How to Find Out When Two Things Will Meet Again

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I need help in finding where 2 people will meet again

• Thread starter Blockade
• First date Sep sixteen, 2015

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  1d motion velocity

Homework Statement

At the instant the traffic light turns green, Stan Speedy starts from balance and accelerates at "a_{due south}". At the same instant Kathy Kool passes Stan with abiding velocity "v_k".
Given: [a_southward, v_k]

How far beyond the starting betoken exercise they meet once more?

Homework Equations

v = v₀ + a*t
10 = x₀ + v₀t + 0.5 * a * t²
v^two = v₀ ² + 2a(10-x₁)

The Attempt at a Solution

Stan:
x₀ = 0 <--- Since he started from rest.
x = ?
a = a_s <--- Given
v₀ = 0 <--- Stan started from balance
5 = v <----- I recollect velocity is equal to itself since a_southward, Stan'south velocity must be constant also.
t = ?

First I used 5 = v₀ + a*t to find the time
v = (0) + (a_s * t)
t = v/a_s

Now I substitute t into the position formula
x = 10₀ + v₀ * t + 0.5 * a * t²
x = 0 + 0 + (0.v * a_s * ^{(v/a_{due south} )2}
x = 0.5 * v²/a_s

And so now I have Stan's position.

Going to list Kathy's variables:
x₀ = ? <--- from the wording of the prompt I assumed that Kathy has already been running before the greenish light.
x = ? <--- somewhere in front of Stan
5 = v_thou <--- Given
v₀ = ?
a = a <--- constant velocity means abiding acceleration
t = ?

And then my plan is to find Kathy'due south position and gear up it equal to Stan'south position and search for the distance of where they see once more. Still, I am pretty sure that my setup so far is wrong. May someone please interject and set me straight?

Answers and Replies

v = v <----- I think velocity is equal to itself since as, Stan's velocity must exist constant too.

Things are always equal to themself. 3=three, five=5, ...
Stan is accelerating, his speed cannot exist constant.

First I used five = five₀ + a*t to find the time

The time of what? How (and why) do yous want to discover the fourth dimension of anything of relevance without because Kathy'south motion here?

a = a <--- abiding velocity ways constant acceleration

It means more that.
Also, meet higher up, a=a is a pointless statement. It is ever true.

So my program is to notice Kathy'south position and set information technology equal to Stan'southward position and search for the altitude of where they meet over again.

That is good, simply as you said, you have to ready the distances first.

Constant velocity means zero acceleration.

Things are ever equal to themself. 3=3, 5=5, ...
Stan is accelerating, his speed cannot be constant.

The time of what? How (and why) do you want to detect the time of annihilation of relevance without because Kathy's motion here?

It means more than than that.
Also, see above, a=a is a pointless statement. It is ever truthful.

That is good, simply equally you said, yous have to fix the distances first.

Thanks, yous've cleared up my confusion on how acceleration cannot exist constant. Then now I am going to try to find Stan's and Kathy's position.

Stan:
x0 = 0 <--- Since he started from remainder.
x = ?
a = as <--- Given
v0 = 0 <--- Stan started from rest
v = ?
t = ?

From here I will endeavour to notice Stan's position:

x = x0 + v0*t + (ane/2)at^2
10 = (0) + (0) + (1/2) at^two
x = (i/two) at^two <------ Stan's Position

Going to list Kathy's variables:
x0 = ? <--- from the wording of the prompt I assumed that Kathy has already been running earlier the green light.
x = ? <--- somewhere in front of Stan
v = vk <--- Given
v0 = ?
a = ? <--- constant velocity means constant acceleration
t = ?

Whatsoever idea where I start from hither? Would information technology be correct if Kathy's x0 = 0m and v0 = 0m/s? I am confused on what to prepare her initial velocity and initial position to since in the prompt sounds similar she was moving before the light even turned greenish.

Thank you, you've cleared up my confusion on how acceleration cannot be abiding. And then at present I am going to try to find Stan's and Kathy's position.

Stan:
x0 = 0 <--- Since he started from rest.
10 = ?
a = equally <--- Given
v0 = 0 <--- Stan started from rest
v = ?
t = ?

From here I will try to detect Stan's position:

x = x0 + v0*t + (i/2)at^2
x = (0) + (0) + (1/2) at^2
x = (1/2) at^2 <------ Stan's Position

The concluding equation gives Stan's position ten from the terminate light as a role of t.

Going to list Kathy's variables:
x0 = ? <--- from the wording of the prompt I assumed that Kathy has already been running earlier the green light.
ten = ? <--- somewhere in front of Stan
v = vk <--- Given
v0 = ?
a = ? <--- abiding velocity means abiding dispatch
t = ?

If Kathy is traveling at a constant velocity of vk, how far has she traveled from the stop light in t seconds? Hint: d = r * t

The concluding equation gives Stan'due south position x from the cease lite every bit a office of t.

If Kathy is traveling at a constant velocity of vk, how far has she traveled from the terminate light in t seconds? Hint: d = r * t

So Kathy'south d = v_g * t, meaning I need to find Kathy'southward time relative to Stan's and plug that into this, and then I tin can get Kathy'due south position and finally set it equal to Stan's position.

Is it correct to detect "t" from Stan'southward position equation and plug it into Kathy's (d = five_k * t)?
Like: d = x = (ane/2) at^2
d = (1/2) at^two
t = sqrt(2d/a)

Kathy's position = 5_m * t
= 5_k * sqrt(2nd/a)

Yous have ii equations and two unknowns, the fashion to solve that is up to yous. Your approach works, but setting the two expressions for the distance equal to each other is probably a chip easier.

Going to list Kathy's variables:
x0 = 0 <--- Kathy'due south initial position is naught because that's when she defenseless up to Stan
x = ? <--- somewhere in front of Stan
5 = vk <--- Given
v0 = ?
a = 0 <--- constant velocity means dispatch = 0
t = T

So, I have made some changes to Kathy's cardinal. I changed her x0 to 0 because that'due south where the origin (pregnant where we'll began the calculations).
I too changed her dispatch to 0 considering it was giving that her velocity "vk" is a constant.

Now, zip has changed near Stan:
Stan:
x0 = 0 <--- Since he started from remainder.
x = ?
a = as <--- Given
v0 = 0 <--- Stan started from rest
v = ?
t = T

At present, what I can see is that Kathy'southward time(t) and initial position (x0) are shared with Stan's.

Where the work begins:

Stan's Last Position:
10 = x0 + v0t + 1/2 * a*t^two
x = 0 + 0 + (one/2) * (as) * T^2
x = (1/2) * (as) * T^2

Kathy'due south Terminal Position:
ten = x0 + v0t + ane/2 * a * t^2
10 = 0 + vk * T + 1/two * (0) * t^2 <--- a = 0 because Kathy's velocity is constant
x = vk * T

Now I will gear up Kathy's and Stan'due south terminal position equal to each other.

vk * T = (1/2) * as * T^2

SOLVING FOR THE TIME:
v = v0 + at
v = 0 + as T
T = v/"every bit" <---- Just a reminder that "every bit" is even so Stan's acceleration.

And so far I am on the correct path? How practice I find how far beyond the starting point do they meet once again? What should I do from here?

Now I will fix Kathy'southward and Stan's final position equal to each other.

vk * T = (ane/2) * as * T^2

Expert. That equation is all you need now. T is the merely unknown, yous tin find T.

Where does that come from at present and what does information technology mean?

Adept. That equation is all you need now. T is the only unknown, you lot can find T.
Where does that come from now and what does information technology mean?

So to find T, I can either apply the equation y'all quoted?

five = v0 + at
vk = 0 + at
t = vk/a

or tin can I

vk * T = (1/ii) * as * T^2

vk = (i/ii) * equally * T
T = 2vk/as

and I am unsure when you asked "Where does that come up from now and what does information technology mean?"

Well...
considering that acceleration is abiding like in this problem,

a = dv/dt

adt = dv

integrat:

a | dt = |dv
at +c = v

and when fourth dimension = 0
v₀ = five_f(0) = a(0) + c

five₀ = c

substitute c with v₀

v = v₀ + at

and it means that its an equation that is relative to time.

Kathy (looking for time):
v_k = v₀ + at
v_k = v₀ + (0)t
v_k = v₀

so I am not sure how to solve for fourth dimension with Kathy's information using v = v₀ + at, unless information technology's valid that I brand her five₀ = 0, simply I am unsure how or if its even valid to practice so since the problem never mentioned her at rest.

Last edited: Sep eighteen, 2015

So to detect T, I tin can either apply the equation you quoted?

five = v0 + at
vk = 0 + at
t = vk/a

or tin can I

vk * T = (1/2) * as * T^ii

vk = (1/ii) * as * T
T = 2vk/as

and I am unsure when you asked "Where does that come from at present and what does it mean?"

Well...
considering that acceleration is constant like in this problem,

a = dv/dt

adt = dv

integrat:

a | dt = |dv
at +c = v

and when time = 0
v₀ = five_f(0) = a(0) + c

5₀ = c

substitute c with 5₀

v = v₀ + at

and information technology ways that its an equation that is relative to time.

Kathy (looking for time):
v_g = v₀ + at
v_k = v₀ + (0)t
v_grand = v₀

so I am non sure how to solve for time with Kathy'south information using 5 = v₀ + at, unless it's valid that I make her v₀ = 0, simply I am unsure how or if its fifty-fifty valid to do so since the trouble never mentioned her at balance.

That's all very confusing because you reuse symbols with different meanings, like v₀, apparently as an initial velocity for both.
Kathy has constant velocity, v_chiliad, so it is also her initial velocity. Y'all tin can write downwards where Kathy is at fourth dimension t (as you already did).
Base on Stan'southward starting from rest and given constant acceleration, you can write downwardly where he is at time t, every bit y'all already did. The final stride is to say that these distances are equal at time T. Y'all do non care what their velocities are at that time.

So to detect T, I can either use the equation you quoted?

v = v0 + at
vk = 0 + at
t = vk/a

or tin I

vk * T = (ane/ii) * equally * T^ii

vk = (ane/2) * equally * T
T = 2vk/every bit

The second arroyo works. The get-go one is randomly mixing equations together, that does non work.

Good, you found the time. Where is Kathy at this fourth dimension T? This will be the concluding answer.

The second approach works. The first ane is randomly mixing equations together, that does not work.

Good, you found the time. Where is Kathy at this time T? This will be the last answer.

Give thanks you very much for your assistance! I demand to do the same trouble also, good thing that this was posted.

Last edited: Sep 19, 2015

Since it asks for "How far across the starting point practice they encounter once more?", I should plug "T" into into Kathy'due south position "10 = v_kt" and that should give me the distance of how far beyond the starting point since I made both Stan's and Kathy's x₀ = 0m?

Right.
You can besides plug it into Stan's position, the result has to exist the same past construction.

Awesome, thanks! Do you mind walking me through a similar problem after I've posted what I've done on it in a piddling scrap?

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