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Consider an object moving along a line with the given velocity v . Assume t is time measured in seconds and velocities have units of \( \frac{\mathrm{m}}{\mathrm{s}} \) 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. \( v(\mathrm{t})=3 \mathrm{t}^{2}-24 \mathrm{t}+21 ;[0,8] \) a. When is the motion in the positive direction? Select the correct choice and, if necessary, fill in the answer box to complete, your choice. A. The motion is in the positive direction for \( t \)-values in the interval(s) (Use a comma to separate answers as needed. Type your answers in interval notation.) B. The motion is never in the positive direction.

Ask by West Ortega. in the United States
Mar 20,2025

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Answer

a. The motion is in the positive direction for \( t \)-values in the interval \([0, 1]\) and never in the positive direction after that.

Solution

Alright, I need to analyze the motion of an object with the given velocity function \( v(t) = 3t^2 - 24t + 21 \) over the interval \([0, 8]\) seconds. The questions are: 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. First, let's tackle part (a). To determine when the motion is in the positive or negative direction, I need to find the intervals where \( v(t) > 0 \) and \( v(t) < 0 \). Given \( v(t) = 3t^2 - 24t + 21 \), I'll start by finding the roots of the equation \( 3t^2 - 24t + 21 = 0 \) to identify the points where the velocity changes direction. Using the quadratic formula: \[ t = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 3 \cdot 21}}{2 \cdot 3} = \frac{24 \pm \sqrt{576 - 252}}{6} = \frac{24 \pm \sqrt{324}}{6} = \frac{24 \pm 18}{6} \] So, the roots are: \[ t = \frac{24 + 18}{6} = 7 \quad \text{and} \quad t = \frac{24 - 18}{6} = 1 \] These roots divide the time axis into three intervals: \( [0, 1] \), \( (1, 7) \), and \( (7, 8] \). Next, I'll test a point in each interval to determine the sign of \( v(t) \): 1. **Interval \([0, 1]\):** Let's choose \( t = 0.5 \) \[ v(0.5) = 3(0.5)^2 - 24(0.5) + 21 = 3(0.25) - 12 + 21 = 0.75 - 12 + 21 = 9.75 > 0 \] So, the motion is in the positive direction on \([0, 1]\). 2. **Interval \((1, 7)\):** Let's choose \( t = 4 \) \[ v(4) = 3(4)^2 - 24(4) + 21 = 3(16) - 96 + 21 = 48 - 96 + 21 = -27 < 0 \] So, the motion is in the negative direction on \((1, 7)\). 3. **Interval \((7, 8]\):** Let's choose \( t = 7.5 \) \[ v(7.5) = 3(7.5)^2 - 24(7.5) + 21 = 3(56.25) - 180 + 21 = 168.75 - 180 + 21 = -0.25 < 0 \] So, the motion is in the negative direction on \((7, 8]\). Therefore, the motion is in the positive direction for \( t \)-values in the interval \([0, 1]\) and never in the positive direction after that.

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To determine when the motion is in the positive direction, we first need to find the critical points where the velocity \( v(t) = 0 \). Setting \( 3t^2 - 24t + 21 = 0 \) and solving gives us the \( t \)-intercepts. We can then analyze the intervals between these points and check the sign of the velocity in each segment. The quadratic formula lets us find that the roots are \( t = 3 + \sqrt{6} \) and \( t = 3 - \sqrt{6} \). By evaluating \( v(t) \) for intervals split by these critical points within the domain \( [0,8] \), you'll find that the motion is in the positive direction in the intervals where \( v(t) > 0 \). To summarize, motion is in the positive direction for the intervals \( [0, 3 - \sqrt{6}) \) and \( (3 + \sqrt{6}, 8] \). To find the displacement over the interval \( [0, 8] \), you can integrate the velocity function: \[ \int_0^8 (3t^2 - 24t + 21) \, dt \] This evaluates to the change in position from time \( t=0 \) to \( t=8 \). Meanwhile, the distance traveled is the integral of the absolute value of the velocity, taking care to account for intervals where the motion changes direction in between the limits. Don’t forget to compute the integrals separately in the segments determined earlier for accurate distance traveled calculations!

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