1. Distance Word Problem

A man "A" sets out from a certain point and traveled at the rate of 6 kph. After "A" had gone two hours , another man "B" set out to overtake "A" and went 4 km the first hour, 5 km the second hour, 6 km the third hour and so gaining 1 km every hour. How many hours after "B" left were they together?

I know the formula D = rt is needed.

Man A:

rate = 6 kmh

time = x + 2

Distance:

6(x + 2)

I am having trouble setting up Man B.

2. Re: Distance Word Problem

Originally Posted by harpazo
A man "A" sets out from a certain point and traveled at the rate of 6 kph. After "A" had gone two hours , another man "B" set out to overtake "A" and went 4 km the first hour, 5 km the second hour, 6 km the third hour and so gaining 1 km every hour. How many hours after "B" left were they together?

I know the formula D = rt is needed.

Man A:

rate = 6 kmh

time = x + 2

Distance:

6(x + 2)

I am having trouble setting up Man B.
Note: The problem doesn't say where Man B starts (tsk! tsk!) so I'll assume both Men A and B start from the same position.

There's probably a more "elegant" solution but we'll just have at it.
Check out where the men are every hour.

For instance, at the end of hour 3 Man A has traveled (6 kph)(3 h) = 18 km. At the end of hour 3 Man A has only traveled for 1 hour so he traveled (4 kph)(1 h) = 4 km.

At the end of hour 4 Man A has traveled 18 km + (6 kph)(1 h) = 18 km + 6 km = 24 km. Man B has traveled 4 km + (5 kph)(1 h) = 4 km + 5 km = 9 km.

Still not at the same distance. So do hour 5, 6, etc. until B catches up with A.

By the way, if you are doing anything involving units I will get on your case if you don't use them correctly. I'm a Physicist... unit analysis is in my DNA.

-Dan

3. Re: Distance Word Problem

Just out of curiosity, does the DNA you are born with determine whether or not you will become a physicist or does your DNA change when you become a physicist?

4. Re: Distance Word Problem

Originally Posted by HallsofIvy
Just out of curiosity, does the DNA you are born with determine whether or not you will become a physicist or does your DNA change when you become a physicist?
It's destiny. Our glorious Physics gifts were given to us by the gods as existence dawned to bless the World with our power and knowledge.

Of course standing in the beam of a particle accelerator might have something to do with it, too.

-Dan

5. Re: Distance Word Problem

Originally Posted by harpazo
A man "A" sets out from a certain point and traveled at the rate of 6 kph.
After "A" had gone two hours , another man "B" set out to overtake "A"
and went 4 km the first hour, 5 km the second hour, 6 km the third hour
and so gaining 1 km every hour.
How many hours after "B" left were they together?
Too busy or tired to draw a "speed = distance/time diagram" ?

Are you expected to solve by trial/error method (as Dan suggests)
or are you expected to solve using formula(s)?
If by formula, you'll need the arithmetic series formula.
The distance will be 12 + 6n representing A's distance.

6. Re: Distance Word Problem

Originally Posted by topsquark
Of course standing in the beam of a particle accelerator might have something to do with it, too.

-Dan
https://en.wikipedia.org/wiki/Anatoli_Bugorski

7. Re: Distance Word Problem

Speaking seriously, I didn't know anyone could survive a proton beam. Amazing.

-Dan

8. Re: Distance Word Problem

Thank you everyone.

9. Re: Distance Word Problem

Originally Posted by DenisB
Too busy or tired to draw a "speed = distance/time diagram" ?

Are you expected to solve by trial/error method (as Dan suggests)
or are you expected to solve using formula(s)?
If by formula, you'll need the arithmetic series formula.
The distance will be 12 + 6n representing A's distance.
Expected? I'm not in a math class with teacher expectations. I simply solve problems to increase my math skills, particularly with respect to word problems. Can you set up the equation(s) for me? I will do and show my work.

10. Re: Distance Word Problem

Arithmetic series formula:
SUM = n[2a + d(n - 1)] / 2
n = number of terms
a = 1st term
d = common difference between terms

With your problem: n = ?, a = 4, d = 1
and the SUM equals A's distance: 6n + 12 (6 miles per n + 12 miles lead).

So:
n[2a + d(n - 1)] / 2 = 6n + 12
Substitute a=4 and d=1, then solve for n.

Good notes.