Multiple choice

A 1.75 m3 capacity tractor loader has a forward loaded speed of 240 m/min, returning unloaded speed of 300 m/min and operates at 80% of the specified speed. It hauls earth over a distance of 60 m with fixed time per trip being 25 seconds. What is its effective cycle time?

  1. 54.25 seconds

  2. 55.50 seconds

  3. 56.75 seconds

  4. 58.75 seconds
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D Correct answer
Explanation

Speed = 80% of specified. Forward speed = 240 * 0.8 = 192 m/min. Return speed = 300 * 0.8 = 240 m/min. Distance = 60m. Time forward = 60/192 = 0.3125 min = 18.75 sec. Time return = 60/240 = 0.25 min = 15 sec. Fixed time = 25 sec. Total = 18.75 + 15 + 25 = 58.75 sec.

AI explanation

Using the formula time equals distance divided by speed, the loaded forward time at 80% of 240 m/min is 60 divided by 192, which equals 0.3125 minutes, and the return time at 80% of 300 m/min is 60 divided by 240, which equals 0.25 minutes. Adding these two travel times to the fixed time of 25 seconds gives a total of 0.5625 minutes plus 25 seconds. Converting 0.5625 minutes to seconds gives 33.75 seconds, and adding the 25 fixed seconds results in an effective cycle time of 58.75 seconds.

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