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Find the distance that the object made during its 6s of moving

The graph provided shows the velocity \( v \) of an object over time \( t \), where velocity is on the y-axis (in meters per second, m/s), and time is on the x-axis (in seconds, s). To find the distance traveled by the object during the 6 seconds, we need to compute the area under the velocity-time graph, as the area under this curve represents the total displacement or distance traveled.

### Step-by-Step Solution:

1. **Analyze the graph:**
   - From \( t = 0 \) to \( t = 2 \), the velocity increases linearly from 0 to 6 m/s.
   - From \( t = 2 \) to \( t = 6 \), the velocity decreases linearly from 6 m/s back to 0 m/s.

2. **Divide the graph into simple shapes:**
   - The graph forms a triangle with a base along the time axis from \( t = 0 \) to \( t = 6 \) and a height of 6 m/s.

3. **Calculate the area of the triangle:**
   The formula for the area of a triangle is:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   - Base \( = 6 \) seconds (from \( t = 0 \) to \( t = 6 \))
   - Height \( = 6 \) m/s (maximum velocity)

   \[
   \text{Area} = \frac{1}{2} \times 6 \times 6 = 18 \, \text{m}
   \]

Thus, the total distance traveled by the object during the 6 seconds is **18 meters**.

Would you like more details or have any questions?

Here are some related questions to deepen understanding:

1. How would the result change if the graph had different shapes, such as rectangles or trapezoids?
2. What does it mean physically if the graph dips below the time axis (negative velocity)?
3. How would you calculate the distance if the object reversed direction halfway through the motion?
4. What is the difference between distance and displacement in this context?
5. How would the total distance change if the velocity was constant instead of varying?

**Tip:** When calculating distance from a velocity-time graph, always remember that the area under the curve (above the time axis) represents distance.

Find the distance that the object made during its 6 seconds of moving.

This problem involves finding the total distance traveled by an object using a velocity-time graph over 6 seconds. The key concept is that the area under the velocity-time curve represents the displacement. Using the formula for the area of a triangle, the total distance calculated is 18 meters.

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