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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?

harsh jain , 6 Years ago
Grade 11
anser 1 Answers
Khimraj

Last Activity: 6 Years ago

x = t2\hat{i} + (t3 -2t)\hat{j}
v = dx/dt = 2t\hat{i} + (3t2 -2)\hat{j}
a = dv/dt = 2\hat{i} + 6t\hat{j}
if v and a are perpendicular then v.a = 0
4t + 18t3 – 12t = 0
2t(9t2 – 4) = 0
so t = 2/3 sec
Hope it clears. If you like answer then please approve my answer.

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