Distance formula question, word problem

[OMGMathPLS]

9:00 AM car A travels north 50 mph
and is 50 miles south of car B
Car B is traveling west at 20 mph

Let (0,0) be initial coordinates of car B on xy plane
units are in miles
Plot locations of each car at 9:00 AM and 11:00 AM

find distance of cars at 11:00 AM.
So I use the midpoint formula but I'm not sure if it matters which one to plug in x1 and x2

I have a 50, another 50 and a 20 but I only have 2 variable spots over a 2.

Can I have some help with this please? Thank you

 

[MarkFL]

You are being asked to find the distance between the cars at 11 am, so you would use the distance formula, not the mid-point formula.

What are the coordinates of both cars at 9 am? and at 11 am?

 

[OMGMathPLS]

Ok, so still the distance.

The coordinates of car a at 9 AM it does not say. i think that's what I'm trying to find out.

 

[MarkFL]

Ok, so still the distance.

The coordinates of car a at 9 AM it does not say. i think that's what I'm trying to find out.

You are told to orient your coordinate axes such that at 9 am, car B is at the origin, which has coordinates $(0,0)$. If car A is 50 miles south of car B at 9 am, then what would its coordinates be?

 

[OMGMathPLS]

You are told to orient your coordinate axes such that at 9 am, car B is at the origin, which has coordinates $(0,0)$. If car A is 50 miles south of car B at 9 am, then what would its coordinates be?

Does it matter if 0, 0 is put in as the first or second?

I think the other pair would be (-50, 20)?

 

[MarkFL]

Does it matter if 0, 0 is put in as the first or second?

I think the other pair would be (-50, 20)?

If car A is due south of car B, then both cars lie along the same vertical line, this means their $x$-coordinates are the same. Now, since car A is south, which I would take as being below on the graph, you want to plot the point 50 units below the origin for the location of car A. What would this point be?

 

[OMGMathPLS]

That's going to be negative -50 because it's -50 south (down) on the number line from 0,0 where car B is.

- - - Updated - - -

I mean the Y line not number line.

 

[MarkFL]

That's going to be negative -50 because it's -50 south (down) on the number line from 0,0 where car B is.

- - - Updated - - -

I mean the Y line not number line.

Correct! :D

So, at 9 am car A is at $(0,-50)$, and car B is at $(0,0)$. Now, given the information on the speed and direction of both cars, what will their coordinates be two hours later, at 11 am?

 

[OMGMathPLS]

View attachment 3317It doesn't matter which order I plug it in as does it?

like if 0, 0 can be set 1 for x1, y1 or x2, y2 does it matter or does it come out the same?

 

[MarkFL]

It doesn't matter which order I plug it in as does it?

like if 0, 0 can be set 1 for x1, y1 or x2, y2 does it matter or does it come out the same?

We're really not ready to plug anything in yet...we first need to determine the coordinates of the two cars at 11 am, since this is the time for which we are asked to find the distance between the cars.

But the order does matter within the pairs, and the coordinates are even referred to as "ordered pairs." The $x$-coordinate is listed first, then the $y$-coordinate for each point, in the form $(x,y)$.

Once you have the two points at 11 am, the suppose car A is at $\left(x_A,y_A\right)$, and car B is at $\left(x_B,y_B\right)$. Then the distance between them would be:

$$d=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2}$$

The order of the two $x$ and two $y$ coordinates doesn't matter, but corresponding coordinates must be paired together. In other words, one term under the radical must be the square of the difference betwee $x$-coordinates and the other term must be the square of the difference between $y$-coordinates.

 

[OMGMathPLS]

We're really not ready to plug anything in yet...we first need to determine the coordinates of the two cars at 11 am, since this is the time for which we are asked to find the distance between the cars.

But the order does matter within the pairs, and the coordinates are even referred to as "ordered pairs." The $x$-coordinate is listed first, then the $y$-coordinate for each point, in the form $(x,y)$.

Once you have the two points at 11 am, the suppose car A is at $\left(x_A,y_A\right)$, and car B is at $\left(x_B,y_B\right)$. Then the distance between them would be:

$$d=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2}$$

The order of the two $x$ and two $y$ coordinates doesn't matter, but corresponding coordinates must be paired together. In other words, one term under the radical must be the square of the difference betwee $x$-coordinates and the other term must be the square of the difference between $y$-coordinates.

I plugged it in. It didn't work. Please see the paper above.

 

[MarkFL]

I plugged it in. It didn't work. Please see the paper above.

You first have to find the coordinates of the cars at 11 am. Until you do that you are not ready to use the distance formula.

Let's look at car A first. It begins at $(0,-50)$, and then travels north at 50 mph for two hours. What will its new coordinates be?

 

[OMGMathPLS]

You first have to find the coordinates of the cars at 11 am. Until you do that you are not ready to use the distance formula.

Let's look at car A first. It begins at $(0,-50)$, and then travels north at 50 mph for two hours. What will its new coordinates be?

Will it be 0, 0?

 

[MarkFL]

Will it be 0, 0?

No, at 50 mph car A will travel 100 miles in 2 hours. We can see this by using the relationship between distance, constant speed and time:

$$d=vt$$

$$d=50\,\frac{\text{mi}}{\text{hr}}\cdot2\text{ hr}=100\text{ mi}$$

What point is 100 units due north of $(0,-50)$?