A particle moves along a path x=y^2+2y+2 in a plane such that y-component of its velocity remains constant at 2 m/s.the magnitude of its acceleration is
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Given:
A particle moves along a path x=y^2+2y+2 in a plane such that y-component of its velocity remains constant at 2 m/s.
To find:
The magnitude of Acceleration of particle ?
Calculation:
Acceleration can be calculated by the 2nd order
derivative of displacement function; since the y component of velocity is constant , the particle will have acceleration only along x axis :
Putting dy/dt = 2 (as given in question)
So, acceleration of particle is 8 m/s².
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