JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 7)
A butterfly is flying with a velocity $$4\sqrt 2 $$ m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is :
$$12\sqrt 2 $$ m
20 m
3 m
15 m
Explanation
The given situation can be represented as
_20th_July_Morning_Shift_en_7_1.png)
In the above figure, v1 is the speed of wind and v21 is the speed of butterfly with respect to wind.
So, v21 can be given as
$${v_{21}} = 4\sqrt 2 \cos 45^\circ \widehat i + 4\sqrt 2 \sin 45^\circ \widehat j$$
$$ = 4\sqrt 2 \times {1 \over {\sqrt 2 }}\widehat i + 4\sqrt 2 \times {1 \over {\sqrt 2 }}\widehat j = 4\widehat i + 4\widehat j$$
and v1 can be given as
$${v_1} = - \widehat j$$
$$\therefore$$ Velocity of butterfly can be given as
$${v_2} = {v_1} + {v_{21}} = 4\widehat i + 4\widehat j - \widehat j = 4\widehat i + 3\widehat j$$
$$\therefore$$ Displacement of butterfly, $$D = {v_2} \times t$$
$$ = (4\widehat i + 3\widehat j) \times 3 = 12\widehat i + 9\widehat j$$
$$\therefore$$ Magnitude of displacement, $$\left| D \right| = \sqrt {{{12}^2} + {9^2}} = 15$$ m
In the above figure, v1 is the speed of wind and v21 is the speed of butterfly with respect to wind.
So, v21 can be given as
$${v_{21}} = 4\sqrt 2 \cos 45^\circ \widehat i + 4\sqrt 2 \sin 45^\circ \widehat j$$
$$ = 4\sqrt 2 \times {1 \over {\sqrt 2 }}\widehat i + 4\sqrt 2 \times {1 \over {\sqrt 2 }}\widehat j = 4\widehat i + 4\widehat j$$
and v1 can be given as
$${v_1} = - \widehat j$$
$$\therefore$$ Velocity of butterfly can be given as
$${v_2} = {v_1} + {v_{21}} = 4\widehat i + 4\widehat j - \widehat j = 4\widehat i + 3\widehat j$$
$$\therefore$$ Displacement of butterfly, $$D = {v_2} \times t$$
$$ = (4\widehat i + 3\widehat j) \times 3 = 12\widehat i + 9\widehat j$$
$$\therefore$$ Magnitude of displacement, $$\left| D \right| = \sqrt {{{12}^2} + {9^2}} = 15$$ m
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