A particle moves on the x-axis so that its acceleration at any time t>0 is given by a(t)= When t=1, v=, and s=.

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Presentation transcript:

A particle moves on the x-axis so that its acceleration at any time t>0 is given by a(t)= When t=1, v=, and s=.

We must find the velocity in terms of t. We know that: Our next step is to solve for the ANTIDERIVATIVE

The antiderivative of A(t) is V(t) so To solve for C we must use the information given to us which says v(1) =

Therefore So...

Does the numerical value of the velocity ever exceed 50? The best way to consider this question is to find the acceleration’s critical values and create a sign study. In the next part of the question we will analyze the velocity function and answer the question...

02A(t) + > V(t) decreasingincreasing We know that acceleration is positive for all values greater than 2. Since acceleration is velocity’s derivative, this tells us that for all values of t>2, velocity is increasing. We can also solve the equation. The only critical value in the given domain is t=2.

Let’s solve the equation for 50. When t = , the velocity function yields 50. Since the function is increasing at this point we know it will exceed 50. We can also see it on velocity’s graph. [0,30] by [-5,55] V(t)

The last step is to find the distance s from the origin at time t=2. We must find the distance formula, also known as the antiderivative of velocity.

Using the given condition that s(1) = we find c to be 1, and the final distance formula is...

In order to find the distance s from the origin at time t=2 we simply evaluate the distance formula at t=2. Therefore, the distance from the origin at time t=2 is.86