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.
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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: \[ \text{Velocity (v)} = \frac{\text{Distance}}{\text{Time}} = \frac{d}{8} \text{ m/s} \]
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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.
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If \( d = 8 \) meters (the total distance in 8 seconds), then: \[ \text{Distance after 10 seconds} = \frac{5 \cdot 8}{4} = 10 \text{ meters} \]
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If \( d = 12 \) meters: \[ \frac{5 \cdot 12}{4} = 15 \text{ meters} \]
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If \( d = 14 \) meters: \[ \frac{5 \cdot 14}{4} = 17.5 \text{ meters} \]
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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.