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Solve this problem - Kinematics - JEE Main

Ship A is sailing towards north-east with velocity \vec{v}= 30 \widehat{i}+50\widehat{j} km/hr where \widehat{i} points east and \widehat{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 :

  • Option 1)

    2.6 hrs.

  • Option 2)

    2.2 hrs.

  • Option 3)

    4.2 hrs.

  • Option 4)

    3.2 hrs.

Answers (1)
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v_{A}=30\widehat{i}+50\widehat{j}

r_{A}=0\widehat{i}+0\widehat{j}

r_{B}=80\widehat{i}+150\widehat{j}

v_{B}=-10\widehat{i}

\vec{v_{r}}=40\widehat{i}+50\widehat{j}

\vec{r_{r}}=-80\widehat{i} - 150\widehat{j}

t_{min}=\frac{\left | \overrightarrow{v_{r}}.\overrightarrow{r_{r}} \right |}{\left | \overrightarrow{v_{r}} \right |}=\frac{(40\times 80+150 \times50)}{4100}

t_{min}= \frac{10700}{4100}=2.6\; hrs.

 


Option 1)

2.6 hrs.

Option 2)

2.2 hrs.

Option 3)

4.2 hrs.

Option 4)

3.2 hrs.

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