A ship A is moving Westwards with a speed of 10 km h -1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h-1. The time after which the distance between them becomes shortest, is:

  • Option 1)

    5\sqrt2\: \: h

  • Option 2)

    10\sqrt2\: \: h

  • Option 3)

    0\: \: h

  • Option 4)

    5 h

 

Answers (1)

As we learnt in 

Case of Relative velocity -

When A & B are moving along with straight line in opposite direction.

vec{V}_{A}= Velocity of object A.

vec{V}_{B}= Velocity of object B.

Relative velocity of A with respect to B is.

vec{V}_{AB}=vec{V}_{A}-(-vec{V}_{B})

-vec{V}_{B}  is because of opposite direction.

vec{V}_{AB}=vec{V}_{A}+(vec{V}_{B})

 

 

- wherein

E.g A body moves in east with speed of 20m-1, And body in west with velocity of 5ms-1.

Find velocity of A with respect to B.

vec{V}_{AB}=vec{V}_{A}+(vec{V}_{B})

= (20+5 ) ms-1

= 25 ms-1

 

 \\sin 45=\frac{PQ}{OQ}\\ PQ=100\times \frac{1}{\sqrt{2}}=50\sqrt{2}m\\ V_{AB}=\sqrt{V_{A}^{2}+V_{B}^{2}}=\sqrt{10^{2}+10^{2}}=10\sqrt{2}

Time taken to reach shortest path is 

t=\frac{PQ}{V_{AB}}=\frac{50\sqrt{2}}{10\sqrt{2}}=5h

Correct option is 4.


Option 1)

5\sqrt2\: \: h

Incorrect

Option 2)

10\sqrt2\: \: h

Incorrect

Option 3)

0\: \: h

Incorrect

Option 4)

5 h

Correct

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