How Long Must the First Arrival Wait in the Los Angeles to San Francisco Drive?

[oldunion]

Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 380mi away. He travels at a steady 45.0mph. Beth leaves Los Angeles at 9:00 a.m. and drives a steady 57.0mph.

they want to know how long does the first to arrive have to wait for the second?

I know the method to go about solving this problem ( i think) but it is unusable unless i were to know how far beth is from san fran. this looks to me like an equation that i don't know, but I am thinking something algebraic to solve for the missing distance.

for alan i said that 380mi/45mph would equal 8 hrs and 20 mins. beth, i don't know

 

[cepheid]

Dude, notice that BOTH Alan and Beth leave for San Francisco *from LA*, so Beth is just as far away from Frisco as Alan is...the information you seek was given to you in the words.

 

[oldunion]

Dude, notice that BOTH Alan and Beth leave for San Francisco *from LA*, so Beth is just as far away from Frisco as Alan is...the information you seek was given to you in the words.

ok which would mean that alan waited for 40 minutes? wrong. or 20 minutes, wrong again. and los angeles is a big frickin city, it could be 380, 390, 370 who knows how many miles. but if it was 380 like alan, and i divided that distance by her speed to get 6.6? and as easy and simple/logical as physics may seem to you, i have been trying to work these problems since yesterday at 3pm.

 

[dextercioby]

The time difference (in hours) is simply
\frac{380M}{45Mph}-\frac{380M}{57Mph}

Compute this number and express it in minutes...

Daniel.

 

[cepheid]

^That's only the difference between their travel times, dexter. But Alan left an hour before Beth, and you're supposed to figure out who arrives first, and how long he/she has to wait for the other.

380/45 = 8.444444444 hours

0.444444444... = 4/9

4/9 hours * 60 min/hour = 26.66666666667 = 27 min

08:00hrs + 08:27hrs = 16:27hrs = 4:27pm <----- Alan's arrival time

I wasn't trying to make fun of you. In the absence of any info regarding their respective locations in LA, it is reasonable to assume that Alan and Beth depart from more or less the same location. It's good that you are thinking above and beyond the problem, and outside the box. Your line of reasoning is realistic. But remember that physics often presents us with idealized cases and reasonable approximations. I agree that LA is very big, so the approximation is not that reasonable here, but they're trying to present you with a simple problem. Purposely witholding information to leave you guessing or force you to introduce new variables would be pretty sadistic.

Beth: 380/57 = 6.66666666667 = 6 2/3 hrs

2/3 hrs = 40 min

09:00hrs + 06:40hrs = 15:40hrs = 3:40pm <---- Beth's arrival time

Beth arrives first. Beth has to wait for Alan for 20 + 27 = 47 minutes.

I hope this answer is correct.

 

[oldunion]

you all have been an outstanding help, the answer was 47.6 which is correct as far as this problem goes. i will surely be back for more questions.

 

[oldunion]

380/45 = 8.444444444 hours

0.444444444... = 4/9

4/9 hours * 60 min/hour = 26.66666666667 = 27 min

08:00hrs + 08:27hrs = 16:27hrs = 4:27pm <----- Alan's arrival time

Your logic on Beth i understand, but i don't follow for Alan. where did .44444 come from. why times 60min/hr

 

[cepheid]

ok...sorry I couldn't get back to you yesterday.

Alan's travel time is 380/45 = 8.444444444444... as I said.

But 8.444444444...hours is an awkward way of expressing the time interval. We'd prefer to keep the eight hours, and express the remaining time in minutes. Just how many minutes is 0.44444... hours? Well, we know that 0.5 hours is 30 minutes, so it has to be less than thirty, but close. Note that 0.44444... = 4/9. So we have four ninths of an hour times sixty minutes PER hour (because that's how many minutes there are in an hour) gives you 26.6666...minutes, which we round up to 27.

So Alan traveled for 8.444444...hours, which expressed more conveniently is approximately 8 hours and 27 minutes.