An object moves according to the equation #y=1/(t+2),y>=0#, how do you find the velocity as a function of t?

Answer 1

#v(t)=(dy)/(dt)=-1/((t+2)^2#

It is apparent that the position of the object is given by #y=1/(t+2)#, where #y>=0#
Therefore velocity is given by #v(t)=(dy)/(dt)=-1/((t+2)^2#
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Answer 2

To find the velocity as a function of time ( t ), you need to differentiate the equation ( y = \frac{1}{t + 2} ) with respect to ( t ). The derivative of ( y ) with respect to ( t ) will give you the velocity function ( v(t) ). So, you differentiate ( y ) with respect to ( t ) using the quotient rule:

[ \frac{dy}{dt} = -\frac{1}{{(t + 2)}^2} ]

So, the velocity function as a function of time ( t ) is ( v(t) = -\frac{1}{{(t + 2)}^2} ).

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Answer from HIX Tutor

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.

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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