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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. \( v(t)=2 \mathrm{t}^{2}-16 \mathrm{t}+30 ;[0,6] \) 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 Ramirez Carrillo. in the United States
Jan 23,2025

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The motion is in the positive direction for \( t \)-values in the intervals \((0, 3)\) and \((5, 6)\).

Solución

To determine when the motion is in the **positive direction**, we need to analyze the velocity function: \[ v(t) = 2t^2 - 16t + 30 \] **Positive direction** corresponds to when the velocity \( v(t) > 0 \). ### Step 1: Find the Critical Points Set \( v(t) = 0 \) to find the critical points: \[ 2t^2 - 16t + 30 = 0 \] Divide the equation by 2: \[ t^2 - 8t + 15 = 0 \] Factor or use the quadratic formula: \[ t = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm 2}{2} \] So, \[ t = 5 \quad \text{and} \quad t = 3 \] ### Step 2: Determine Intervals Where \( v(t) > 0 \) Since the coefficient of \( t^2 \) is positive, the parabola opens upwards. Therefore, \( v(t) > 0 \) outside the roots. For the interval \([0, 6]\): - \( v(t) > 0 \) when \( t < 3 \) and \( t > 5 \). ### Conclusion The motion is in the **positive direction** for \( t \)-values in the intervals: \[ (0, 3) \quad \text{and} \quad (5, 6) \] **Therefore, the correct choice is:** **A.** The motion is in the positive direction for \( t \)-values in the intervals \((0, 3)\) and \((5, 6)\).

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To determine when the motion is in the positive direction, we need to analyze the velocity function \( v(t) = 2t^2 - 16t + 30 \). First, we will find the critical points by setting the velocity equal to zero: \[ 2t^2 - 16t + 30 = 0 \] Dividing the entire equation by 2 gives: \[ t^2 - 8t + 15 = 0 \] Factoring the quadratic, we get: \[ (t - 3)(t - 5) = 0 \] Thus, the solutions are \( t = 3 \) and \( t = 5 \). We can now test the intervals \( [0, 3) \), \( (3, 5) \), and \( (5, 6] \) to determine the sign of \( v(t) \). - For \( t < 3 \) (e.g., \( t = 0 \)): \[ v(0) = 30 \quad \text{(positive)} \] - For \( t = 4 \) (between 3 and 5): \[ v(4) = 2(4^2) - 16(4) + 30 = 32 - 64 + 30 = -2 \quad \text{(negative)} \] - For \( t > 5 \) (e.g., \( t = 6 \)): \[ v(6) = 2(6^2) - 16(6) + 30 = 72 - 96 + 30 = 6 \quad \text{(positive)} \] From this analysis, we find that the object is moving in the positive direction during the intervals \( [0, 3) \) and \( (5, 6] \). So the correct choice is: A. The motion is in the positive direction for \( t \)-values in the interval(s) \( [0, 3) \cup (5, 6] \). --- Time travel back to the 1700s! In the glorious era of the Scientific Revolution, renowned mathematicians began to lay the groundwork for calculus, with Gottfried Wilhelm Leibniz and Isaac Newton leading the charge. Their exploration of rates of change and motion was groundbreaking and paved the way for future physicists to understand dynamics—transforming the abstract world of mathematics into practical applications! Now for a fun twist! Ever wondered how velocity plays into everyday life? Think about driving a car: when you're accelerating toward a red light, your velocity may become negative as you decelerate. Understanding when your speed is positive or negative is essential for safe driving and can even keep you out of trouble with the law! So next time you hit the road, remember to check your “v” for velocity!

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