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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?


2 years ago

Khimraj
3007 Points
							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 = 04t + 18t3 – 12t = 02t(9t2 – 4) = 0so t = 2/3 secHope it clears. If you like answer then please approve my answer.

2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions