An object's two dimensional velocity is given by #v(t) = ( cost , -t^3 +4t )#. What is the object's rate and direction of acceleration at #t=3 #?

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Answer 1 · 2017-11-21T06:51:28

The rate of acceleration is #==23ms^-2# in the direction #=269.7^@#anticlockwise from the x-axis.

The acceleration is the derivative of the velocity

#v(t)=(cost,-t^3+4t)#

#a(t)=v'(t)=(-sint,-3t^2+4)#

When #t=3#

#a(3)=(-sin(3), -3^3+4)=(-0.14,-23)#

The rate of acceleration is

#=||a(3)||=sqrt((-0.14)^2+(-23)^2)=23ms^-2#

The direction is

#theta=arctan(-23/-0.14)=269.7^@# anticlockwise from the x-axis.

Answer 2 · 2017-11-21T06:57:13

#23# units and #4.71# radian

As two dimensional velocity of object is given by #v(t)=(cost,-t^3+4t)#

accelaration is given by #(dv)/(dt)# i.e.

#a(t)=(-sint,-3t^2+4)#

and at #t=3#, this is #-sin3hati+(-3*3^2+4)hatj#

or #a(t=3)=-0.14112hati-23hatj#

and its magnitude is #sqrt((0.14112)^2+23^2)=sqrt529.02=23#

and direction is #pi+tan^(-1)23/0.14112=pi+1.569=4.71# radian