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

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Susmita · Answer

x=y 2 +2y+2 Differentiate with respect to time t to obtain velocity components along x and y direction respectively. v x =2yv y +2v y where v x =dx/dt and v y =dy/dt. Given v y =5m/s. So v x =10y+10 Differentiate again with respect to time t to obtain acceleration. a x =10v y =50m/s 2 Please approve the answer if helped.

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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

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

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