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

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Answer 1 · 2016-04-23T09:43:51

#vec a=-0,85 hati+4 hatj#
#theta=-78^o#

#vec a(t)=vec a hat i+ vec a hat j#

#vec a(t)=d/(d t)(sqrt(t^2-1)-2t)hat i+d/(d t)(t^2)hat j#

#vec a(t)=((2t)/(2*sqrt(t^2-1))-2)hat i+(2t)hat j#

#vec a(2)=((2*2)/(2*sqrt(2^2-1))-2)hat i+(2*2)hat j#

#vec a(2)=(2/sqrt3-2)hat i+4hatj#

#vec a(2)=((2-2sqrt3)/sqrt3)hat i+4 hatj#

#vec a=((2sqrt3-6)/3)hati+4 hatj#

#vec a=-0,85 hati+4 hatj#

#||vec a ||=sqrt((-0,85)^2+4^2)" magnitude of a"#

# a=4,09 " "("unit")/s^2#

#tan theta=4/(-0,85)#

#theta=-78^o#