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

Question

1156 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-19T06:57:46

The rate of acceleration is #=4.26ms^-2# in the direction of #=76.4º# clockwise from the x-axis

The acceleration is the derivative of the velocity.

#v(t)=(t-2, 2cos(pi/2t)-t)#

#a(t)=v'(t)=(1, -pisin(pi/2t)-1)#

Therefore,

#a(5)=(1, -pi-1)=(1,-4.14)#

The rate of acceleration is

#||a(4)||=sqrt(1+4.14^2)#

#=sqrt18.15#

#=4.26ms^-2#

The direction is

#theta=arctan(-(pi+1))#

#theta# lies in the 4th quadrant

#theta=-76.4º#