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A particle move along the parabolic path x = y²+2y+2 in such a way that the y-component of velocity vector remain 5m/s during the motion. The magnitude of the acceleration of the particle is
x=y2+2y+2Differentiate with respect to time t to obtain velocity components along x and y direction respectively.vx=2yvy+2vywhere vx=dx/dt and vy=dy/dt.Given vy=5m/s.So vx=10y+10Differentiate again with respect to time t to obtain acceleration.ax=10vy=50m/s2Please approve the answer if helped.
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