John Johnson set a world record for the "100.0" m run with a time of 9.86 s. If, after crossing the finish line, John walked directly back to his starting point in 90.9 s, what is the magnitude of his average velocity for the 200.0m?
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To find the magnitude of John Johnson's average velocity for the 200.0 m, we need to first determine his total displacement and the time it took for him to complete the 200.0 m.
Since John ran 100.0 m in 9.86 s, it means his velocity during the run was 100.0 m / 9.86 s = 10.15 m/s.
After crossing the finish line, John walked directly back to his starting point. Walking directly back means that his displacement would be the same as the distance he ran, but in the opposite direction. So his displacement is -100.0 m (negative because he walked in the opposite direction).
John walked back to his starting point in 90.9 s, covering a distance of 100.0 m. From this information, we can calculate his walking velocity as follows:
Walking velocity = Distance / Time = 100.0 m / 90.9 s = 1.10 m/s
Since John walked in the opposite direction, his walking velocity would also be negative. So his walking velocity is -1.10 m/s.
To find his average velocity for the 200.0 m, we add up the running and walking velocities:
Average velocity = (Running velocity + Walking velocity) / 2
Average velocity = (10.15 m/s + (-1.10 m/s)) / 2
Average velocity = 9.05 m/s / 2
Average velocity = 4.53 m/s
Therefore, the magnitude of John Johnson's average velocity for the 200.0 m is 4.53 m/s.