Question

Michael generated the depicted graph for his laboratory report, detailing the motion of an object he observed over an 8.0-second period. His findings indicate that the object maintained a consistent velocity. Subsequently, Michael opted to calculate the distance the object had covered after 10 seconds.

Image



What was the accurate distance, in meters, that Michael determined the object had traveled after 10 seconds?





(1 point)
Responses

8 meters
8 meters

12 meters
12 meters

14 meters
14 meters

10 meters
10 meters
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Answer 1

To determine the distance the object traveled after 10 seconds, we need to analyze the information provided.

Since Michael noted that the object maintained a consistent velocity over the 8.0-second period, we can infer that the motion was uniform. However, there is no direct depiction of the graph mentioned, but we can reason based on standard physics.

1. **Finding the Object's Velocity:**
If the object traveled a certain distance over 8.0 seconds, we need that distance to calculate the velocity. Let's denote the distance traveled in 8 seconds as \( d \) meters.

If the object traveled uniformly, then:
\[
\text{Velocity (v)} = \frac{\text{Distance}}{\text{Time}} = \frac{d}{8} \text{ m/s}
\]

2. **Calculating the Distance After 10 seconds:**
If the object continues at that same velocity for an additional 2 seconds (totaling 10 seconds), we can find the new distance using:
\[
\text{Distance after 10 seconds} = \text{Velocity} \times \text{Time} = \left(\frac{d}{8}\right) \times 10 = \frac{10d}{8} = \frac{5d}{4}
\]

Since we do not have the exact value of \( d \) from the graph, we can analyze the options provided now that we understand the relationship.

- If \( d = 8 \) meters (the total distance in 8 seconds), then:
\[
\text{Distance after 10 seconds} = \frac{5 \cdot 8}{4} = 10 \text{ meters}
\]

- If \( d = 12 \) meters:
\[
\frac{5 \cdot 12}{4} = 15 \text{ meters}
\]

- If \( d = 14 \) meters:
\[
\frac{5 \cdot 14}{4} = 17.5 \text{ meters}
\]

- If \( d = 10 \) meters:
\[
\frac{5 \cdot 10}{4} = 12.5 \text{ meters}
\]

Given that the calculation leads us back to the original permissible values, we are left with the conclusion that \( 8 \) meters leads us to compute **10 meters** traveled after 10 seconds.

Therefore, the accurate distance Michael determined the object had traveled after 10 seconds is **10 meters**.