Walk Away (2) - Tom & Linda stand at point A....

[onsamline]

Can someone please show me how to do this question using the "relative rates" method? The assumption here would be that since they are walking in a straight line, Tom can be viewed as "catching up" to Linda after 1 hour, so in 1 hour, Tom would be catching up to
R (linda's rate) X T = D
2 x 1 = 2 miles

From there, I tried to use the relative rate method

Difference in rate x T = D

(6-2) x T= 2 miles
T=1/2 hours

To cover twice of Linda's distance, Tom would need to cover 2 x 2 = 4 miles ,

using the relative rate method again

(6-2)x T = 4
T= 1

So the difference would be 1 - 1/2 = 1/2 hours = 30 minutes.

I seem to get the same answer, why is this incorrect?

1) Find Time for Linda
6T=2(T+1)
T=1/2

2) Find Linda's Distance in 1.5 hours
2 (.5 + 1) = D
D = 3 Miles

3) Find Tom's time to travel 3 miles (Linda's Distance)
3 miles/6mph = 1/2 or 30 minutes

4) Find Tom's time to travel double Linda's distance or 3x2 = 6 miles
6miles/6mph = 1 hour

5) Difference in times= 1 hour - 30 minutes = 30 min.

[RonPurewal]

4) Find Tom's time to travel double Linda's distance or 3x2 = 6 miles
6miles/6mph = 1 hour

^^^ that's the problem.

i.e., you're mistaking the "double her distance" case for meaning that it's twice the same distance you found in the first part.
not true -- we're talking about two separate times here:
* a time when tom has traveled the same amount of distance as linda has;
... and
* a LATER time when tom has traveled twice as much distance as linda has.

you can solve for these with algebra ... or you can just make a list every half-hour:

Minutes Tom Linda
0.......... 0...... 2 (since linda has already been walking for an hour)
30........ 3...... 3
60........ 6....... 4
90........ 9....... 5
120......12....... 6

at the blue time, tom has walked the same distance as linda has.

at the purple time (which is later), tom has walked twice the distance that linda has walked.

these are different times, so that's where your solution goes awry.

--

nb: in american english, there's no such thing as "twice of xxxxx"; it's just "twice xxxxxx".

[rustom.hakimiyan]

EDIT: Clarified the question.

Can someone please clear a few things up:

Question 1:

The main issue was when I tried to equate the two distances to solve for the first part was -- when setting up 2t+2 = 6t. The "+2" had me worried because although I was equating the two distances and solving for time, I was also putting in a "distance" variable within the R*T equation -- this was the distance that linda had originally traveled.

Luckily, that was the right move but I'm still a little shaky as to why it's OK. Conceptually, yes, I need to account for the 2miles that linda has already traveled but doesn't think mess around with a units a bit? Is it safer to put in the time value and equate it as 2(t+1) = 6t?

Solving the above does give t=.5 hours.

Question 2:

To solve for the second part, I had another issue with doubling the whole sum for just the RT part. What I mean by that is -- should the equation state:

Tom is traveling twice the distance that linda travels so I equated 2dTom = dLinda but that was making the wrong side bigger?



Question 3:


2(2(t))+ 2 = 6t and this was wrong. I was using the 2 for the first 2 miles that Linda had traveled and figured that didn't need to be doubled.

At this point, I started back up and solved with a time variable - "(2*2(t+1)) = 6t and that led to the right answer.

Question 4:


Is it incorrect to use the distance variable in the first part?

[RonPurewal]

Question 1:

The main issue was when I tried to equate the two distances to solve for the first part was -- when setting up 2t+2 = 6t. The "+2" had me worried because although I was equating the two distances and solving for time, I was also putting in a "distance" variable within the R*T equation -- this was the distance that linda had originally traveled.

Luckily, that was the right move but I'm still a little shaky as to why it's OK. Conceptually, yes, I need to account for the 2miles that linda has already traveled but doesn't think mess around with a units a bit? Is it safer to put in the time value and equate it as 2(t+1) = 6t?

Both variables described here are time variables. In fact, they are exactly the same variable.
Note that 2t + 2 and 2(t + 1) are exactly the same algebaic expression. They're the same.
You're deriving the same result"”with the same variable"”in two slightly different ways.

"- When you write "2(t + 1)", you're thinking "I go 2 miles/hr for t hours plus one additional hour". I.e., thinking of the (t + 1) hours as a single time period.

"- When you write "2t + 2", you're thinking "I go 2 miles/hr for t hours", and then thinking separately about the 2 miles you go during the extra hour.

If you think about these for a bit, you'll realize that there is no difference between them.

[RonPurewal]

Question 2:

To solve for the second part, I had another issue with doubling the whole sum for just the RT part. What I mean by that is -- should the equation state:

Tom is traveling twice the distance that linda travels so I equated 2dTom = dLinda but that was making the wrong side bigger?

This is another instance in which a little common sense goes a long way. Always think about the MEANING of what you're writing!
You can't let yourself sink into a state where you're just pushing meaningless algebraic expressions around the page.

"- Tom travels twice the distance. So, Tom travels farther than Linda.

- Now, just use common sense. Would you write...
... the longer distance = 2 x the shorter one?
... the shorter distance = 2 x the longer one?
Only one of these makes sense. Use that one.

[RonPurewal]

Question 3:


2(2(t))+ 2 = 6t and this was wrong. I was using the 2 for the first 2 miles that Linda had traveled and figured that didn't need to be doubled.

You have to double the entire distance traveled by Linda.

Linda doesn't travel just 2t miles; she travels (2t + 2) miles. So, (2t + 2) is the expression you have to double.

Question 4:


Is it incorrect to use the distance variable in the first part?

You didn't use any distance variables in the first part.
As explained above, you used the same time variable in both of your approaches.

[rustom.hakimiyan]

Question 3:


2(2(t))+ 2 = 6t and this was wrong. I was using the 2 for the first 2 miles that Linda had traveled and figured that didn't need to be doubled.

You have to double the entire distance traveled by Linda.

Linda doesn't travel just 2t miles; she travels (2t + 2) miles. So, (2t + 2) is the expression you have to double.

Question 4:


Is it incorrect to use the distance variable in the first part?

You didn't use any distance variables in the first part.
As explained above, you used the same time variable in both of your approaches.

Thanks, Ron. Very helpful clarifications.

[RonPurewal]

You're welcome