# What is the instantaneous velocity of an object with position at time t equal to # f(t)= ((t-2)^2,5t-3) # at # t=2 #?

Instantaneous velocity is

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To find the instantaneous velocity at ( t = 2 ) for the object with position given by ( f(t) = ((t - 2)^2, 5t - 3) ), we first differentiate the position function to get the velocity function. The velocity function ( v(t) ) is given by:

[ v(t) = \left( \frac{d}{dt}((t - 2)^2), \frac{d}{dt}(5t - 3) \right) ]

Differentiating each component separately:

[ v(t) = \left( 2(t - 2) \cdot \frac{d}{dt}(t - 2), 5 \cdot \frac{d}{dt}t \right) ]

[ v(t) = \left( 2(t - 2) \cdot 1, 5 \cdot 1 \right) ]

[ v(t) = \left( 2t - 4, 5 \right) ]

Now, to find the instantaneous velocity at ( t = 2 ), substitute ( t = 2 ) into the velocity function:

[ v(2) = \left( 2(2) - 4, 5 \right) ]

[ v(2) = \left( 0, 5 \right) ]

So, the instantaneous velocity of the object at ( t = 2 ) is ( (0, 5) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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