Question

A man holding a flag is running in north-eas
direction with speed 10 m/s. Wind is blowing
east direction with speed 5/2 m/s. Find th
direction in which flag will flutter.
(1) South
(2) North
(3) East
(4) West
A ship A is moving Westwards with a speed c
10 kmh- and a ship B 100 km South of A.
moving Northwards with a speed of 10 kmh- the
shortest distance between them is​

Answer 1

Answer:

the answer is may be south

Explanation:

because logically, the velocity of man is in direction of North-east so the direction of flag will be in south-southwest. The direction of wind is in East so the resultant will be towards the velocity of man because his velocity is maximum then wind. So the answer is south.

Answer 2

Answer:

(I). The direction of flag is south.

(a) is correct option.

(II). The shortest distance between them is​ 100 km

Explanation:

Given that,

Speed of man = 10 m/s

Speed of wind $v_{w}= 5\sqrt{2}\ m/s$

We need to calculate the velocity of man

$v_{m}=10\cos\theta i+10\sin\theta j$

$v_{m}=5\sqrt{2} i+5\sqrt{2} j$

$v_{w}=5\sqrt{2}i+0$

We need to calculate the resultant

$v_{mw}=v_{w}-v_{m}$

$v_{mw}= 5\sqrt{2}i-5\sqrt{2} i-5\sqrt{2} j$

$v_{mw}=-5\sqrt{2} j$

Here, -j shows the direction of south.

So, The direction of the flag is south.

(II). The shortest distance between ship A and ship B when the ship B will be at origin.

We need to calculate the time of Ship B

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100}{10}$

$t=10\ hr$

We need to calculate the shortest distance

$d =v\times t$

Put the value into the formula

$d=10\times10$

$d =100\ km$

Hence, (I). The direction of flag is south.

(II). The shortest distance between them is​ 100 km