Ship $A$ is located $4 km$ north and $3 km$ east of ship $B$ . Ship $A$ has a velocity of $20 {kmh}^{- 1}$ towards the south and ship $B$ is moving at $40 {kmh}^{- 1}$ in a direction $37 °$ north of east. $X$ and $Y$ -axes are along east and north directions, respectively

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Velocity of $A$ relative to $B$ is $(- 32 \hat{i} - 44 \hat{j}) km / h$

Position of $A$ relative to $B$ as a function of time is given by $(32 t \hat{i} - 44 \hat{j}) km$

Position of $A$ relative to $B$ as a function of time is given by $r_{A B} = [(3 - 32 t) \hat{i} + (4 - 44 t) \hat{j}] km$

Velocity of $A$ relative to $B$ is $(32 \hat{i} - 44 \hat{j}) km / h$

answer is A, B.

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Detailed Solution

$r_{B} = 0$ at $t = 0$

$r_{A} = 3 \hat{i} + 4 \hat{j}$ \\
At $t = 0$
$v_{A} = (- 20 \hat{j})$

$v_{B} = (40 \cos 37 °) \hat{i} + (40 \sin 37 °) \hat{j}$
$= (32 \hat{i} + 24 \hat{j})$
$v_{A B} = (- 32 \hat{i} - 44 \hat{j}) km / h$

At time $t = 0$

$r_{A B} = r_{A} - r_{B} = (3 \hat{i} + 4 \hat{j})$

At time $t = t$ ,
$r_{A B} = (r_{A B}$ at $t = 0) + v_{A B} t$
$= (3 - 32 t) \hat{i} + (4 - 44 t) \hat{j}$

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