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Question 986425: An ant climbs to the very end of the second hand on a wall mounted clock at exactly 9:15:00. The second hand is 13.5 cm long. When the time is exactly 9:17:30, (a) what distance did the ant travel? (b) What is the ant's average velocity at this time? (e) At what time would the ant have traveled a distance of 100. m (give your answer to the nearest second).
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! At 9:17.30, the second hand has traveled 900 degrees. I am going to assume the minute is defined by the second hand's being at the 12 o'clock position. The distance traveled is 2.5 circumferences, and each circumference is 2*pi*13.5 cm=84.823 (I won't round yet) cm. Therefore, the total distance is 212.058 (rounded) cm
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The average velocity was 212.058 cm (units in cm/sec)/150 sec, or 1.414 cm/sec or 0.01414 m/sec
It would require 100m/0.01414 m/sec (units in seconds)=7073.55 sec or 117.89 minutes which is 117m53s. That would be at 11:12.53. Rounding earlier will change the time.
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rough check.
In an hour, the second hand goes around 60 times and that is 120*pi*13.5 cm=5089 cm or 50.89m
In two hours, it would be 101.78m, but this is 2 minutes fewer than 2 hours, and in a minute, the circumference is 84 cm, so in two minutes it would be 168 cm. The answer makes sense. Seven additional seconds is 10 cm.
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