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. $$ v(\mathrm{t})=2 \mathrm{t}^{2}-16 \mathrm{t}+30 ;[0,6] $$ 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) $$ (3,5) $$. (Use a comma to separate answers as needed. Type your answers in interval notation.) B. The motion is never in the negative direction. b. The displacement of the particle is 36 m . c. The total distance traveled by the particle is m .

Question

UpStudy AI · Accepted Answer

To determine the **total distance traveled** by the particle over the interval $$[0, 6]$$ seconds, we'll analyze the velocity function and calculate the necessary integrals.

## Given:
$$ v(t) = 2t^2 - 16t + 30 \quad \text{m/s} $$
$$ \text{Interval: } [0, 6] \, \text{seconds} $$

## Steps to Find Total Distance Traveled:

1. **Determine When the Velocity is Zero:**
   
   To find when the particle changes direction, set $$ v(t) = 0 $$:
   $$
   2t^2 - 16t + 30 = 0 \
   t^2 - 8t + 15 = 0 \quad (\text{Divided by } 2)
   $$
   Solving the quadratic equation:
   $$
   t = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm 2}{2} \
   t = 5 \, \text{seconds} \quad \text{and} \quad t = 3 \, \text{seconds}
   $$
   
   Thus, the velocity changes sign at $$ t = 3 $$ seconds and $$ t = 5 $$ seconds.

2. **Determine Intervals of Positive and Negative Velocity:**
   
   - **Positive Velocity:** $$ [0, 3) $$ and $$ (5, 6] $$ seconds
   - **Negative Velocity:** $$ (3, 5) $$ seconds
   
   *(This aligns with your selections in part (a))*

3. **Calculate Displacement in Each Interval:**
   
   The displacement is the integral of velocity over time. However, for total **distance**, we take the absolute value of displacement in each interval.

- **From $$ t = 0 $$ to $$ t = 3 $$:**
     $$
     \int_{0}^{3} v(t) \, dt = \int_{0}^{3} (2t^2 - 16t + 30) \, dt = 36 \, \text{meters}
     $$
   
   - **From $$ t = 3 $$ to $$ t = 5 $$:**
     $$
     \int_{3}^{5} v(t) \, dt = -\frac{8}{3} \, \text{meters} \quad (\text{Negative, so distance is } \frac{8}{3} \, \text{meters})
     $$
   
   - **From $$ t = 5 $$ to $$ t = 6 $$:**
     $$
     \int_{5}^{6} v(t) \, dt = \frac{8}{3} \, \text{meters}
     $$

4. **Sum the Absolute Displacements:**
   $$
   \text{Total Distance} = 36 + \frac{8}{3} + \frac{8}{3} = 36 + \frac{16}{3} = \frac{124}{3} \, \text{meters}
   $$
   $$
   \frac{124}{3} \, \text{meters} = 41\frac{1}{3} \, \text{meters}
   $$

## **Answer to Part (c):**
The total distance traveled by the particle over the interval $$[0, 6]$$ seconds is **$$41\frac{1}{3}$$ meters**.
The total distance traveled by the particle over the interval $$[0, 6]$$ seconds is $$41\frac{1}{3}$$ meters.

Real Tutor Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions