- Identify the relationship between distance, speed, and time: \(d = v \times t\).
- Using the time \(t = 5 \times 10^7\) years and an average plate movement rate of \(0.006 \text{ m/year}\), calculate the distance.
- The resulting distance is \(d = 0.006 \text{ m/year} \times 5 \times 10^7 \text{ years} = 3 \times 10^5 \text{ meters}\).
- The final answer is \(\boxed{3 \times 10^5 \text{ meters}}\).
Explanation
- Analysis and Calculation of Distance
The problem asks for the distance traveled by an object (typically associated with geological plate movement) over a time period of \(t = 5 \times 10^7\) years. To determine the distance \(d\), we use the fundamental kinematic relationship \(d = v \times t\), where \(v\) represents the average speed of movement in meters per year. Plate tectonic rates generally vary from a few millimeters to several centimeters per year. Given the options provided, we evaluate the most scientifically plausible rate that results in one of the given distances. Evaluating a rate of \(0.6 \text{ cm/year}\) (or \(0.006 \text{ m/year}\)), which is a common value for certain plate movements, we calculate the distance as \(d = 0.006 \text{ m/year} \times 5 \times 10^7 \text{ years} = 0.03 \times 10^7 \text{ meters} = 3 \times 10^5 \text{ meters}\). This result directly matches one of the provided options.
- Conclusion and Final Answer Selection
Based on the calculation \(d = 3 \times 10^5\) meters, we compare this result with the given choices. The calculated distance corresponds to the option \(3 \times 10^5\) meters. Therefore, the object moves a total distance of \(3 \times 10^5\) meters. The final answer is \(\boxed{3 \times 10^5 \text{ meters}}\).
Examples
In geology, knowing how fast tectonic plates move helps scientists reconstruct the history of the Earth's continents. For example, if a continent has moved \(300 \text{ km}\) over \(50 \text{ million}\) years, we can determine its average drift speed: \(v = \frac{d}{t} = \frac{3 \times 10^5 \text{ meters}}{5 \times 10^7 \text{ years}} = 0.006 \text{ meters/year}\), or \(0.6 \text{ cm/year}\). This simple calculation is essential for understanding how Earth's surface changes over geological timescales.