It's silly that I've gone years and years and I still don't understand it...but could someone explain distance problems to me?

- anonymous

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Please pose a specific distance problem...
Such as...?

- anonymous

I don't have an example off the top of my head... but just like, Uhm...say two trains are heading toward each other. One is going 40 MPH and the other is going 25MPH. How long would it take for them to meet?

- anonymous

Distance divided by time=speed. Time times speed=Distance

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Insufficient info. So, what you ought to do is find a problem you want to work and post your question. Just knowing the speed won't tell you enough to figure out the time. what else do you need? What if one train is pointed east in SF and the other is pointed West in NYC? or what if one is in Missouri? See?

- anonymous

Oh! I see. Terribly sorry. Let's say, to begin with, they are 170 MPH apart. One heading north, one heading south.

- anonymous

MPH? Sorry- I mean miles.

- anonymous

Try an easier problem. Your car is at home. The grocery store is 10 miles away. If your route is a straight line at constant speed of 10 miles per hour, how long will it take you to travel to the store?

- anonymous

One hour. I understand that, and I understand the formula, but it's when there are two different objects involved. But thank you none-the-less.

- anonymous

Okay, moving on up...

- anonymous

So, with two objects moving you just have two separate parts of the problem just like going to the store.

- anonymous

What if the store was very special, a mobile store in a van. It is also moving toward you and departing at the same time as you, also at 10 MPH. How long till you meet the store on your path?

- anonymous

Half an hour.

- anonymous

You got it.
Now, you can change the speeds of the objects and figure that out.

- anonymous

See where this is going?

- anonymous

Okay, thank you lots! I do indeed. I just never knew where to begin and how to set it up.

- anonymous

Let's say now that you drive at 20 MPH and the store moves toward you at 10 MPH. When will you meet?

- anonymous

Oh my... Would it be fifteen minutes?

- anonymous

You can get a feel for it by trial and error... You will eventually find the moment of meeting, and you will get it!

- anonymous

try the distance formula that kandybabii gave above to confirm your answer.

- anonymous

hint: use it twice. Once for the store, and once for your car.

- anonymous

And so, once I get the 1/2hour for me and the 1hour for the van, what do I do with them?

- anonymous

Use the formula d/t = s
rearrange it to st = d
after 15 minutes, you see there's a gap still?

- anonymous

20 MPH x 1/4 hour = 5 miles
10 MPH x 1/4 hour = 2.5 miles.
Still 2.5 miles apart.
Need more time.
Lather rinse and repeat until they meet at the same time, is the way to get the hang of it.

- anonymous

Or rather, adjust the time until the distances sum to 10.

- anonymous

Okay. Hmm.

- anonymous

I just don't understand...still. :/

- anonymous

try 17 minutes. Use 17/60 ths hours instead of 1/4. See? and just try until the distances of each add to 10, the original distance apart. 10 has to be covered regardless of who goes fast and who goes slow in order for them to meet. That's the problem, more generally.

- anonymous

Okay. I think my problem is I'm a visual learner...so it's more difficult with math. I think in the morning, I'm going to get my friend to help. In my old math class we used to set up a chart, and I was going to do that, but soon realised I didn't remember where the information went.

- anonymous

20mph * x hours + 10mph * x hours = 10 miles

- anonymous

OHH. That helps!

- anonymous

20x makes a line from point a
10x makes a line from point b
change x until they connect in between point a and point b at point M for Meeting Point.

- anonymous

With word problems, you have to peel the story away to be left with numbers and units.

- anonymous

Here's a guy who has a website with tons of them on video, and he's great and trustworthy.
http://www.khanacademy.org/video/two-passing-bicycles-word-problem?playlist=Algebra

- anonymous

Yeah, I can tell. So, with that equation, you'd have to insert actual numbers for 'x' before you can multiply and say 20x + 10x = 10 and then 30x = 10 and just get 1/3?

- anonymous

And thank you for the video :)

- anonymous

You are right.

- anonymous

1/3 hours is 20 minutes.

- anonymous

So, you know how to do distance problems after all. Good Job!

- anonymous

Oh yay!

Looking for something else?

Not the answer you are looking for? Search for more explanations.