# What is the instantaneous velocity of an object moving in accordance to # f(t)= (t^3/(t-2),1/t) # at # t=1 #?

The instantaneous velocity of the object is 1/4 (distance unit)/(time unit)

Hence, The instantaneous velocity of the object is 1/4 (distance unit)/(time unit)

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To find the instantaneous velocity of an object moving according to the function ( f(t) = \left( \frac{t^3}{t - 2}, \frac{1}{t} \right) ) at ( t = 1 ), we need to find the derivative of the position function ( f(t) ) with respect to time ( t ) and then evaluate it at ( t = 1 ).

Given ( f(t) = \left( \frac{t^3}{t - 2}, \frac{1}{t} \right) ), the derivative of ( f(t) ) with respect to ( t ) is:

[ f'(t) = \left( \frac{d}{dt} \left( \frac{t^3}{t - 2} \right), \frac{d}{dt} \left( \frac{1}{t} \right) \right) ]

[ = \left( \frac{(3t^2)(t - 2) - t^3(1)}{(t - 2)^2}, -\frac{1}{t^2} \right) ]

[ = \left( \frac{3t^3 - 6t^2 - t^3}{(t - 2)^2}, -\frac{1}{t^2} \right) ]

[ = \left( \frac{2t^3 - 6t^2}{(t - 2)^2}, -\frac{1}{t^2} \right) ]

Now, we'll evaluate ( f'(t) ) at ( t = 1 ):

[ f'(1) = \left( \frac{2(1)^3 - 6(1)^2}{(1 - 2)^2}, -\frac{1}{(1)^2} \right) ]

[ = \left( \frac{2 - 6}{1}, -1 \right) ]

[ = \left( -4, -1 \right) ]

Therefore, the instantaneous velocity of the object at ( t = 1 ) is ( (-4, -1) ).

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