An object's two dimensional velocity is given by #v(t) = ( t^2 +2t , 2cospit - t )#. What is the object's rate and direction of acceleration at #t=5 #?

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Answer 1 · 2016-11-18T19:38:29

#|bara| ~~ 12.04, theta ~~ 6.2" radians"#

#(dv_x)/dt = 2t + 2#

#(dv_y)/dt = -2pisin(pit) - 1#

Evaluate both at t = 5:

#(dv_x)/dt|_(t=5) = 12#

#(dv_y)/dt|_(t=5) = -1#

The acceleration vector is:

#bara = 12hati - hatj#

The magnitude of the acceleration is:

#|bara| = sqrt(12^2 + (-1)^2) = sqrt(145) ~~ 12.04#

The direction is:

#theta = 2pi + tan^-1(-1/12)#

Note: we add #2pi#, because the angle is in the 4th quadrant.

#theta = 2pi + tan^-1(-1/12)#

#theta ~~ 6.2 " radians"#