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`        A particle moves to two dimensional orbit defined by x(t) =A(2at-sinat) y(t)=A(1-cosat) A) Find the tangential acceleration a1 and normal acceleration an as function of time where the tangential and normal components are taken with respect to the velocity. B) Determine at what times in the orbit an has a maximum.`
one year ago

Vikas TU
11137 Points
```							(1)  Tangential acceleration a1 = dv/dt = d^2x(t)/dt^2 dx/dt = A(2a –acos(at))dv/dt = A(a^2sin(at))a1 = A(a^2sin(at)) Normal acceleration would be: d^2y(t)/dt^2 dy/dt = Aasin(at)d^2y/dt^2 = Aa^2cos(at) an = Aa^2cos(at)  (2) For an to be maximum, d(an)/dt = 0Aa^3sin(at) = 0at = npit = n*pi/a it will be maximum.
```
one year 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