onsider the following problem.
Find the distance traveled in 25 seconds by an object traveling at a velocity of v(t) = 20 + 6cos(t) feet per second.
Decide whether the problem can be solved using precalculus, or whether calculus is required.
The problem can be solved using precalculus.
The problem requires calculus to be solved.    

If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, use a graphical or numerical approach to estimate the solution. (Round your answer to the nearest foot.)

To determine whether the problem can be solved using precalculus or requires calculus, let's examine the given information:

- The velocity function is \( v(t) = 20 + 6 \cos(t) \) feet per second.
- We need to find the distance traveled over 25 seconds.

### Step 1: Analyze the problem
The velocity function, \( v(t) \), is not constant; it involves a cosine term. To calculate the total distance traveled over time when the velocity varies, we need to compute the **displacement**. Displacement for variable velocity is given by the integral of the velocity function over the given time interval.

### Step 2: Conclusion on method
Because the velocity is a function of time and not constant, **calculus is required**. Specifically, we need to integrate the velocity function to find the total distance traveled. This is not a situation that can be handled purely with precalculus (which works with constant velocities).

### Step 3: Set up the integral
The distance traveled from \( t = 0 \) to \( t = 25 \) seconds is the integral of the velocity function over this interval:

\[
\text{Distance} = \int_0^{25} v(t) \, dt = \int_0^{25} \left( 20 + 6 \cos(t) \right) \, dt
\]

### Step 4: Solve the integral
Let's compute the integral step by step:
\[
\int (20 + 6 \cos(t)) \, dt = \int 20 \, dt + \int 6 \cos(t) \, dt
\]

- \( \int 20 \, dt = 20t \)
- \( \int 6 \cos(t) \, dt = 6 \sin(t) \)

So, the total integral is:

\[
20t + 6 \sin(t)
\]

### Step 5: Evaluate the integral at the bounds
Now, we need to evaluate the expression \( 20t + 6 \sin(t) \) from \( t = 0 \) to \( t = 25 \):

\[
\left[ 20(25) + 6 \sin(25) \right] - \left[ 20(0) + 6 \sin(0) \right]
\]

At \( t = 25 \):

- \( 20(25) = 500 \)
- \( 6 \sin(25) \) (numerical value: approximately \( 6 \times (-0.132) \approx -0.792 \))

At \( t = 0 \):

- \( 20(0) = 0 \)
- \( 6 \sin(0) = 0 \)

Thus, the total distance is approximately:

\[
500 - 0.792 = 499.208 \text{ feet}
\]

### Step 6: Round to the nearest foot
Rounding the result to the nearest foot gives a total distance of approximately:

\[
\boxed{499 \text{ feet}}
\]

### Final Answer:
The distance traveled by the object in 25 seconds is approximately **499 feet**.

---
Would you like further details, or do you have any other questions?

### Relative questions:
1. How do you interpret the integral in terms of displacement or distance?
2. What would happen if we integrated a velocity function that had both positive and negative values?
3. How does the presence of the cosine function in the velocity affect the overall distance?
4. Can we use numerical methods like Riemann sums to approximate this integral?
5. What are other real-life applications of finding distances using calculus?

### Tip:
When solving for distance using velocity, remember that velocity is the derivative of position, so distance requires integrating the velocity function.

Find the distance traveled in 25 seconds by an object traveling at a velocity of v(t) = 20 + 6cos(t) feet per second. Decide whether the problem can be solved using precalculus, or whether calculus is required.

Solve a calculus problem to find the distance traveled in 25 seconds by an object moving at a velocity of v(t) = 20 + 6cos(t) feet per second. Learn how to use definite integrals to compute total distance and apply the Fundamental Theorem of Calculus to solve. Suitable for Grades 11-12 or college students.

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