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

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Answer 1 · 2017-05-06T20:15:14

#|veca| ~~ 4.09"ms"^-2 #
#theta ~~ 1.779"rad or "101.9^@#

Here we have velocity along #x# direction:

#v_x(t)=sqrt (t^2-1)-2t#

#:.# acceleration along #x# direction:

#a_x(t)=(dv_x)/(dt)=t/(sqrt (t^2-1))-2#

At #t=2s#:

#a_x(2"s")=(2/ sqrt3-2)"ms"^-2#

Similarly

#a_y(t)=2t#

At #t=2s#:

#a_y(2"s")=4"ms"^-2#

The acceleration vector at #t = 2#, is #veca = 2(sqrt3/3-1)hati+ 4hatj#

The acceleration is the magnitude of this vector:

#|veca| = sqrt((2(sqrt3/3-1))^2+4^2)#

#|veca| ~~ 4.09"ms"^-2 #

The direction is:

#theta = tan^-1(4/(2(sqrt3/3-1)))+ pi#

Please notice that we have added #pi# because the vector points to the 2nd quadrant.

#theta ~~ 1.779"rad or "101.9^@#