# What is the position of a particle at time #t=2# if a particle moves along the x axis so that at any time #t>0#, the velocity is given by #v(t)=4-6t^2#, and at position #x=7# at #t=1#?

so:

So the law becomes:

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To find the position of the particle at ( t = 2 ), we first integrate the velocity function to get the displacement function.

[ v(t) = 4 - 6t^2 ]

Integrate ( v(t) ) with respect to ( t ) to get the displacement function ( x(t) ):

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

[ x(t) = 4t - 2t^3 + C ]

Given that the particle is at position ( x = 7 ) at ( t = 1 ), we can use this information to find the constant ( C ):

[ x(1) = 4(1) - 2(1)^3 + C = 4 - 2 + C = 7 ] [ C = 5 ]

Now, we have the displacement function:

[ x(t) = 4t - 2t^3 + 5 ]

To find the position of the particle at ( t = 2 ), substitute ( t = 2 ) into the displacement function:

[ x(2) = 4(2) - 2(2)^3 + 5 ]

[ x(2) = 8 - 16 + 5 ]

[ x(2) = -3 ]

Therefore, the position of the particle at ( t = 2 ) is ( x = -3 ).

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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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