Question

The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are the magnitudes of the velocity and acceleration of the collar at t = 4 s.

Answer 1

Answer with explanation:

Part a)

`v_(radial)=(dr)/(dt)=(d(2+0.7t^(2)))/(dt)\\\\v_(radial)=1.4t\\\\\therefore v_(radial)|_(t=4)=1.4* 4=5.6ft/s\\\\v_(angular)=r|_(t=4)* (d\theta )/(dt)=13.2(3.5t)/(dt)=46.2fts^(-1)\\\\\therefore v=\sqrt{v_(radial)^(2)+v_(angular)^(2)}\\\\v=46.53ft/s`

Part b)

`a_(radial)=(d^(2)r)/(dt^(2))=(d^(2)(2+0.7t^(2)))/(dt^(2))\\\\a_(radial)=1.4ft/s^(2)\\\\a_(angular)=r* (d^(2)\theta )/(dt^(2))\\\\a_(angular)=r* (d^(2)(3.5t) )/(dt^(2))\\\\\therefore a_(angular)=0\\\\\therefore Accleration=1.4ft/s^(2)`