Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units of

StartFraction m Over s EndFractionms.

**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)equals=22tsquared2negative 24−24tplus+7070​;

​[0,88​]

Question content area bottom

Part 1

**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)

enter your response here.

​(Use a comma to separate answers as needed. Type your answers in interval​ notation.)

B.

The motion is never in the positive direction.

To solve this problem, let’s examine each part step-by-step. The given velocity function is:

\[
v(t) = 22t^2 - 24t + 70
\]

and the interval is \([0, 8]\). Now, let’s go through each question in detail.

---

### Part (a): Determine when the motion is in the positive and negative directions.
To determine when the object is moving in the positive or negative direction, we need to analyze the sign of \( v(t) \):
- The motion is in the **positive direction** when \( v(t) > 0 \).
- The motion is in the **negative direction** when \( v(t) < 0 \).

#### Step 1: Set up the inequality \( v(t) > 0 \).
\[
22t^2 - 24t + 70 > 0
\]

This is a quadratic inequality. To solve it, we first find the roots of the equation \( v(t) = 0 \) by solving:
\[
22t^2 - 24t + 70 = 0
\]

#### Step 2: Solve the quadratic equation.
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 22 \), \( b = -24 \), and \( c = 70 \):
\[
t = \frac{-(-24) \pm \sqrt{(-24)^2 - 4 \cdot 22 \cdot 70}}{2 \cdot 22}
\]

Calculating the discriminant:
\[
(-24)^2 - 4 \cdot 22 \cdot 70 = 576 - 616 = -40
\]

Since the discriminant is negative, there are no real roots, meaning the quadratic expression \( 22t^2 - 24t + 70 \) does not cross zero.

#### Step 3: Determine the sign of \( v(t) \).
Since there are no real roots, \( v(t) \) does not change sign over the interval. Checking \( v(0) \):
\[
v(0) = 22(0)^2 - 24(0) + 70 = 70 > 0
\]

Thus, \( v(t) > 0 \) for all \( t \) in \([0, 8]\). This means the motion is always in the positive direction over the given interval.

**Answer for Part (a):** The motion is in the positive direction for \( t \) in \([0, 8]\).

---

### Part (b): Find the displacement over the interval \([0, 8]\).
Displacement is the change in position over an interval, and it is found by integrating the velocity function over that interval.

\[
\text{Displacement} = \int_0^8 v(t) \, dt = \int_0^8 (22t^2 - 24t + 70) \, dt
\]

#### Step 1: Integrate \( v(t) \).
\[
\int (22t^2 - 24t + 70) \, dt = \frac{22t^3}{3} - \frac{24t^2}{2} + 70t = \frac{22t^3}{3} - 12t^2 + 70t
\]

#### Step 2: Evaluate the integral from 0 to 8.
\[
\int_0^8 v(t) \, dt = \left[ \frac{22t^3}{3} - 12t^2 + 70t \right]_0^8
\]

Calculating each term at \( t = 8 \):
\[
\frac{22 \cdot 8^3}{3} - 12 \cdot 8^2 + 70 \cdot 8
\]

Breaking it down:
1. \( \frac{22 \cdot 512}{3} = \frac{11264}{3} \approx 3754.67 \)
2. \( -12 \cdot 64 = -768 \)
3. \( 70 \cdot 8 = 560 \)

Adding these together:
\[
3754.67 - 768 + 560 = 3546.67
\]

So, the displacement over \([0, 8]\) is approximately \( 3546.67 \) meters.

**Answer for Part (b):** The displacement over \([0, 8]\) is approximately \( 3546.67 \) meters.

---

### Part (c): Find the distance traveled over the interval \([0, 8]\).
Since \( v(t) > 0 \) throughout \([0, 8]\), the object does not change direction, and the distance traveled is the same as the displacement.

**Answer for Part (c):** The distance traveled over \([0, 8]\) is approximately \( 3546.67 \) meters.

---

Would you like further details on any specific part of the solution?

---

### Related Questions:
1. How would the solution change if \( v(t) \) had real roots within \([0, 8]\)?
2. What would be the effect on displacement if \( v(t) \) included a constant term?
3. How can we interpret the displacement if the interval was extended beyond \([0, 8]\)?
4. What is the physical meaning of the integral of velocity in real-world applications?
5. How does the calculation of distance change if \( v(t) \) had a negative interval?

### Tip:
When dealing with motion problems, always check if the velocity function has any roots within the interval to identify changes in direction.

Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units of m/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) = 22t^2 - 24t + 70; [0,8]

Learn how to analyze the direction of motion, calculate displacement, and determine distance traveled for an object moving with a velocity function v(t) = 22t^2 - 24t + 70 over the interval [0,8]. Explore calculus concepts and use the quadratic equation and definite integrals to solve this kinematics problem, suitable for advanced high school or introductory college calculus students.

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