How do you find the acceleration for the function #s(t)=t^3+3t^2# ?

Answer 1
Let us first find the velocity function #v(t)#. #v(t)=s'(t)=3t^2+6t#
Let us now find the acceleration function #a(t)#. #a(t)=v'(t)=s''(t)=6t+6#
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Answer 2

To find the acceleration for the function ( s(t) = t^3 + 3t^2 ), you need to take the second derivative of the function with respect to time (( t )). The second derivative represents the rate of change of acceleration with respect to time. So, compute the second derivative of ( s(t) ) by taking the derivative of ( s'(t) ), which is the first derivative of ( s(t) ).

The first derivative of ( s(t) ) is ( s'(t) = 3t^2 + 6t ).

Then, take the derivative of ( s'(t) ) to find the second derivative, denoted as ( s''(t) ).

( s''(t) = \frac{d}{dt}(3t^2 + 6t) )

( s''(t) = 6t + 6 )

Therefore, the acceleration function ( a(t) ) for ( s(t) = t^3 + 3t^2 ) is ( a(t) = 6t + 6 ).

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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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