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. $$ \mathrm{v}(\mathrm{t})=2 \mathrm{t}^{2}-16 \mathrm{t}+30 ;[0,6] $$ 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) [ 0,3$$ ) $$, $$ (5,6] $$. (Use a comma to separate answers as needed. Type your answers in interval notation.) B. The motion is never in the positive direction. When is the motion in the negative direction? Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. The motion is in the negative 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 negative direction.

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To solve the problem, we will follow the steps outlined: 1. **Extract known conditions:** - The velocity function is given by $$ v(t) = 2t^2 - 16t + 30 $$. - The interval for time $$ t $$ is $$[0, 6]$$. ### Part a: Determine when the motion is in the positive and negative direction. The motion is in the positive direction when $$ v(t) > 0 $$ and in the negative direction when $$ v(t) < 0 $$. #### Step 1: Find the roots of the velocity function. To find when the velocity is zero, we will solve the equation $$ v(t) = 0 $$: $$ 2t^2 - 16t + 30 = 0 $$ We can simplify this equation by dividing all terms by 2: $$ t^2 - 8t + 15 = 0 $$ Now, we will use the quadratic formula $$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $$ a = 1, b = -8, c = 15 $$. #### Step 2: Calculate the discriminant and roots. Let's calculate the discriminant: $$ D = b^2 - 4ac = (-8)^2 - 4 \cdot 1 \cdot 15 = 64 - 60 = 4 $$ Now, we can find the roots: $$ t = \frac{8 \pm \sqrt{4}}{2 \cdot 1} = \frac{8 \pm 2}{2} $$ Calculating the two roots: $$ t_1 = \frac{10}{2} = 5, \quad t_2 = \frac{6}{2} = 3 $$ #### Step 3: Test intervals around the roots. We will test the intervals $$ (0, 3) $$, $$ (3, 5) $$, and $$ (5, 6) $$ to determine the sign of $$ v(t) $$. - For $$ t \in (0, 3) $$, choose $$ t = 1 $$: $$ v(1) = 2(1)^2 - 16(1) + 30 = 2 - 16 + 30 = 16 > 0 $$ - For $$ t \in (3, 5) $$, choose $$ t = 4 $$: $$ v(4) = 2(4)^2 - 16(4) + 30 = 32 - 64 + 30 = -2 < 0 $$ - For $$ t \in (5, 6) $$, choose $$ t = 5.5 $$: $$ v(5.5) = 2(5.5)^2 - 16(5.5) + 30 = 60.5 - 88 + 30 = 2.5 > 0 $$ #### Conclusion for Part a: - The motion is in the positive direction for $$ t $$ in the intervals $$ [0, 3) $$ and $$ (5, 6] $$. - The motion is in the negative direction for $$ t $$ in the interval $$ (3, 5) $$. ### Part b: Find the displacement over the given interval. Displacement is calculated by integrating the velocity function over the interval $$[0, 6]$$: $$ \text{Displacement} = \int_0^6 v(t) \, dt $$ Let's calculate this integral: $$ \int_0^6 (2t^2 - 16t + 30) \, dt $$ Calculating the integral: $$ = \left[ \frac{2}{3}t^3 - 8t^2 + 30t \right]_0^6 $$ Calculating at the bounds: $$ = \left( \frac{2}{3}(6)^3 - 8(6)^2 + 30(6) \right) - \left( 0 \right) $$ Calculating each term: $$ = \frac{2}{3}(216) - 8(36) + 180 $$ Calculating: $$ = 144 - 288 + 180 = 36 $$ ### Part c: Find the distance traveled over the given interval. Distance is the integral of the absolute value of the velocity function. We need to split the integral at the points where the velocity changes sign, which are $$ t = 3 $$ and $$ t = 5 $$: $$ \text{Distance} = \int_0^3 v(t) \, dt + \int_3^5 -v(t) \, dt + \int_5^6 v(t) \, dt $$ Calculating each integral: 1. **From 0 to 3:** $$ \int_0^3 (2t^2 - 16t + 30) \, dt $$ 2. **From 3 to 5:** $$ \int_3^5 -(2t^2 - 16t + 30) \, dt $$ 3. **From 5 to 6:** $$ \int_5^6 (2t^2 - 16t + 30) \, dt $$ Let's calculate these integrals: 1. **From 0 to 3:** $$ = \left[ \frac{2}{3}t^3 - 8t^2 + 30t \right]_0^3 $$ $$ = \left( \frac{2}{3}(27) - 8(9) + 90 \right) - 0 = 18 - 72 + 90 = 36 $$ 2. **From 3 to 5:** $$ = -\left[ \frac{2}{3}t^3 - 8t^2 + 30t \right]_3^5 $$ $$ = -\left( \left( \frac{2}{3}(125) - 8(25) + 150 \right) - \left( 18 - 72 + 90 \right) \right) $$ $$ = -\left( \frac{250}{3} - 200 + 150 - 36 \right) = -\left( \frac{250}{3} - 86 \right) = -\left( \frac{250 - 258}{3} \right) = \frac{8}{3} $$ 3. **From 5 to 6:** \[ = \left[ \frac{2}{3}t^3 - **Part a:** - **Positive Direction:** $$ t $$ in the intervals $$[0, 3)$$ and $$(5, 6]$$ - **Negative Direction:** $$ t $$ in the interval $$(3, 5)$$ **Part b:** - **Displacement:** 36 meters **Part c:** - **Distance Traveled:** $$ \frac{8}{3} + 36 + \frac{2}{3} = 40 $$ meters

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