1. When Kyle saunters to class at an average speed of 48 meters per minute, he arrives 9 seconds late. When he hustles to class at an average speed of 64 meters per minute, he arrives 9 seconds early. At what average speed
should Kyle travel in order to arrive at class exactly on time?
He should get to class early clean teh greenboard for his teacher and maybe she'll runaway to Mexico with him.
To get teh teacher to give him a good grade he must do 6inches per second per second for at least 11 minutes. Sleeping in teh wet spot earns extra credit.
56m/m
I get 54.957.
54.857 or 384/7 to be exact.
Would one of the last two posters please explain your answer? I got 54.945. I figured it was wrong because of how uncertain I was, but I know it's close.
54.857
Your right, it is 54.857. I made a typo.
I used the method of distance = velocity * time. The distance is the same whether he is late or early. Thus:
48(t+9)=64(t-9)
Rearrange and solve for t, and you should get t=63. Now solve for the distance:
48(63+9)=3456
and the speed would be distance over time: 3456/63.
I'm sure there are more elegant solutions, but that works.
only confusing due to the phrasing of the question. try: Two people walk to class at 48 m/m and 64 m/m and get there eighteen seconds apart, how far did they walk?
classic trick question.
chaz22 wrote:
only confusing due to the phrasing of the question. try: Two people walk to class at 48 m/m and 64 m/m and get there eighteen seconds apart, how far did they walk?
Answering this question would get you a distance. The original question asks for an answer in terms of speed which requires solving for distance and for time and then for speed. Your question is phrased more simply because it is a more simple question.
d/48=t+9
d/64=t-9
Add:
d(1/48+1/64)=2t
d/t=2*16*12/7=384/7
More importantly, someone needs to ask this teacher how walking at 25 minute 1600m pace (64m/m) could possibly be considered hustling.
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