# What is the instantaneous velocity of an object with position at time t equal to # f(t)= (tsqrt(t+2),t^2-2t) # at # t=1 #?

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If you calculate the function of distance in time

The instantaneous velocity of an object is the derivative of distance as a function of time.

The distance traveled by the object can be taken as its the distance between (0,0) and the current position of the object.

I think?

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To find the instantaneous velocity of an object at ( t = 1 ), we need to find the derivative of the position function with respect to time, and then evaluate it at ( t = 1 ).

Given the position function ( f(t) = (t\sqrt{t + 2}, t^2 - 2t) ), we need to find ( f'(t) ) and then evaluate it at ( t = 1 ).

To find ( f'(t) ), we differentiate each component of the position function with respect to time ( t ).

The derivative of ( t\sqrt{t + 2} ) with respect to ( t ) is: [ \frac{d}{dt} (t\sqrt{t + 2}) = \sqrt{t + 2} + \frac{t}{2\sqrt{t + 2}} ]

The derivative of ( t^2 - 2t ) with respect to ( t ) is: [ \frac{d}{dt} (t^2 - 2t) = 2t - 2 ]

Now, we evaluate these derivatives at ( t = 1 ):

For the first component: [ \sqrt{1 + 2} + \frac{1}{2\sqrt{1 + 2}} = \sqrt{3} + \frac{1}{2\sqrt{3}} ]

For the second component: [ 2(1) - 2 = 0 ]

So, the instantaneous velocity of the object at ( t = 1 ) is ( (\sqrt{3} + \frac{1}{2\sqrt{3}}, 0) ).

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