A student walks east 100 meters in minutes. She then walks north 200 meters in 4 minutes. Which statement is true?

1 answer

To analyze the situation, let's break it down based on the student’s movements:

  1. The student walks 100 meters east.
  2. Then, the student walks 200 meters north in 4 minutes.

To determine which statement about her movement could be true, we can look at a few key points:

  • Total distance traveled: The total distance the student walked is \( 100 \text{ m (east)} + 200 \text{ m (north)} = 300 \text{ meters} \).

  • Direction of displacement: Her final position would be northeast of her starting point, since she walked east first and then north.

  • Time taken for the eastward walk: If the student took \( x \) minutes to walk 100 meters east, then her speed during this segment would be \( \frac{100}{x} \) m/min.

  • Velocity during the northward walk: During the northward walk of 200 meters in 4 minutes, her average speed would be \( \frac{200 \text{ m}}{4 \text{ min}} = 50 \text{ m/min} \).

Without knowing the specifics of which statements are provided, some true statements we can infer might include:

  • The total distance walked is 300 meters.
  • The student walked a total of 200 meters north after walking 100 meters east.
  • The final position is northeast of the start point.

If any specific statements were given, please provide them, and I can analyze which one is true based on the details we've discussed.