|v_A-v_B|=\left(\sqrt{{v^2}_A+{v_B}^2}\right)=\sqrt{{10}^2+{10}^2}=10\sqrt2\frac{km}{h}

Thus, time is taken to reach the shortest path:


Or, t=\frac{50\sqrt2}{10\sqrt2}=5\ h               

Posted by

Ajit Kumar Dubey

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