a ship streams due north at 16km/hr and a ship B due west at 12 km/hr.At a certain instant B is 10 km north east of A .Find the velocity of A relative to B.Find also the nearest distance of approach of ships

Dear Student,

Please find below the solution to the asked query:

Given that the ship A is moving towards north 16 kmph. Ship B moves towars west with 12 kmph​. Let after t hr time the ship A appeard 10 km towards north - east. Then,

The distance travelled by car A is,

$x_A=16 t km$

The distance travelled by car B is,

$x_B=12t km$

Therefore,

$$\begin{gathered} 10=\sqrt{{\left(16t\right)}^2+{\left(12t\right)}^2}\Rightarrow 256t^2+144t^2=100 \\ \Rightarrow 400t^2=100\Rightarrow t^2=\frac{1}{4} \\ \Rightarrow t=0.5 hr \end{gathered}$$

The relative velocity is along the line joining the two ships.

Let the line joining the two ships makes an angle θ with the horizontal at that moment, then,

$$\begin{gathered} V_{a,b}=v_a-v_b=16 \sin \theta - 12 cos \theta =16\times \frac{8}{10}-12\times \frac{6}{10} \\ \Rightarrow V_{a,b}=\frac{128-72}{10}=\frac{56}{10} \\ \Rightarrow V_{a,b}=5.6 kmph \end{gathered}$$

The distance between the ships at any time t is,

It is not mentioned from where the two ships are starting. Because, if the two ships are starting simultaneously from origin, then the nearest distance is Zero.

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