To estimate how far the car travels from $t = 0$ to $t = 10$ seconds, we can apply the trapezoidal rule using the given speed values at specific time intervals.

Trapezoidal Rule

The trapezoidal rule helps estimate the area under the curve, which in this case represents the distance traveled over time, as distance is the integral of velocity.

Step-by-Step Calculation:

The table gives the velocity $v(t)$ at specific times $t$:

We use the trapezoidal rule for each interval:

$\text{Distance} \approx \sum \frac{1}{2} \times (t_{\text{end}} - t_{\text{start}}) \times (v_{\text{end}} + v_{\text{start}})$

Intervals:

• From $t = 0$ to $t = 2$: $\frac{1}{2} \times (2 - 0) \times (60 + 40) = 100 \, \text{ft}$
• From $t = 2$ to $t = 4$: $\frac{1}{2} \times (4 - 2) \times (40 + 30) = 70 \, \text{ft}$
• From $t = 4$ to $t = 6$: $\frac{1}{2} \times (6 - 4) \times (30 + 20) = 50 \, \text{ft}$
• From $t = 6$ to $t = 8$: $\frac{1}{2} \times (8 - 6) \times (20 + 40) = 60 \, \text{ft}$
• From $t = 8$ to $t = 10$: $\frac{1}{2} \times (10 - 8) \times (40 + 45) = 85 \, \text{ft}$

Total Distance:

$\text{Total Distance} = 100 + 70 + 50 + 60 + 85 = 365 \, \text{ft}$

Thus, the best estimate of how far the car travels from $t = 0$ to $t = 10$ seconds is 365 ft.

The correct answer is (C) 365 ft.

Would you like more details on the trapezoidal rule or have any other questions?

Here are 5 related questions:

1. How can you apply the trapezoidal rule in cases with unequal time intervals?
2. What is the physical meaning of the integral of velocity over time?
3. How does the midpoint rule differ from the trapezoidal rule in numerical integration?
4. Can the trapezoidal rule always guarantee accurate results? Under what conditions might it fail?
5. How would the estimate change if we had more data points for the speed?

Tip: When working with numerical integration, dividing the interval into more sub-intervals improves accuracy.