Suppose that a slug moves along a 1 meter elastic band and, after each minute, t
ID: 2944566 • Letter: S
Question
Suppose that a slug moves along a 1 meter elastic band and,after each minute, the elastic band is uniformly stretched by 1 additional
meter. If the slug moves 1 centimeter per minute, (i) will the slug ever reach
the end of the elastic band, and (2) if so, how many minutes will it take?
Hint: To solve this problem, rst nd the ratio of the distance travelled by
the slug to the total length of the elastic band each minute. Try to evaluate
the sum of these terms on a computer; how important is roundo error? (It
will be very dicult to accurately compute the sum!)
Explanation / Answer
THE SLUG IS AT A ..
HENCE THE DISTANCE TO TRAVEL = AB0=100 CM.
REMAINING DISTANCE IN TERMS OF FRACTION OF TOTAL LENGTH=100/100=1
2. AT T=1 MT.
IF B1 IS THE END OF THE BAND AT THIS TIME THEN AS PER GIVEN DATA
AB1 =100+100=2*100=200 CM.
THE SLUG IS AT A1 WHERE
AA1= 1 CM.AS PER INITIAL BAND POSITION ..
BUT AS THE BAND EXPANDED UNIFORMLY BY (2*100)/100=2 TIMES,THE POSITION IS A1' AFTER STRECH
AA1'=2 CM.
THE DISTANCE REMAINING TO TRAVEL = 200-2=198 CM.
REMAINING DISTANCE IN TERMS OF FRACTION OF TOTAL LENGTH=198/200=0.99
3. AT T=2 MT.
IF B2 IS THE END OF THE BAND AT THIS TIME THEN AS PER GIVEN DATA
AB2 =100+100+100=3*100=300 CM.
THE SLUG IS AT A2 WHERE
A1'A2= 1 OR AA2=AA1'+A1'A2=3 CM.AS PER PREVIOUS BAND POSITION ..
BUT AS THE BAND EXPANDED UNIFORMLY BY (3*100)/200=3/2 TIMES,THE POSITION IS A2' AFTER STRECH
AA2'=1.5*3=4.5 CM.
THE DISTANCE REMAINING TO TRAVEL = 300-4.5=295.5 CM.
REMAINING DISTANCE IN TERMS OF FRACTION OF TOTAL LENGTH=295.5/300
4. AT T=3 MT.
IF B3 IS THE END OF THE BAND AT THIS TIME THEN AS PER GIVEN DATA
AB3 =100+100+100+100=4*100=400 CM.
THE SLUG IS AT A3 WHERE
A2'A3= 1 OR AA3=AA2'+A2'A3=4.5+1=5.5 CM.AS PER PREVIOUS BAND POSITION ..
BUT AS THE BAND EXPANDED UNIFORMLY BY (4*100)/300=4/3 TIMES,THE POSITION IS A3' AFTER STRECH
AA3'=5.5*4/3=22/3 CM.
THE DISTANCE REMAINING TO TRAVEL = 400-22/3=1118/3 CM.
REMAINING DISTANCE IN TERMS OF FRACTION OF TOTAL LENGTH=1118/1200
5. AT T=4 MT.
IF B4 IS THE END OF THE BAND AT THIS TIME THEN AS PER GIVEN DATA
AB4 =100+100+100+100+100=5*100=500 CM.
THE SLUG IS AT A4 WHERE
A3'A4= 1 OR AA4=AA3'+A3'A4=(22/3)+1=25/3 CM.AS PER PREVIOUS BAND POSITION ..
BUT AS THE BAND EXPANDED UNIFORMLY BY (5*100)/400=5/4 TIMES,THE POSITION IS A4' AFTER STRECH
AA4'=(5/4)*(25/3)=125/12 CM.
THE DISTANCE REMAINING TO TRAVEL = 500-125/3=1375/3 CM.
REMAINING DISTANCE IN TERMS OF FRACTION OF TOTAL LENGTH=1375/1500
5.AT T=5 MT.
IF B5 IS THE END OF THE BAND AT THIS TIME THEN AS PER GIVEN DATA
AB5 =100+100+100+100+100+100=6*100=600 CM.
THE SLUG IS AT A5 WHERE
A4'A5= 1 OR AA5=AA4'+A4'A5=(125/12)+1=137/12 CM.AS PER PREVIOUS BAND POSITION ..
BUT AS THE BAND EXPANDED UNIFORMLY BY (6*100)/500=6/5 TIMES,THE POSITION IS A4' AFTER STRECH
AA4'=(6/5)*(137/12)=822/60 CM.
THE DISTANCE REMAINING TO TRAVEL = 600-822/60=35178/60 CM.
REMAINING DISTANCE IN TERMS OF FRACTION OF TOTAL LENGTH=35178/36000
IT IS NOT A CONVERGING SEQUENCE
HENCE THE SLUG CAN NEVER REACH THE END IN ANY FINITE TIME
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