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A ship 'A' is moving westwards with a speed of 10 km/h and a ship B 100 km south of 'A' is moving northwards with a speed of 10 km/h. The time after which the distance between them becomes shortest is?

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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

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manish

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