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1st Equation of Motion

Net force act...

Question

Net force acting on a particle of mass 2 kg is 10 N in north direction. At t = 0, particle was moving eastwards with 10 m/s. Find displacement and velocity of particle after 2 s.

10 \sqrt{5}

m at

{cot}^{- 1} (2)

from east towards north,

10 \sqrt{2} m s^{- 1} a t 45^{\circ}

from east towards north.

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10 \sqrt{2}

m at

{cot}^{- 1} (2)

from east towards north,

10 \sqrt{2} m s^{- 1} a t 45^{\circ}

from east towards north.

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10 \sqrt{5}

m at

{cot}^{- 1} (2)

from east towards north,

10 \sqrt{5} m s^{- 1} a t 45^{\circ}

from east towards north.

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10 \sqrt{2}

m at

{cot}^{- 1} (2)

from east towards north,

10 \sqrt{5} m s^{- 1} a t 45^{\circ}

from east towards north.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $10 \sqrt{5}$ m at ${cot}^{- 1} (2)$ from east towards north, $10 \sqrt{2} m s^{- 1} a t 45^{\circ}$ from east towards north.
$a = 5 m / s^{2} a l o n g n o r t h, t h u s, v e l o c i t y a l o n g e a s t w o n^{'} t c h a n g e t = 2 s u = 10 m / s a l o n g e a s t d i s p l a c e m e n t a l o n g n o r t h = \frac{1}{2} a t^{2} = 10 m d i s p l a c e m e n t a l o n g e a s t = u t = 20 m t o t a l d i s p l a c e m e n t = \sqrt{10^{2} + 20^{2}} = 10 \sqrt{5} m v e l o c i t y a l o n g n o r t h = a t = 10 m / s f i n a l v e l o c i t y = 10^i + 10^j$

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