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the I.Sc lecture theatre of a college is 40 ft wide and has a door at a corner. A teacher enters at 12.00 noon through the door and makes 10 rounds along the 40 ft wall back and fortg during the period and finally leaves the classroom at 12.50 p.m. through the same door. compute his average speed and average velocity.
Total time for which the teacher moves within the room, t = 50 min = 3000 sIn one round the teacher covers = 40 + 40 = 80 ftSo, the total distance covered by the teacher in 10 rounds is = 10 × 80 = 800 ftNow, average speed = total distance/total time = 800/3000 = (4/15) ft/sNow, for average velocity, the net displacement of the teacher is zero. Since, he leaves the room through the same door through which he entered, the average velocity will be zero.
Total time for which the teacher moves within the room, t = 50 min = 3000 s
In one round the teacher covers = 40 + 40 = 80 ft
So, the total distance covered by the teacher in 10 rounds is = 10 × 80 = 800 ft
Now, average speed = total distance/total time = 800/3000 = (4/15) ft/s
Now, for average velocity, the net displacement of the teacher is zero. Since, he leaves the room through the same door through which he entered, the average velocity will be zero.
In one round teacher cover 40 +40 =80 ft Doin 10 rounds it will cover 80×10 =800 Avg speed =800÷50=16ft/minSince there was no displacmdisp therefore avg velocity is zero
total speed travelled in 50mon = 800 ft.average speed = 800/50 ft/min = 16 ft/min.At 12:00 noon he is at the door and at 12:50 he is again at the same door.therefore the displacementduring the 50 mins is 0.Average velocity is also equal to 0
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