# What is the intersection point of two objects on a position vs time graph?

• nmnna
In summary: One way to interpret part 3 is to assume that the brown dots represent the starting points of the two objects.As noted by BVU in the second post in this thread, the book answer for part 3 is correct if (as stated) the 2nd object's line is changed to go through (0,0), and also the original intersection between the objects' lines (12,11) stays the same.As shown below: green has done 8, red has done 11.
nmnna
Homework Statement
Given a position vs time graph.
1)Find ##\frac{v_{2}}{v_{1}}##
2)At what time the objects will meet?
3)Find ##\frac{\Delta{x_2}}{\Delta{x_1}}##, if the second object started moving from the origin.
Relevant Equations
##v = \frac{\Delta{x}}{t}##
The graph:

1) $$v_1 = \frac{\Delta{x}}{t} = \frac{5 - 3}{3} = \frac{2}{3}$$
$$2 = \frac{\Delta{x}}{t} = \frac{4 - 0}{5 - 1} = 1$$
$$\frac{v_2}{v_1} = \frac{1}{2/3} = \frac{3}{2} = 1.5$$

2)Points of intersection of the lines with the x-axis: ##I## (0; 3) and ##II## (0; -1), thus
$$\frac{2}{3}t + 3 = t - 1$$
$$t=\frac{-4}{(\frac{2}{3} - 1)} = 12 \ \text{seconds after starting motion}$$

I'm confused about the 3rd part.
If the second object starts moving from the origin then the graph should look something like this

We can clearly see that the objects met when 9 seconds passed and calculations lead to the same value as well:
$$\frac{2}{3}t + 3 = t$$
We know that the velocity of the first one is ##\frac{2}{3} \frac{\text{m}}{\text{sec}}##, we can find the distance it passed in 9 seconds, which is ##6## meters, for the second one I got ##9## meters. So the ratio ##\frac{\Delta{x_2}}{\Delta{x_2}} = \frac{9}{6} = 1.5##.
But the answer in my textbook is 1.375
Where I went wrong?
This is a translation from another language so if something is not clear please tell me.

Last edited:
nmnna said:
I'm confused about the 3rd part.
So am I . The only explanation I can give that comes out at the book answer is:
3) Assume the point of intersection is the same as in part 2 -- but object 2 started from the origin.

##\ ##

hmmm27 and Steve4Physics
BvU said:
So am I . The only explanation I can give that comes out at the book answer is:
3) Assume the point of intersection is the same as in part 2 -- but object 2 started from the origin.
I thought about this situation too, but by solving this i got ##\frac{12}{2/3\times12+3} = \frac{12}{11} \approx 1.09##, which is not the expected answer...

What is the position of the intersection point in 2) ?

##\ ##

BvU said:
What is the position of the intersection point in 2) ?

##\ ##

You can get their answer if you assume that object 2 starts at the position of the origin of the x-axis. (Don't change the original graph for object 2.) Assume object 1 starts at x = 3. The problem should have explicitly stated this if this is what they intended. With this interpretation, the two objects do not start their motions at the same time.

Last edited:
TSny said:
You can get their answer if you assume that object 2 starts at the position of the origin of the x-axis. (Don't change the original graph for object 2.) Assume object 1 starts at x = 3. The problems should have explicitly stated this if this is what they intended. With this interpretation, the two objects do not start their motions at the same time.
As I understand from your answer my solution should look something like this, right?
$$\frac{\Delta{x_2}}{\Delta{x_1}} = \frac{(2/3\times12) + 3}{(9\times1) - 1} = \frac{11}{8}$$
 This is wrong

I think in order to get the answer you want, you have to consider that the objects will meet at t=12sec. but the second object will have a different velocity. In that case you get : 11/8=1.375. The velocity of the second object has to be 11/12 so that : 12*(11/12)=11=(2/3)*12+3.

nmnna said:
As I understand from your answer my solution should look something like this, right?
$$\frac{\Delta{x_2}}{\Delta{x_1}} = \frac{(2/3\times12) + 3}{(9\times1) - 1} = \frac{11}{8}$$
PS This is wrong
I don't understand your calculation. The numerator appears to be calculating the position of object 1 at t = 12, but you're using this for ##\Delta x_2##.

One way to interpret part 3 is to assume that the brown dots represent the starting points of the two objects.

nmnna
As noted by BVU in the second post in this thread, the book answer for part 3 is correct if (as stated) the 2nd object's line is changed to go through (0,0), and also the original intersection between the objects' lines (12,11) stays the same.

Last edited:
nmnna
As shown below: green has done 8, red has done 11.

##\ ##

nmnna

## What is a position vs time graph?

A position vs time graph is a visual representation of an object's position over a period of time. It shows how the object's position changes over time, with time being plotted on the horizontal axis and position on the vertical axis.

## How do you interpret a position vs time graph?

To interpret a position vs time graph, you can look at the slope of the line. A steeper slope indicates a faster rate of change in position, while a flatter slope indicates a slower rate of change. The direction of the line also tells you the direction of the object's motion - a positive slope indicates motion in the positive direction, while a negative slope indicates motion in the negative direction.

## What can cause a curved line on a position vs time graph?

A curved line on a position vs time graph can indicate that the object is accelerating. This means that the object's rate of change in position is changing over time. The direction of the curve can also tell you the direction of the acceleration - a concave up curve indicates positive acceleration, while a concave down curve indicates negative acceleration.

## How can you calculate velocity from a position vs time graph?

Velocity can be calculated from a position vs time graph by finding the slope of the line at any given point. The slope represents the object's instantaneous velocity at that point in time. To find the average velocity over a period of time, you can find the slope of the line between two points on the graph.

## What is the difference between a position vs time graph and a velocity vs time graph?

A position vs time graph shows an object's position over time, while a velocity vs time graph shows an object's velocity over time. The slope of a position vs time graph represents velocity, while the slope of a velocity vs time graph represents acceleration. Additionally, a position vs time graph can have a curved line, while a velocity vs time graph will always have a straight line.

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