Thread: Distance,Time,Rate Story Problem

Hello,

Another fine day for math isn't it? Well thank you all for helping me ahead of time!

Story Problem:

You can jog 8 miles in the same time that it takes to travel 16 miles by car. IF the car's rate is 9 miles per hour faster than your jogging rate, find the average rate of each.

Things we know:

Jogging rate = X
Cars Rate = X+9
Distance jogging = 8 Miles
Distance Car = 16


We don't know over all time and general rate...

I was thinking I could take the time and set it equal to 1 and go from there so:

Jogging Rate:

1=(1/8) / X : So X=1/8

Cars Rate:

(1/8) + 9 = 9.125


Is that correct? .125 and 9.125?

It REALLY helps to write complete definitions.

Distance is in Miles
Time is in Hours

X = Jogging Rate in mph -- This would be a little better.

X mph = Jogging Rate -- I would prefer this

X mph + 9 mph = Car Driving Rate

Now, WRITE IT DOWN!!

Distance = Rate * Time

8 Mile = X mph * Time
16 Mile = (X mph + 9 mph) * Time

We know already that "Time" is the same in each.

Now, what on earth is that 1/8 mph business? First, it takes 64 hours to get 8 miles at this speed. Second, the units aren't even mph. What are they? It is all very confusing. Never do that. Third, you're going to get a lot more than 16 miles at 9.125 mph for 64 hours. You do need to make sure your answers make sense.

Originally Posted by TKHunny

It REALLY helps to write complete definitions.

Distance is in Miles
Time is in Hours

X = Jogging Rate in mph -- This would be a little better.

X mph = Jogging Rate -- I would prefer this

X mph + 9 mph = Car Driving Rate

Now, WRITE IT DOWN!!

Distance = Rate * Time

8 Mile = X mph * Time
16 Mile = (X mph + 9 mph) * Time

We know already that "Time" is hte same in each.

Now, what on earth is that 1/8 mph business? First, it takes 64 hours to get 8 miles at this speed. Second, the units aren't even mph. What are they? It is all very confusing. Never do that. Third, you're going to get a lot more than 16 miles at 9.125 mph for 64 hours. You do need to make sure your answers make sense.

Okay..... Well......... I'm sorry that I didn't reach your expectation on how to format it? I'm not a math instructor? And as for the 1/8 it was taking what I knew from another problem and trying to apply it to this problem.
So if you don't mind; There is no need for the all CAPS! Because as fair as I can see- I'm back at square one because I don't know what time is, nor the X in rate to figure it out.

If I take

Dist = Rate * Time

That gets me to (Let X = Rate for Jogging)

8= X * T or ( Let X = Rate for Jogging + 9 for Car )

16 = (x+9) * T


So now I'm really confused....

Problem: You can jog 8 miles in the same time that it takes to travel 16 miles by car. IF the car's rate is 9 miles per hour faster than your jogging rate, find the average rate of each.
Let x be the rate of "you" and y be the rate of the car.
Distance=rate*time
Distance/rate=time
8/x=16/y
x+9=y
-------------------------------
solve:
8y=16x
8(x+9)=15x
8x+72=15x
72=7x
x=72/7 mph
y=1152/56=288/14=144/7mph

Originally Posted by cpuboye11

I'm sorry that I didn't reach your expectation on how to format it?

It's not an expectation. It's a helpful hint. LEARN how to structure so that the notation helps you. I am NOT talking about formatting. I'm talking about organizing in a useful way, to your advantage.

as for the 1/8 it was taking what I knew from another problem and trying to apply it to this problem.

That's nice, but why do you believe the 1/8 has any transferability? CLEAR definitions will help you with such decisions.

There is no need for the all CAPS!

Too bad you just ignored me. I guess I should use DOUBLE CAPS!

8= X * T or ( Let X = Rate for Jogging + 9 for Car )

16 = (x+9) * T

Confused about what? You have it! Now solve it.

Seriously, your life will be improved if you carry the units, rather than just hoping that you have them right.

Hello, cpuboye11!

You can jog 8 miles in the same time that it takes to travel 16 miles by car.
If the car's rate is 9 mph faster than your jogging rate, find the average rate of each.

Here's a back-door solution . . .

Let = your jogging speed.
Then = your driving speed.

We are told: you can drive twice as far as you can jog.
Hence, your driving speed must be twice your jogging speed.

There is an equation . . .


By the way, don't get your shorts in a knot.
Tkhunny used caps in one brief phase.
It was for emphasis, not to insult or offend.

If you are offended by italics, disregard this post.

You guys act like math is nothing. But for some people such as my self its like a knife being shoved in to my head. So I might be over acting but when you sitting in front of a math book for 2 hrs working on 1 out of 50 problems and the book doesn't do crap for you and you think, maybe someone can explain this to you a little bit better then the book. And then when you ask and you get a bunch of smart a$$ responses in an already stressful situation- the human mind tends to react as such.

So my apologies, I will keep my under-formatted and apparently dumb questions to my self. Thank you.

Request for thread closed.

What you are saying is that you come here asking for help but don't actually want any help?

We know what it's like to stare and struggle. We know how to overcome these struggles and stop staring.
Any hint we give is not designed to offend. It actually will help.
Notation is important. It will help you start knowing and stop staring.
Learning to communicate ideas clearly is part of the study of mathematics. The more you learn, the better it will be for you.
There are no contributors here who are offended by bad notation or lack of understanding. It's why we're here, to help you improve on these things.

Recommendation: Let's ignore any impressions of offense and learn some mathematics.