Ship A is sailing towards north-east with vel | vedclass

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(C) Let the position of ship $A$ be at the origin $(0,0)$.The initial position vectors are $\vec{r}_A = 0\hat{i} + 0\hat{j}$ and $\vec{r}_B = 80\hat{i

Vedclass · Accepted Answer

$2.6$

Vedclass · Answer

(C) Let the position of ship $A$ be at the origin $(0,0)$.The initial position vectors are $\vec{r}_A = 0\hat{i} + 0\hat{j}$ and $\vec{r}_B = 80\hat{i} + 150\hat{j}$.The velocity vectors are $\vec{v}_A = 30\hat{i} + 50\hat{j}$ and $\vec{v}_B = -10\hat{i}$.The relative position vector is $\vec{r}_{BA} = \vec{r}_B - \vec{r}_A = 80\hat{i} + 150\hat{j}$.The relative velocity vector is $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = (-10\hat{i}) - (30\hat{i} + 50\hat{j}) = -40\hat{i} - 50\hat{j}$.The time $t$ at which the distance is minimum is given by $t = -\frac{\vec{r}_{BA} \cdot \vec{v}_{BA}}{|\vec{v}_{BA}|^2}$.Substituting the values:$t = -\frac{(80\hat{i} + 150\hat{j}) \cdot (-40\hat{i} - 50\hat{j})}{(-40)^2 + (-50)^2}$$t = -\frac{(80 	imes -40) + (150 	imes -50)}{1600 + 2500}$$t = -\frac{-3200 - 7500}{4100} = \frac{10700}{4100} = \frac{107}{41} \approx 2.61\,	ext{hrs}$.Rounding to the nearest given option,the time is $2.6\,	ext{hrs}$.

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