Question

Ship A is moving towards south with a speed of 10km/h.Another ship B is moving towards east with a speed of 10km/h.At a certain instant the ship B is due south of ship A.Find the shortest distance between the ships

Answer 1

Answer:

As we learnt in

Case of Relative velocity -

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

Velocity of object A.

Velocity of object B.

Relative velocity of A with respect to B is.

is because of opposite direction.

- 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.

= (20+5 ) ms-1

= 25 ms-1

Time taken to reach shortest path is

Correct option is 4.

Option 1)

Incorrect

Option 2)

Incorrect

Option 3)

Incorrect

Option 4)

5 h

Correct

Answer 2

The time after which the distance between the ships becomes shortest is 5hrs

Explanation:

• Given -Ship A is moving towards south with a speed of 10km/h
• Another ship B is moving towards east with a speed of 10km/h

Condition is At a certain instant the ship B is due south of ship A.

It is clear from the diagram that the shortest distance between the ship A and B is PQ.

Angle formed will be 45 degree when the ships moves

Here,  $sin45^{0}$=$\frac{PQ}{OQ}$

$PQ=100$×$\frac{1}{\sqrt{2} }$

$PQ =50\sqrt{2}$

=$\sqrt{Va^{2} +Vb^{2} }$

$=\sqrt{10^{2} +10^{2} }$

= $10\sqrt{2$

So, the time taken to reach shortest path is

$t=\frac{PQ}{Vab}$

$t=\frac{50\sqrt{2} }{10\sqrt{2} }$

$t=5hrs$

The time after which the distance between the ships becomes shortest is 5hrs

#SPJ3