A particle moves in the xy plane and at time T at the point whose coordinates are (t^2,t^3-2t). Then at what instant of time will its velocity and acceleration vectors will be perpendicular to each other?

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Khimraj · Answer

x = t 2 + (t 3 -2t) v = dx/dt = 2t + (3t 2 -2) a = dv/dt = 2 + 6t if v and a are perpendicular then v.a = 0 4t + 18t 3 – 12t = 0 2t(9t 2 – 4) = 0 so t = 2/3 sec Hope it clears. If you like answer then please approve my answer.

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A particle moves in the xy plane and at time T at the point whose coordinates are (t^2,t^3-2t). Then at what instant of time will its velocity and acceleration vectors will be perpendicular to each other?

A particle moves in the xy plane and at time T at the point whose coordinates are (t^2,t^3-2t). Then at what instant of time will its velocity and acceleration vectors will be perpendicular to each other?

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A particle moves in the xy plane and at time T at the point whose coordinates are (t^2,t^3-2t). Then at what instant of time will its velocity and acceleration vectors will be perpendicular to each other?

A particle moves in the xy plane and at time T at the point whose coordinates are (t^2,t^3-2t). Then at what instant of time will its velocity and acceleration vectors will be perpendicular to each other?

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