# The velocity of an object with a mass of #3 kg# is given by #v(t)= 6 t^2 -4 t #. What is the impulse applied to the object at #t= 3 #?

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To find the impulse applied to the object at ( t = 3 ), we first need to find the change in momentum of the object between ( t = 0 ) and ( t = 3 ). The impulse is equal to the change in momentum.

Given that the velocity function is ( v(t) = 6t^2 - 4t ), we need to find the momentum function by integrating the velocity function with respect to time:

[ p(t) = \int v(t) , dt = \int (6t^2 - 4t) , dt ]

[ p(t) = 2t^3 - 2t^2 + C ]

Where ( C ) is the constant of integration.

Now, to find the change in momentum between ( t = 0 ) and ( t = 3 ), we subtract the momentum at ( t = 0 ) from the momentum at ( t = 3 ):

[ \Delta p = p(3) - p(0) = (2(3)^3 - 2(3)^2) - (2(0)^3 - 2(0)^2) ]

[ \Delta p = (54 - 18) - (0 - 0) = 36 , \text{kg m/s} ]

Therefore, the impulse applied to the object at ( t = 3 ) is ( 36 , \text{kg m/s} ).

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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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