Question

Ship $A$ is sailing towards north-east with velocity $\vec{v} = 30 \hat{i} + 50 \hat{j} $ km/hr where $\hat{i}$ points east and $\hat{j}$, north. Ship $B$ is at a distance of $80\, km$ east and $150\, km$ north of Ship $A$ and is sailing towards west at $10\, km/hr$. A will be at minimum distance from $B$ in :

• A. 4.2 hrs
• B. 2.2 hrs
• C. 3.2 hrs
• D. 2.6 hrs

Answer 1

Answer: (D)

If we take the position of ship 'A' as origin then positions and velocities of both ships can be given as : $\vec{V}_{A} = \left(30\hat{i} +50 \hat{j}\right)km /hr $
$ \vec{v}_{B} = - 10\hat{i} km /hr $
$\vec{r}_{A} = 0\hat{i} + 0 \hat{j} $
$ \vec{r}_{B} = \left(80\hat{i} + 150\hat{j}\right) km$
Time after which distance between them will be minimum
$ t = - \frac{\vec{r}_{BA} . \vec{v}_{BA}}{\left|\vec{v}_{BA}\right|^{2}} $
where $ \vec{r}_{BA} = \left(80 \hat{i} +150 \hat{j}\right) km $
$ \vec{v}_{BA} =- 10 \hat{i} - \left(30\hat{i} +50\hat{j}\right) $
$\left( - 40 \hat{i} - 50 \hat{j}\right) km/hr $
$\therefore t = - \frac{\left(80 \hat{i} + 150 \hat{j}\right) . \left(-40\hat{i} - 50 \hat{j}\right)}{\left|- 40\hat{i} - 50\hat{j}\right|^{2}} $
$= \frac{3200+7500}{4100} hr = \frac{10700}{4100} hr = 2.6 hrs $