1. ## Two Cars

Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car's average speed is 10 mph more than the other's. The faster car arrives st Wildwood at 11:00 am, 1/2 hour before the other car. What was the average speed of each car? How far did each travel?

Car 1: rate = x + 10; time = 3 hours

Car 2: rate = x; time = 3.5 hours

Equation to find the average speed of each car is:

3/(x + 10) = x/(3.5)

After finding x, I can use D = rt TWICE to find how far each car travelled.

Correct?

2. ## Re: Two Cars Originally Posted by harpazo Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car's average speed is 10 mph more than the other's. The faster car arrives st Wildwood at 11:00 am, 1/2 hour before the other car. What was the average speed of each car? How far did each travel?

Car 1: rate = x + 10; time = 3 hours

Car 2: rate = x; time = 3.5 hours

Equation to find the average speed of each car is:

3/(x + 10) = x/(3.5)
x is a rate with units of, say, miles/hour while "3" and "3.5" are times, in hours. The fraction on the left has units of hours/(miles per hour)= hours^2/miles while the fraction on the right has units of (miles/hour)/hours= miles/hour^2. They can't be equal!

After finding x, I can use D = rt TWICE to find how far each car travelled.

Correct?
No, your equation makes no sense. If the first car has speed x+ 10 and the distance from Commercial Boulevard to Wildwood is "d" then the time taken is d/(x+ 10). The time taken by the second car, at speed x is d/x. Saying that the first car arrived at 11:00 means that d/(x+ 10)= 3. Saying the first car arrived 1/2 an hour ahead of the second means that d/x= 3.6. Solve those equations for x and d.

Exactly how are you interpreting " How far did each travel?" In what time span? The "d" above is the distance the first car travels in 3 hours. Are you interpreting this as asking how far the second car traveled in that same time?

3. ## Re: Two Cars

Next time you get one of these, DRAW A DIAGRAM (as is, you seem to be guessing!)

(8.00am)@x.............3.5x.............>11.30am

(8.00am)@x+10.......3(x+10)........>11.00am

3(x+10) = 3.5x

4. ## Re: Two Cars Originally Posted by HallsofIvy x is a rate with units of, say, miles/hour while "3" and "3.5" are times, in hours. The fraction on the left has units of hours/(miles per hour)= hours^2/miles while the fraction on the right has units of (miles/hour)/hours= miles/hour^2. They can't be equal!

No, your equation makes no sense. If the first car has speed x+ 10 and the distance from Commercial Boulevard to Wildwood is "d" then the time taken is d/(x+ 10). The time taken by the second car, at speed x is d/x. Saying that the first car arrived at 11:00 means that d/(x+ 10)= 3. Saying the first car arrived 1/2 an hour ahead of the second means that d/x= 3.6. Solve those equations for x and d.

Exactly how are you interpreting " How far did each travel?" In what time span? The "d" above is the distance the first car travels in 3 hours. Are you interpreting this as asking how far the second car traveled in that same time? Originally Posted by DenisB Next time you get one of these, DRAW A DIAGRAM (as is, you seem to be guessing!)

(8.00am)@x.............3.5x.............>11.30am

(8.00am)@x+10.......3(x+10)........>11.00am

3(x+10) = 3.5x
I am selecting questions from the textbook that the author did not give an example for. This is what matters to me. Selection of questions that the author has fully explained in the chapter example section is too easy. My goal is to rationalize my way through word problems not given as examples at the beginning of each chapter.

5. ## Re: Two Cars Originally Posted by DenisB Next time you get one of these, DRAW A DIAGRAM (as is, you seem to be guessing!)

(8.00am)@x.............3.5x.............>11.30am

(8.00am)@x+10.......3(x+10)........>11.00am

3(x+10) = 3.5x
HallsofIvy.

6. ## Re: Two Cars

3(x+10) = 3.5x

3x + 30 = 3.5x

3x - 3.5x = -30

-0.5x = -30

x = -30/-0.5

x = 60

Distance travelled by car 1:

d = 3x + 30

d = 3(60) + 30

d = 180 + 30

d = 210 miles

Distance travelled by car 2:

d = 3.5x

d = 3.5(60)

d = 210 miles

Average speed for car 1:

speed = distance/time

speed = 210/3

Speed = 70 mph

Average speed for car 2:

Speed = distance/time

Speed = 210/3.5

Speed = 60 mph

7. ## Re: Two Cars

For these sorts of problems, ie when you have three variables (eg speed, distance, time) and know a relationship between them (eg s=d/t), I always like to set up a table such as: and develop your equation from there.

Since Distance = Speed x time, we get d =3.5x for Car1 and d=3(x+10) for Car2.

Therefore:

3.5x = 3(x+10) and go from there.

8. ## Re: Two Cars Originally Posted by harpazo Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car's average speed is 10 mph more than the other's. The faster car arrives st Wildwood at 11:00 am, 1/2 hour before the other car. What was the average speed of each car? How far did each travel?
Car 1: rate = x + 10; time = 3 hours
Car 2: rate = x; time = 3.5 hours
Equation to find the average speed of each car is:
3/(x + 10) = x/(3.5)
Next time you try something similar, equate the DISTANCE instead:
3(x + 10) = 3.5x

Much easier...

9. ## Re: Two Cars Originally Posted by Debsta For these sorts of problems, ie when you have three variables (eg speed, distance, time) and know a relationship between them (eg s=d/t), I always like to set up a table such as: and develop your equation from there.

Since Distance = Speed x time, we get d =3.5x for Car1 and d=3(x+10) for Car2.

Therefore:

3.5x = 3(x+10) and go from there. Originally Posted by DenisB Next time you try something similar, equate the DISTANCE instead:
3(x + 10) = 3.5x

Much easier...
See my full reply above. Is it right?

10. ## Re: Two Cars Originally Posted by harpazo See my full reply above. Is it right?
Ya. But way too long...
On a timed test, you'd be in trouble!

We're trying to lend you an electric shovel to dig a hole,
but you keep using a kitchen fork 11. ## Re: Two Cars Originally Posted by DenisB Ya. But way too long...
On a timed test, you'd be in trouble!

We're trying to lend you an electric shovel to dig a hole,
but you keep using a kitchen fork DenisB:

I am 53 years old. What timed test are you talking about? I ended my school days in the Fall semester 1993.

12. ## Re: Two Cars

Ya...but if you do start tutoring, you'll need to know the "quicker ways"
since students are given "timed tests"...like 1/2 hour for 6 problems.

Your age 53 is young...I'm 77 !

13. ## Re: Two Cars

See my full reply above. Is it right?

Yes it is!

14. ## Re: Two Cars Originally Posted by DenisB Ya...but if you do start tutoring, you'll need to know the "quicker ways"
since students are given "timed tests"...like 1/2 hour for 6 problems.

Your age 53 is young...I'm 77 !
I had no idea that you are in your 70s. You are young too compared to other members. I recall Soroban telling me back in 2006 that he was in his 80s. BTW, what happened to MHF member Soroban? He was my online tutor at the time.

15. ## Re: Two Cars

YES: I remember Soroban well; inactive for a few years now. Quite the guy!!