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To find the displacement from 13 to 15 seconds using the velocity vs. time graph, we can calculate the area under the curve between these time points.

1. The graph shows that from 13 to 15 seconds, the velocity is increasing from 0 m/s to approximately 17 m/s. The shape formed in this interval is a triangle.

2. The formula for the area of a triangle is:
   \[
   \text{Area (Displacement)} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   - Base (time) = \( 15 \, \text{s} - 13 \, \text{s} = 2 \, \text{seconds} \)
   - Height (velocity) = 17 m/s (the final velocity at 15 seconds)

3. Using the formula:
   \[
   \Delta x = \frac{1}{2} \times 2 \, \text{s} \times 17 \, \text{m/s} = 17 \, \text{meters}
   \]

So, the displacement from 13 to 15 seconds is approximately **17 meters**, which closely matches the option **Δx = 16 m**.

Would you like further details or explanations? Here are some follow-up questions:

1. How would you calculate displacement for other time intervals?
2. Can you explain how the graph's shape affects the displacement calculation?
3. What happens when velocity is negative on a velocity-time graph?
4. How is displacement different from distance traveled?
5. What role does the area under a velocity-time graph play in finding displacement?

**Tip:** The area under a velocity-time graph always gives you the displacement, while the total area (including negative values) gives the total distance traveled.

What is the displacement from 13-15 seconds based on the given velocity vs. time graph?

Learn how to calculate displacement from a velocity vs. time graph for the time interval 13-15 seconds. This problem involves key kinematics concepts and requires applying the formula for the area of a triangle to determine the displacement. Suitable for students in Grades 9-12.

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