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.

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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.