An object's two dimensional velocity is given by #v(t) = ( 3t^2 - 5t , -t^3 +4t )#. What is the object's rate and direction of acceleration at #t=1 #?

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Answer 1 · 2016-05-04T17:59:44

#a(1)=sqrt2#
#alpha=45^o#

#a_x(t)=d/(d t) v_x(t)" ; "a_y(t)=d/(d t) v_y(t)#

#a_x(t)=d/(d t)(3t^2-5t)=6t-5#

#a_x(1)=6*1-5=1#

#a_y(t)=d/(d t) (-t^³+4t)=-3t^2+4#

#a_y(1)=-3*1^2+4#

#a_y(1)=-3+4#

#a_y(1)=1#

#a(1)=sqrt(a_x(1)^2+a_y(1)^2)#

#a(1)=sqrt(1^2+1^2)#

#a(1)=sqrt2#

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#tan alpha=(a_y(1))/(a_x(1))=1/1=1" ; "alpha=45^o#