Story Math Question. Yes, it is homework.

Last night three of us could not solve a story question in our son’s 7th grade math book.

Person A leaves home at 9am and he is walking 4 mph.

Person B leaves home a half hour later in the same direction running 8.5 mph.

How long will it take for B to catch A?

What would be the algebraic expression to figure this out?

(We cheated and went to the back of the book and got the answer, which was 9 something, but we could not figure out how they came up with this answer. )

Our son went to school with this unanswered, so I/we are not violating the TOS stuff.

You want to set up 2 equations for the positions of the walker/runner.

Starting with t=0 when the runner left the house I set up function 1 for the position of the walker wrt time. And then set up function 2 for the position of the runner wrt time.

Since when the runner catches up to the walker they will be at the same position you can set function 1 = function 2 and solve for t.

Assuming they’re leaving from the same house I got a time around 26 minutes.

x multiplied by 4 = (x-0.5) multiplied by 8.5

Here’s a way to think about it. At 9:30, when person B has just left, person A has already traveled 2 miles. The difference in speed between the two people is 4.5 mph. So how long does it take for person B to go 2 miles at 4.5 mph?

That won’t do it. We can’t assume that person A stopped walking after 30 minutes and is waiting for person B to catch up.

He hasn’t stopped walking - but the relative velocity is 4.5 between the two - so if you reduce the runner’s speed to 4.5, then the walker has relatively stopped.

I like this solution.

Joe

This method does work (ie. gives the same answer as my method :D). 4.5 mph is the speed of the runner relative to the walker.

Oh yeah, I hadn’t thought of it that way.

My husband and his boss ( an engineer with 30 years experience) both futz over this and came up with this:
Brother A leaves at 9:00am at 4 mph

Brother B leaves at 9:30am at 8.5 mph in the same direction

Distance = Rate x Time

We need to find out what time is (T)?

4mph x T = 8.5mph x ( T - 30 min) since the one brother left 30 min later

4T = 8.5T - 255

255 = 4.5T

255/4.5 = T

56.66 = T

56.66 minutes after brother A leaves, brother B will be next to him.

At 9:56am they will be side by side!

Naturally, I said, it would be sooner if there were a Starbucks in the path of A.

Thanks everyone.

**The Lurker Above **is using this equation:

4(t+30) = 8.5t

which gives 26.66 minutes (A is 30 minutes ahead of B).
ShirleyUjest is using this equation:

4t = 8.5(t-30),

which gives 56.66 minutes (B is 30 minutes behind A).

They give different answers because they’re answering different questions. Using the first equation, you’re answering the question: once B leaves*, ***how long will it take to catch A? Well, B leaves 30 minutes later, and it will take 26.66 minutes once he does, so 26.66 + 30 = 56.66.

The second equation finds the answer directly.

This is the same equation in my post .

Sorry! Sorry!

B can never catch A. It’s been proved.

At 9:30, when B sets out, A has walked 2 miles.
It will take 0.23529 hours for B to reach that point.
However, in that time, A has walked another 0.94196 miles.
To run that extra distance will take B another 0.11082 hours.
But in that time, A will have walked another 0.22163 miles.
And so on. The sequence never converges, so A and B never meet.

The problem statement says “How long will it take for B to catch A?”. That’s why I set t=0 for when B left the house. 26.7 minutes after B leaves and 56.7 minutes after A leaves are indeed the same time; but 26.7 minutes is the direct answer to the problem statement as I read it.

I’d mark both correct, so long as the answer statement specified which.

I suppose if showing our work were important, we should have converted time to hours first. We then would have gotten 0.944 hours, which is 56 minutes.

Thanks, Zeno.

Furthermore, at no point are A or B actually moving.

Good old Zeno…

I thought that only worked for tortoises.