An object's two dimensional velocity is given by #v(t) = ( sqrt(t-2)-t , t^2)#. What is the object's rate and direction of acceleration at #t=7 #?

1 Answer
Apr 27, 2016

Answer:

#a(7)_x=-0,76#
#a(7)_y=2*7=14#
#a(7)=14,02#
#alpha=86,89^o#

Explanation:

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#a(t)_x=d/(d t) (sqrt(t-2)-t)=1/(2*sqrt(t-2))-1#

#a(7)_x=1/(2*sqrt(7-2))-1#

#a(7)_x=1/(2*sqrt5)-1#

#a(7)_x=(1-2*sqrt5)/(2*sqrt5)#

#a(7)_x=((1-2*sqrt5)*(2*sqrt5))/((2*sqrt5)*(2*sqrt5))#

#a(7)_x=(2*sqrt5-20)/20#

#a(7)_x=(sqrt5-10)/10=(2,37-10)/10=-0,76#

#a(t)_y=d/(d t) (t^2)=2*t#

#a(7)_y=2*7=14#

#a(7)=sqrt((a(x)_7)^2+(a(7)_y)^2)#

#a(7)=sqrt(((-0,76)^2+14^2)#

#a(7)=sqrt(0,58+196)#

#a(7)=14,02#

#tan alpha=(a(7)_y)/(a(7)_x)#

#tan alpha=14/(-0,76)#

#tan alpha=-18,42#

#alpha=86,89^o#