To estimate the distance traveled using the velocities at the end of each time period, we can apply the trapezoidal rule to approximate the total distance. The velocity values are given, and the intervals are spaced by 10 seconds. We'll estimate the distance in miles based on these velocity values.

Formula:

The trapezoidal rule for approximating distance over time is given by:

$\text{Distance} \approx \frac{\Delta t}{2} \left( v_1 + 2v_2 + 2v_3 + \dots + 2v_{n-1} + v_n \right)$

Where:

• $\Delta t = 10$ seconds (which is $\frac{10}{3600}$ hours, as there are 3600 seconds in an hour to convert seconds to hours)
• $v_i$ represents the velocity at each point.

Step-by-Step Calculation:

1. Time Intervals: $\Delta t = \frac{10}{3600}$ hours (since the velocities are given in miles per hour).

2. Velocities: The velocities at the end of each time period are: $v_1 = 168.0, \, v_2 = 107.6, \, v_3 = 98.8, \, v_4 = 125.5, \, v_5 = 177.1, \, v_6 = 175.6$

3. Trapezoidal Rule Application: $\text{Distance} \approx \frac{10}{3600} \times \frac{1}{2} \times \left( 168.0 + 2(107.6 + 98.8 + 125.5 + 177.1) + 175.6 \right)$

Let's calculate this.The estimated distance traveled using the velocities at the end of the time periods is approximately 3.305 miles.

Would you like any further explanation on the calculation process or the trapezoidal rule?

Here are 5 related questions to expand on this concept:

1. How does the trapezoidal rule compare to other numerical integration methods, such as Simpson's rule?
2. What happens to the accuracy of the trapezoidal rule if time intervals are uneven?
3. How would this estimate change if we used the average velocities at each interval instead of the end velocities?
4. How can this approach be used to estimate acceleration or deceleration from velocity data?
5. What are other practical applications of the trapezoidal rule in physics or engineering?

Tip: When using numerical methods like the trapezoidal rule, smaller time intervals typically result in more accurate estimates of distance or area.