At a harbour, a boat is standing and wind is blowing at a speed of $\sqrt{2}$ m/sec. due to which, the flag on the boat flutters along northeast. Now the boat enters in to river, which is flowing with a velocity of 2 m/sec. due north. The boat starts with zero velocity relative to the river and its constant acceleration relative to the river is $0.2 m / s e c^{2}$ due east. In which direction will the flag flutter at 10 seconds?

south-east

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south-west

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30^{0}

south of west

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west

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Solution

The correct option is B south-west
$V_{w} = 1^i + 1^j$
$V = a t$
$V = (0.2) 10 = 2 m / s e c .$
$V_{b o a t} = 2^i + 2^j$
$V_{w / b o a t} = V_{w} - V_{b o a t}$
$V_{w / b o a t} = (1^i + 1^j) - (2^i + 2^j) = - 1^i - 1^j$
So, the flag will flutter towards south-west.

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The correct option is B south-westVw=1^i+1^jV=atV=(0.2)10=2m/sec.Vboat=2^i+2^jVw/boat=Vw−VboatVw/boat=(1^i+1^j)−(2^i+2^j)=−1^i−1^jSo, the flag will flutter towards south-west.

The correct option is B south-west
$V_{w} = 1^i + 1^j$
$V = a t$
$V = (0.2) 10 = 2 m / s e c .$
$V_{b o a t} = 2^i + 2^j$
$V_{w / b o a t} = V_{w} - V_{b o a t}$
$V_{w / b o a t} = (1^i + 1^j) - (2^i + 2^j) = - 1^i - 1^j$
So, the flag will flutter towards south-west.