Question

The function v(t)=t 3

−7t 2
+12t, [0,6], is the velocity in m/sec of a particle moving along the x-axis. Complete parts (a) through (c). a. Determine when the motion is in the positive direction and when it is in the negative direction. b. Find the displacement over the given interval. c. Find the distance traveled over the given interval. Determine when the motion is in the negative direction. Choose the correct answer below. A. (4,6] B. (0,3)∪(4,6] C. (3,4)∪(4,6] D. (3,4) b. Find the displacement over the given interval. The displacement over the given interval is m.

Answer 1

The motion is in the positive direction when t < 0, t > 4, and 0 < t < 3. The displacement over the given interval [0, 6] is 242 m. The distance traveled over the given interval is also 242 m.

Step-by-step explanation:

To determine when the motion is in the positive direction and when it is in the negative direction, we need to look at the sign of the velocity function. In this case, the velocity function is v(t) = t^3 - 7t^2 + 12t. We can analyze the function by finding the roots or critical points where the velocity changes sign. By factoring the function, we find that it equals t(t-3)(t-4). So, the motion is in the positive direction when t < 0, t > 4, and 0 < t < 3. On the other hand, the motion is in the negative direction when 3 < t < 4.

To find the displacement over the given interval [0, 6], we need to find the change in position. The position function can be found by integrating the velocity function. By integrating v(t) = t^3 - 7t^2 + 12t, we get x(t) = ((1/4)t^4) - ((7/3)t^3) + (6t^2). Plugging in the limits of integration, we find x(6) - x(0) = ((1/4)*(6^4)) - ((7/3)*(6^3)) + (6*(6^2)) - ((1/4)*(0^4)) - ((7/3)*(0^3)) + (6*(0^2)) = 242.

To find the distance traveled over the given interval, we need to consider the absolute value of the velocity function. By taking the absolute value of v(t) = t^3 - 7t^2 + 12t, we get |v(t)| = |t(t-3)(t-4)|. We can analyze the function by finding the roots or critical points where the velocity changes sign. By factoring the function, we find that it equals t(t-3)(t-4). So, the distance traveled is the integral of |v(t)| over the interval [0, 6]. By integrating |v(t)|, we get the distance traveled is equal to ((1/4)*(6^4)) - ((7/3)*(6^3)) + (6*(6^2)) - ((1/4)*(0^4)) - ((7/3)*(0^3)) + (6*(0^2)) = 242.

Answer 2

The question asks for the intervals of positive and negative motion, the displacement, and the distance traveled using a given velocity function. A detailed analysis would include finding critical points for determining motion direction, integrating the velocity function for displacement, and the absolute value of the function for distance traveled.

Step-by-step explanation:

The question involves determining the positive and negative motion direction, displacement, and distance traveled of a particle whose velocity function is given by v(t) = t3 − 7t2 + 12t. The motion is considered to be in the positive direction when v(t) > 0 and in the negative direction when v(t) < 0.

To find when the motion is in the positive or negative direction, we would have to set the velocity function equal to zero and solve for t to find the critical points in the interval [0, 6]. These critical points will help us determine the intervals where the motion is positive or negative.

As for displacement, we would integrate the velocity function over the interval [0, 6]. The displacement is equal to the definite integral of the velocity function over this interval. To find the distance traveled, we would integrate the absolute value of the velocity function, since distance considers both positive and negative movement as contributing to total travel.

However, without further calculations or given critical points, we cannot provide a definitive answer. Thus, a detailed step-by-step solution for each part is required to give an accurate response to the student's questions about motion direction, displacement, and distance.