Asked by Gino
From t = 0 to t = 3 min 25 seconds, a man stands still, and from t = 3 min 25 seconds to t = 7 mins 35 s, he walks briskly in a straight line at a constant speed of 1.65 m/s. What is the magnitude of his average velocity vavg over this entire time interval?
Answers
Answered by
Henry
T = 3 25/60 + (7 35/60-3 25/60)
T = 3.417 + (7.583-3.417) = 7.583 Min.
= 455 s.
t = (7.583-3.417)min * 60s/min = 250 s.
d = V*t = 1.65m/s * 250s = 412.4 m.
V = d/T = 412.4m/455s = 0.906 m/s.
T = 3.417 + (7.583-3.417) = 7.583 Min.
= 455 s.
t = (7.583-3.417)min * 60s/min = 250 s.
d = V*t = 1.65m/s * 250s = 412.4 m.
V = d/T = 412.4m/455s = 0.906 m/s.
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