An object's two dimensional velocity is given by #v(t) = ( t^3, t-t^2sin(pi/8)t)#. What is the object's rate and direction of acceleration at #t=3 #?

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Answer 1 · 2017-04-01T15:53:05

The rate of acceleration is #=27.63ms^-1# in the direction of #=167.69#º

The velocity is

#v(t)=(t^3,t-t^2sin((pi/8)t))#

Acceleration is the first derivative of the velocity

#a(t)=(dv(t))/dt=(3t^2,1-2tsin(pi/8)t-t^2(pi/8)cos(pi/8)t)#

When #t=3#

#a(3)=(dv(3))/dt=(27,1-6sin(3/8pi)-(9/8pi)cos(3/8pi))#

#=(27,1-5.54-1.35)#

#=(27,-5.89)#

The object rate of acceleration is

#=sqrt((27^2)+(-5.89)^2)#

#=27.63ms^-2#

The direction is

#=arctan(-5.89/27)#

The angle is in the 2nd quadrant and is

#=167.69#º