a particle starts to move from rest along X axis with an acceleration of 2m/s 2 after 4 sec its acceleration is also towards -ve y axis of 2m/s 2 what will be its quadrant after 8 sec's?

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Vikas TU · Answer

at 4 seconds, v = 0 + 2*4icap => 8 icap after then, v2 = 8 icap – 2jcap *8 => 8icap -16jcap angle = arctan(-16/8) = arctan(-2) = – 64.3 degree. that is in fourth quadrant.

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a particle starts to move from rest along X axis with an acceleration of 2m/s 2 after 4 sec its acceleration is also towards -ve y axis of 2m/s 2 what will be its quadrant after 8 sec's?

a particle starts to move from rest along X axis with an acceleration of 2m/s² after 4 sec its acceleration is also towards -ve y axis of 2m/s² what will be its quadrant after 8 sec's?

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a particle starts to move from rest along X axis with an acceleration of 2m/s 2 after 4 sec its acceleration is also towards -ve y axis of 2m/s 2 what will be its quadrant after 8 sec's?

a particle starts to move from rest along X axis with an acceleration of 2m/s2 after 4 sec its acceleration is also towards -ve y axis of 2m/s2 what will be its quadrant after 8 sec's?

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a particle starts to move from rest along X axis with an acceleration of 2m/s² after 4 sec its acceleration is also towards -ve y axis of 2m/s² what will be its quadrant after 8 sec's?