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Composition Setup

Excerpt
hiddentrue

Two people moving in one dimension with constant speed are destined to meet – but where?


Two people have decided to use a mountain trail to get some exercise. They start out from the parking lot at the bottom of the trail at the same time. Person A runs the trail at a constant speed |vA| = 5.0 m/s. Person B walks the trail at a constant speed |vB|=1.0 m/s. Given that the people must return along the same path they climbed up, and given that the summit of the trail is d = 3.0 km from the parking lot, how far from the summit will the people be when they meet going in opposite directions? (Assume neither person pauses.)

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idsys
Systems:
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idsys

Each person will be treated as a point particle.

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idint
Interactions:
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idint

Not relevant to this model.

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idmod
Model:
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idmod

applies to each person separately. Depending upon how you visualize the problem, the model may have to be applied twice to the runner (person A). We will suggest two possible methods by which to apply this model in the Approach.

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idapp
Approach:

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idapp

This problem stretches the definition of One-Dimensional Motion with Constant Velocity. Even if we assume the path is perfectly straight, the runner must reverse direction at the summit, and so it would seem that person A's velocity changes its mathematical sign within the problem. We will present two ways to deal with this issue. The first is more straightforward conceptually, but is slightly more tedious. The second requires deeper physical reasoning, but is slightly faster.

Info

Even though the dynamics of the motions described in this problem are very different if the path is curvy instead of straight, the kinematics are mathematically equivalent. It is mathematically possible to parameterize the motion along a non-self-intersecting path as a one-dimensional motion. Since this problem only deals with kinematics, our conclusions are valid for a curvy path as well.

Deck of Cards
idmethdeck
}{composition-setup} {excerpt:hidden=true}Two people moving in one dimension with constant speed are destined to meet -- but where?{excerpt} Two people have decided to use a mountain trail to get some exercise. They start out from the parking lot at the bottom of the trail at the same time. Person A runs the trail at a constant speed |_v_~A~| = 5.0 m/s. Person B walks the trail at a constant speed |_v_~B~|=1.0 m/s. Given that the people must return along the same path they climbed up, and given that the summit of the trail is _d_ = 3.0 km from the parking lot, how far from the summit will the people be when they meet going in opposite directions? (Assume neither person pauses.) {toggle-cloak:id=sys} *Systems:* {cloak:id=sys}Each person will be treated as a point particle.{cloak} {toggle-cloak:id=int} *Interactions:* {cloak:id=int}Not relevant to this model.{cloak} {toggle-cloak:id=mod} *Model:* {cloak:id=mod}[1-D Motion (Constant Velocity)] applies to each person separately. Depending upon how you visualize the problem, the model may have to be applied twice to the runner (person A). We will suggest two possible methods by which to apply this model in the Approach.{cloak} {toggle-cloak:id=app} *Approach:* {cloak:id=app} This problem stretches the definition of One-Dimensional Motion with Constant Velocity. Even if we assume the path is perfectly straight, the runner must reverse direction at the summit, and so it would seem that person A's velocity changes its mathematical sign within the problem. We will present two ways to deal with this issue. The first is more straightforward conceptually, but is slightly more tedious. The second requires deeper physical reasoning, but is slightly faster. {info}Even though the _dynamics_ of the motions described in this problem are very different if the path is curvy instead of straight, the _kinematics_ are mathematically equivalent. It is mathematically possible to parameterize the motion along a non-self-intersecting path as a one-dimensional motion. Since this problem only deals with kinematics, our conclusions are valid for a curvy path as well.{info} {deck:id=methdeck} {card:label=Method 1: Split the Motion} h3. Method 1: Split the Motion {toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color} {cloak:id=diag} One way to be sure that each person has constant velocity is to split the problem into two parts. The point of division is when person A reaches the summit and turns around. If we set up a one-dimensional coordinate system as shown below, then during the first part of the problem person A moves with velocity _v_~A1~ = + 5.0 m/s and person B moves with _v_~B1~ = + 1.0 m/s. During the second part of the problem, person A moves with _v_~A2~ = -- 5.0 m/s while person B still moves with _v_~B2~ = + 1.0 m/s. !meet.png|width=700! {note}Can you think of a reason that it might have been a good idea to put _x_ = 0 m at the summit instead of the parking lot? We will encounter one such reason at the end of this method.{note} {cloak:diag} {toggle-cloak:id=math} {color:red} *Mathematical Representation* {color} {cloak:id=math} For our chosen model, there is only one Law of Change: {latex}
Card
labelMethod 1: Split the Motion

Method 1: Split the Motion

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iddiag
Diagrammatic Representation

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iddiag

One way to be sure that each person has constant velocity is to split the problem into two parts. The point of division is when person A reaches the summit and turns around. If we set up a one-dimensional coordinate system as shown below, then during the first part of the problem person A moves with velocity vA1 = + 5.0 m/s and person B moves with vB1 = + 1.0 m/s. During the second part of the problem, person A moves with vA2 = – 5.0 m/s while person B still moves with vB2 = + 1.0 m/s.

Image Added

Note

Can you think of a reason that it might have been a good idea to put x = 0 m at the summit instead of the parking lot? We will encounter one such reason at the end of this method.

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idmath
Mathematical Representation

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idmath

For our chosen model, there is only one Law of Change:

Latex
\begin{large} \[ x = x_{\rm i} + vt\]\end{large}
{latex}

Because

we

have

divided

the

problem,

however,

we

must

apply

this

one

law

a

total

of

_

four

times

_

(person

A

during

part

1,

person

B

during

part

1,

person

A

during

part

2,

and

person

B

during

part

2).

Thus,

we

have:

{

Latex
}
\begin{large} \[ x_{\rm A1}=x_{\rm A1,i}+v_{\rm A1}t_{1}\]\[ x_{\rm B1}=x_{\rm B1,i}+v_{\rm B1}t_{1}\]\[ x_{\rm A2}=x_{\rm A2,i}+v_{\rm A2}t_{2}\]\[ x_{\rm B2}=x_{\rm B2,i}+v_{\rm B2}t_{2}\] \end{large}
{latex} {note}It is important that although persons A and B will potentially have different positions and velocities during each part, they share the same time. When part 1 ends, the same amount of time has elapsed for each person. Thus, in the equations, _t_ is only labeled with "1" or "2", not "A1" or "A2".{note} {cloak:math} {toggle-cloak:id=up} {color:red} *Split the Problem -- Runner on the way up* {color} {cloak:id=up} Four equations seems intimidating, but in this case a systematic approach gives quick results. Begin with part 1. In part 1, both persons start in the parking lot, meaning that (in our coordinate system) _x_~1A,i~ = _x_~1B,i~ = 0 m. This means: {latex}
Note

It is important that although persons A and B will potentially have different positions and velocities during each part, they share the same time. When part 1 ends, the same amount of time has elapsed for each person. Thus, in the equations, t is only labeled with "1" or "2", not "A1" or "A2".

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math

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idup
Split the Problem – Runner on the way up

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Four equations seems intimidating, but in this case a systematic approach gives quick results. Begin with part 1. In part 1, both persons start in the parking lot, meaning that (in our coordinate system) x1A,i = x1B,i = 0 m. This means:

Latex
\begin{large} \[x_{\rm A1} = v_{\rm A1}t_{1}\]\[x_{\rm B1}=v_{\rm B1}t_{1}\]\end{large}
{latex}

We

have

already

determined

the

velocities,

but

we

still

have

two

unknowns

in

each

equation.

To

solve

either

one,

we

must

remember

how

we

defined

the

parts

of

our

problem.

If

we

recall

that

part

1

ends

when

person

A

reaches

the

summit,

then

we

must

have _x_~A1~ = 3000 m. We can use this to solve for _t_~1~: {latex}

have xA1 = 3000 m. We can use this to solve for t1:

Latex
\begin{large} \[t_{1} = \frac{x_{\rm A1}}{v_{\rm A1}} = 600 \:{\rm s}\] \end{large} 
{latex}

Now,

because

_

t

_~1~

1 is

in

person

B's

equation

also,

we

find:

{

Latex
}
\begin{large} \[ x_{\rm B1} = x_{\rm A1} \frac{v_{\rm B1}}{v_{\rm A1}} = 600 \:{\rm m}\] \end{large}
{latex} {tip}If you are evaluating each expression as you go, think about whether the numbers make sense. Given that person B walks at 1
Tip

If you are evaluating each expression as you go, think about whether the numbers make sense. Given that person B walks at 1 m/s,

we

certainly

do

expect

that

_

t

_~1~

1 and

_

x

_~B1~

B1 will

be

the

same.

{tip} {cloak:up} {

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up
up

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:

id

=down} {color:red} *Split the Problem -- Runner on the way back* {color} {cloak:id=down} With part 1 understood, we move on to part 2. The first important realization here is that part 2 ends when the two persons meet. Thus, we must have _x_~A2~ = _x_~B2~. From our four original equations, that means: {latex}

down
Split the Problem – Runner on the way back

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With part 1 understood, we move on to part 2. The first important realization here is that part 2 ends when the two persons meet. Thus, we must have xA2 = xB2. From our four original equations, that means:

Latex
\begin{large} \[x_{\rm A2,i}+v_{\rm A2} t_{2} = x_{\rm B2,i}+v_{\rm B2}t_{2}\]\end{large}
{latex}

The

second

important

realization

is

that

part

2

begins

where

part

1

ends.

This

relationship

is

expressed

mathematically

by

writing:

{

Latex
}
\begin{large} \[x_{\rm A2,i} = x_{\rm A1} \]\[x_{\rm B2,i} = x_{\rm B1} = x_{\rm A1}\frac{v_{\rm B1}}{v_{\rm A1}}\]\end{large}
{latex}

which

means:

{

Latex
}
 \begin{large}\[t_{2} = \frac{x_{\rm A1} \left(1-\frac{v_{\rm B1}}{v_{\rm A1}}\right)}{v_{\rm B2} - v_{\rm A2}} = \frac{x_{\rm A1}}{v_{\rm A1}}\frac{v_{\rm A1} - v_{\rm B1}}{v_{\rm B2} - v_{\rm A2}} = 400\:{\rm s}\]\end{large}
{latex} {warning}Notice that _v_~B2~ - _v_~A2~ is
Warning

Notice that vB2 - vA2 is 6.0

m/s,

*

not

*

-4.0

m/s,

since

_

v

_~A2~

A2 is

*

negative

*

5.0

m/s.

It

is

easy

to

make

mistakes

with

negative

signs.

In

this

case,

however,

your

final

answer

for

the

meeting

location

will

clearly

indicate

there

has

been

a

mistake

if

you

subtract

incorrectly.

(Try

it

and

see

what

happens.)

{warning} With _t_~2~ in hand, we know the time of the meeting. We still do not know the position, however. To get the position, we have to substitute our answer for _t_~2~ into either the equation for _x_~A2~ or that for _x_~B2~. Selecting the equation for person A: {latex}

With t2 in hand, we know the time of the meeting. We still do not know the position, however. To get the position, we have to substitute our answer for t2 into either the equation for xA2 or that for xB2. Selecting the equation for person A:

Latex
\begin{large}\[ x_{\rm A2} = x_{\rm A1} + v_{\rm A2}\frac{x_{\rm A1}}{v_{\rm A1}}\frac{v_{\rm A1}-v_{\rm B1}}{v_{\rm B2}-v_{\rm A2}} \]\end{large}
{latex}

A

complicated

equation

like

this

is

worth

simplifying

to

see

if

we

can

make

some

sense

of

it.

Some

algebra

will

enable

us

to

get

rid

of

the

compound

fractions:

{

Latex
}
\begin{large} \[ x_{\rm A2} = x_{\rm A1} \left(\frac{v_{\rm B2}-v_{\rm A2} + \frac{v_{\rm A2}}{v_{\rm A1}}(v_{\rm A1}-v_{\rm B1})}{v_{\rm B2}-v_{\rm A2}}\right) = \frac{x_{\rm A1}}{v_{\rm A1}}\left(\frac{v_{\rm B2}v_{\rm A1}-v_{\rm A2}v_{\rm B1}}{v_{\rm B2}-v_{\rm A2}}\right)\]\end{large}
{latex} {tip}Some checks of this expression are possible. Substituting _t_~2~ into the expression for _x_~B2~ should give the same answer. If the walker's speed is zero or the runner's speed is infinite, the meeting should take place in the parking lot
Tip

Some checks of this expression are possible. Substituting t2 into the expression for xB2 should give the same answer. If the walker's speed is zero or the runner's speed is infinite, the meeting should take place in the parking lot (x=0).

{tip}

It

is

important

to

recognize

that

we

are

not

finished

yet.

The

problem

asks

for

the

distance

of

the

meeting

from

the

summit.

We

have

found

the

position

of

the

meeting.

To

find

the

distance

requested,

we

must

calculate:

{

Latex
}
\begin{large} \[ |x_{\rm summit} - x_{\rm meeting}| = x_{\rm A1} - \frac{x_{\rm A1}}{v_{\rm A1}}\left(\frac{v_{\rm B2}v_{\rm A1}-v_{\rm A2}v_{\rm B1}}{v_{\rm B2}-v_{\rm A2}}\right) = \frac{x_{\rm A1}}{v_{\rm A1}} \frac{v_{\rm A2}v_{\rm B1}-v_{\rm A1}v_{\rm A2}}{v_{\rm B2}-v_{\rm A2}} = 2000 \;{\rm m} \]\end{large}
{latex} {tip}This equation too can be checked. Now the limit as the walker's speed goes to zero or the runner's to infinity should be x = 3000 m. Also, once you have calculated the value, it is a good idea to check that the runner has traveled 5 times as far as the walker (is that true for our answer?).{tip} {cloak:down} {card} {card:label=Method 2: Total Distance} h3. Method 2: Total Distance {toggle-cloak:id=diag2} {color:red} *Diagrammatic Representation* {color} {cloak:id=diag2} Another approach to this problem is to avoid the issue of the direction switch for the runner by thinking in terms of distance alone. The key to such a restructuring of the problem is to consider the distance covered by _both_ persons. Sketching a graph like the one below might help with this. In the graph the red line represents the runner while the blue line represents the walker. By the time they meet, the runner has already been to the summit, covering a distance _d_ = 3000 m in the process. At this point, we do not know how far back down the mountain the runner has made it, nor do we know how far up the mountain the walker has come before they meet. Consider, however, that the distance the runner has come down the mountain _plus_ the distance the walker has come up the mountain must add to equal the distance to the summit. Again, the sketch might help you to see this. !meetgraph.png|width=650! {cloak:diag2} {toggle-cloak:id=math2} {color:red} *Mathematical Representation* {color} {cloak:id=math2} With this important realization, we can say that the total distance covered by the runner in the time from leaving the parking lot _plus_ the total distance covered by the walker since leaving the parking lot is equal to twice the summit distance, 2 _d_ = 6000 m. Now, we simply construct an equation that says the same thing: {latex}
Tip

This equation too can be checked. Now the limit as the walker's speed goes to zero or the runner's to infinity should be x = 3000 m. Also, once you have calculated the value, it is a good idea to check that the runner has traveled 5 times as far as the walker (is that true for our answer?).

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down
down

Card
labelMethod 2: Total Distance

Method 2: Total Distance

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iddiag2
Diagrammatic Representation

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iddiag2

Another approach to this problem is to avoid the issue of the direction switch for the runner by thinking in terms of distance alone. The key to such a restructuring of the problem is to consider the distance covered by both persons. Sketching a graph like the one below might help with this. In the graph the red line represents the runner while the blue line represents the walker. By the time they meet, the runner has already been to the summit, covering a distance d = 3000 m in the process. At this point, we do not know how far back down the mountain the runner has made it, nor do we know how far up the mountain the walker has come before they meet. Consider, however, that the distance the runner has come down the mountain plus the distance the walker has come up the mountain must add to equal the distance to the summit. Again, the sketch might help you to see this.

Image Added

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diag2
diag2

Toggle Cloak
idmath2
Mathematical Representation

Cloak
idmath2

With this important realization, we can say that the total distance covered by the runner in the time from leaving the parking lot plus the total distance covered by the walker since leaving the parking lot is equal to twice the summit distance, 2 d = 6000 m. Now, we simply construct an equation that says the same thing:

Latex
\begin{large} \[ 2d = |v_{\rm A}|t + |v_{\rm B}|t \] \end{large}
{latex} {note}This equation is found using the definition of [distance] and [speed]. It does not follow directly from the Laws of Change of any of our models. Going outside the usual models requires confidence with the material. If you are studying physics for the first time, you will likely find Method 1 more understandable.{note} Solving for _t_ gives: {latex}
Note

This equation is found using the definition of distance and speed. It does not follow directly from the Laws of Change of any of our models. Going outside the usual models requires confidence with the material. If you are studying physics for the first time, you will likely find Method 1 more understandable.

Solving for t gives:

Latex
 \begin{large} \[ t = \frac{2d}{|v_{\rm A}|+|v_{\rm B}|} = 1000 \;{\rm s}\] \end{large
}{latex
}

This

equation

is

simple,

but

we

have

to

be

careful

about

the

meaning

of

the

result.

We

have

not

set

up

equations

for

the

position

of

either

person

yet.

We

have

to

do

that

now.

The

easiest

equation

to

use

is

that

of

the

walker,

who

does

not

change

direction.

The

Law

of

Change

for

our

model

tells

us:

{

Latex
}
\begin{large} \[ x_{\rm B} = x_{\rm B,i} + v_{B} t \] \end{large}
{latex}

If

we

assume

the

same

coordinate

system

as

we

defined

in

Method

1

(see

the

figure

above)

then

the

walker

begins

at

_

x

_~B~

B=0

and

we

find:

{

Latex
}
\begin{large} \[ x_{\rm B} = v_{B} \frac{2d}{|v_{\rm A}| + |v_{\rm B}|} = 1000 \:{\rm m} \] \end{large}
{latex}

As

in

method

1,

we

must

remember

that

this

is

the

position

of

the

walker

at

the

meeting,

not

the

distance

from

the

summit.

To

find

the

distance

from

the

summit

we

calculate:

{color} {latex}

Latex
\begin{large} \[ |x_{\rm summit} - x_{\rm meeting}| = d - v_{\rm B} \frac{2d}{|v_{\rm A}| + |v_{\rm B}|} = d \frac{|v_{\rm A}|+|v_{\rm B}|-2v_{\rm B}}{|v_{\rm A}| + |v_{\rm B}|} = 2000 \:{\rm m} \] \end{large}
{latex} {tip}Is this expression equivalent to that found in method 1?{tip} {cloak:math2} {card} {deck} {cloak:app}
Tip

Is this expression equivalent to that found in method 1?

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math2

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app