# What is the instantaneous velocity of an object moving in accordance to # f(t)= (sqrt(t+2),t+4) # at # t=1 #?

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Instantaneous velocity is given by the vector:

# vec(v) = << sqrt(3)/6,1 >> # , or,#sqrt(3)/6 hat(i)+hat(j)# , or,#( (sqrt(3)/6),(1) )#

We have:

Then;

And so the instantaneous velocity is given by the vector:

If we want the instantaneous speed , it is given by:

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The instantaneous velocity of an object moving according to the function ( f(t) = (\sqrt{t+2}, t+4) ) at ( t = 1 ) can be found by taking the derivative of the position function with respect to time and then evaluating it at ( t = 1 ).

The derivative of ( f(t) = (\sqrt{t+2}, t+4) ) with respect to ( t ) is ( f'(t) = \left(\frac{1}{2\sqrt{t+2}}, 1\right) ).

Substituting ( t = 1 ) into the derivative, we get ( f'(1) = \left(\frac{1}{2\sqrt{1+2}}, 1\right) = \left(\frac{1}{2\sqrt{3}}, 1\right) ).

Therefore, the instantaneous velocity of the object at ( t = 1 ) is ( \left(\frac{1}{2\sqrt{3}}, 1\right) ).

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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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