# How do you find the total displacement for the particle whose position at time #t# is given by #s(t)=3t^2-5t# on the interval #0<=t<=2#?

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To find the total displacement of the particle on the interval (0 \leq t \leq 2), we need to calculate the change in position from the initial time to the final time. The total displacement is given by the difference in position at the final time and the initial time.

First, we find the position of the particle at (t = 0): [s(0) = 3(0)^2 - 5(0) = 0]

Next, we find the position of the particle at (t = 2): [s(2) = 3(2)^2 - 5(2) = 3(4) - 5(2) = 12 - 10 = 2]

The total displacement is then: [ \text{Total displacement} = s(2) - s(0) = 2 - 0 = 2 ]

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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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