An employee drove to work on Monday at the rate of 45 mph...

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KingAce

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PROBLEM: An employee drove to work on Monday at the rate of 45 mph and arrived one minute early. On Tuesday, he left home at the same time and drove at 40mph. He arrived at work one minute late.

a.) How far is it to work?
b.) At what speed would he have to travel to arrive 5 minutes early?

I assume I'd have to solve this problem by finding an equation where the time is y value and the x is the rate or speed (because time depends on speed). What type of equation would this be, and how can I solve this problem? Thanks so much. :)
 
Not quite, Jonboy: 1 minute = 1/60 hour.
Also, easier to use time DIFFERENCE = 2 minutes = 1/30 hour.

45 = d / t [1]

40 = d / (t + 1/30) [2]
 
Hi. I'm a little confused; do I set 45=d/t and 40=d/(t+1/30) equal to each other and solve? Or do I set 45(t-1) and 40(t+1) equal to each other and solve? Or do I do something entirely different? Please help- thanks!
 
KingAce said:
Hi. I'm a little confused; do I set 45=d/t and 40=d/(t+1/30) equal to each other and solve? Or do I set 45(t-1) and 40(t+1) equal to each other and solve? Or do I do something entirely different? Please help- thanks!
Click to expand...
Hmmm....you don't seem ready for this, OH King; you got these:
45 = d / t [1]
40 = d / (t + 1/30) [2]

[1]: d = 45t
[2]: d = 40(t + 1/30)

SO: 45t = 40(t + 1/30)
If you can't solve that for t, you need classroom/teacher help.
 
jonboy said:
Hmm I though that as long as the rate is constant in \(\displaystyle d\,=\,rt\) then it will work.
Click to expand...
Yes, BUT you can't mix hours with minutes: you were using hours in the speed
and minutes in the time...
 
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