Question

A particle moves along a straight line. For 2≤ ≤5, the velocity of the particle is given by ()=4cos(^2/8). The position of the particle at time =2 is (2)=6. For 2≤≤5, what is the position of the particle when it is farthest to the right?

Answer 1

To find the position of the particle when it is farthest to the right, we use integration to find the position function and solve for the constant C using the initial condition. We then find the maximum value of the position function by taking the derivative, finding the critical point, and evaluating the second derivative. Finally, we substitute the critical point into the position function to find the position of the particle when it is farthest to the right.

Step-by-step explanation:

To find the position of the particle when it is farthest to the right, we need to find the maximum value of the position function. The position function can be found by integrating the velocity function with respect to time. Given that the velocity is given by v(t) = 4cos^2(t/8) for 2 ≤ t ≤ 5, we can integrate it to find the position function. The position function is given by p(t) = 16sin(t/8) - 16(t/8) + C, where C is a constant.

Using the initial condition p(2) = 6, we can solve for the constant C. Plugging in t = 2 and p(t) = 6, we get 6 = 16sin(2/8) - 16(2/8) + C. Solving for C, we find that C = 2 + 4sin(1).

To find the position when the particle is farthest to the right, we need to find the maximum value of the position function. Taking the derivative of the position function and setting it equal to zero, we can find the critical points. The maximum value occurs when the second derivative is negative at the critical point. After evaluating the second derivative, we find that the position is farthest to the right at t = 5. Substituting t = 5 into the position function, we find that the position of the particle when it is farthest to the right is p(5) = 16sin(5/8) - 16(5/8) + (2 + 4sin(1)).

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