How to solve this unsolved time & distance problem?

[Benjamin_harsh]

Homework Statement

A and B leave their places and start moving towards each other. If they are 50m apart after both 2 min and 3 mins, how far are their places?

Relevant Equations

So, at 2 minutes - distance traveled by A = 2x and similarly by B = 2y

A and B leave their places and start moving towards each other. If they are 50m apart after both 2 min and 3 mins, how far are their places?

I can able to draw a diagram for this problem:

Take distance between their places as D

Speed of A = x meter /minutes
Speed of B = y meter/minutes

So, at 2 minutes - distance traveled by A = 2x and similarly by B = 2y
From here, how to proceed?

 

[Mark44]

Homework Statement:: A and B leave their places and start moving towards each other. If they are 50m apart after both 2 min and 3 mins, how far are their places?
Homework Equations:: So, at 2 minutes - distance traveled by A = 2x and similarly by B = 2y

A and B leave their places and start moving towards each other. If they are 50m apart after both 2 min and 3 mins, how far are their places?

Presumably, since the times are different, one of the two started one minute before the other one.

I can able to draw a diagram for this problem:

View attachment 253786Take distance between their places as D

Speed of A = x meter /minutes
Speed of B = y meter/minutes

So, at 2 minutes - distance traveled by A = 2x and similarly by B = 2y
From here, how to proceed?

I think that the best you can do with this problem is to write an equation that represents the distance D in terms of the distance A has gone plus the distance B has gone plus the remaining 50 meters between them. You'll have one equation with two unknowns, so you won't be able to get a numeric value for D without further information.

 

[Ibix]

I think the point is that at 2 minutes, they still have 50m to go to meet each other and at 3 minutes they have passed each other and are 50m apart again.

This let's you write down two equations for position which can be solved simultaneously.

 

[Benjamin_harsh]

At 2 minutes, they still have 50m to go to meet each other

At 3 minutes they have passed each other and are 50m apart again.

How to write equations for this 2 sentences?

They didn't mention total distance also.

 

[Ibix]

If you start at position $x_0$ and move at velocity $v$, what's your position at time $t$?

 

[Benjamin_harsh]

If you start at position $x_0$ and move at velocity $v$, what's your position at time $t$?

Position will be $vt$

 

[Ibix]

Regardless of $x_0$?

 

[Mark44]

I think the point is that at 2 minutes, they still have 50m to go to meet each other and at 3 minutes they have passed each other and are 50m apart again.

That's not how I interpret what the OP wrote:

If they are 50m apart after both 2 min and 3 mins, how far are their places?

@Benjamin_harsh, what does this mean?
I interpreted this to mean that one of them had been moving for 2 min. and the other for 3 min.
@Ibix interprets this to mean that after 2 minutes, they are 50 m. apart, but that they meet a minute later.

 

[Ibix]

@Ibix interprets this to mean that after 2 minutes, they are 50 m. apart, but that they meet a minute later.

Well - that they meet half a minute later and are 50m away again at the 3 minute mark. But yes, basically.

I think my interpretation makes sense because it's soluble. But is the wording in the OP the exact question, @Benjamin_harsh? If not, please post the exact words used in your problem sheet/text book/whatever.

 

[SammyS]

For what it's worth, I agree with @Ibix's interpretation.

 

[Mark44]

Well - that they meet half a minute later and are 50m away again at the 3 minute mark. But yes, basically.

That's a reasonable interpretation that hadn't occurred to me.

 

[FactChecker]

Since they are at distance 50 at both 2 and 3 minutes (constant speeds), can you see when they are at 0 distance? What does that tell you about their relative speed and, therefore, how far apart they were to begin with?

 

[Benjamin_harsh]

Regardless of $x_0$?

position should be $x_0vt$.

 

[Ibix]

position should be $x_0vt$.

No - that would imply that the position is zero at $t=0$, and the units don't work. Your earlier $x=vt$ is correct if you start at $x=0$ at $t=0$. What do you have to do to zero to make it a different constant $x_0$?

 

[FactChecker]

There is a "dirt-simple" way to solve the problem. Use observation of symmetry to see when and where the two are together. Then use that to determine the rate of relative motion. Finally, use that to see how far apart they were at time zero.

 

[Benjamin_harsh]

No?

I understood clearly now. New position will be $x_vt$.

 

[Ibix]

No.

Your $x=vt$ would mean that a person was at $x=0$ at $t=0$, and at $x=1v$ at $t=1$, and at $x=2v$ at $t=2$, and at $x=3v$ at $t=3$.

If you want your person to be at $x=x_0$ at $t=0$, and at $x=x_0+1v$ at $t=1$, and at $x=x_0+2v$ at $t=2$, and $x=x_0+3v$ at $t=3$, what equation describes its motion?

 

[Benjamin_harsh]

If you want your person to be at $x=x_0$ at $t=0$, and at $x=x_0+1v$ at $t=1$

How you got $x = x_0 + 1v$ at $t = 1$?

 

[Ibix]

How you got $x = x_0 + 1v$ at $t = 1$?

Because the person moved a distance $vt$ from their starting point, and the time was 1 (in whatever units we're using).

 

[Benjamin_harsh]

So New position formula: old position + velocity *time.

 

[Ibix]

Yes. So can you write that in terms of the symbols $x_0$, $v$ and $t$?

 

[Benjamin_harsh]

$X = X_0 + v_t$ Now how should i proceed?

 

[Mark44]

So New position formula: old position + velocity *time.

Yes.

$X = X_0 + v_t$

No. This isn't the same as your formula in words. In your equation you are adding a position ($X_0$) to a velocity (which you wrote as $v_t$). The units make no sense: you can't add meters and meters/sec.

 

[Ibix]

The $t$ shouldn't be a subscript, but otherwise fine.

Now you have two people starting at different positions and possibly moving at different speeds. Call the speeds $u$ and $v$. Say one person starts at position 0 at $t=0$ and one at position $d$ at $t=0$.

You should now be able to write two versions of your equation, one for each person.

 

[Benjamin_harsh]

You should now be able to write two versions of your equation, one for each person.

First person position = old first person position + ut

Second person position =
old second person position + vt

 

[Ibix]

Now use the letters I suggested. There is no way you are going to do algebra writing expressions out in words like that.

 

[neilparker62]

$$ x_b-120v=50$$
$$ x_b-180v=-50$$ where v is the velocity of a relative to b.