Question

$A$ ship is moving in the westward direction with a speed of $10kmph$ and a ship $B100km$ south of $A$, is moving northward with a speed oh $10kmph$. The time after which the distance between them becomes shortest, is

• A. $15h$
• B. $5\surd 2h$
• C. $10\surd 2h$
• D. $5h$

Answer 1

The correct option is D $5h$

Given that,

Speed of ship A $v_A=10km/h$

Speed of ship B$v_B=10km/h$

Distance $d=10km$

Now, according to diagram

the shortest distance between the ship $A$ and $B$ is $PQ$

$\sin {45}^0=\frac{PQ}{OQ}$

$PQ=100\times \frac{1}{\sqrt{2}}$

$PQ=50\sqrt{2}km$

Now, the velocity is

$v_{AB}=\sqrt{v_A^2+v_B^2}$

$v_{AB}=\sqrt{{\left(10\right)}^2+{\left(10\right)}^2}$

$v_{AB}=10\sqrt{2}km/h$

Now, the time taken is

$t=\frac{50\sqrt{2}}{10\sqrt{2}}$

$t=5h$

Hence, the time after to reach shortest path is $5h$