Train A, traveling 70 miles per hour (mph), leaves Westford heading toward Eastford, 260 miles away. At the same time Train B, traveling 60 mph, leave -

[LolRaccoon]

leaves Eastford heading toward Westford. When do the two trains meet? How far from each city do they meet?

 

[Medical Hawaii]

260

 

[Least Concern]

not enough information. are the tracks straight between the two cities? are the trains traveling on those tracks rather than trying to go towards each other by going around the other side of the globe? do they accelerate to top speed instantly?

 

[Driftwood]

Section 3-3 : Complex Roots​

In this section we will be looking at solutions to the differential equation

ay′′+by′+cy=0ay″+by′+cy=0
in which roots of the characteristic equation,

ar2+br+c=0ar2+br+c=0
are complex roots in the form r1,2=λ±μir1,2=λ±μi.

Now, recall that we arrived at the characteristic equation by assuming that all solutions to the differential equation will be of the form

y(t)=erty(t)=ert
Plugging our two roots into the general form of the solution gives the following solutions to the differential equation.

y1(t)=e(λ+μi)tandy2(t)=e(λ−μi)ty1(t)=e(λ+μi)tandy2(t)=e(λ−μi)t
Now, these two functions are “nice enough” (there’s those words again… we’ll get around to defining them eventually) to form the general solution. We do have a problem however. Since we started with only real numbers in our differential equation we would like our solution to only involve real numbers. The two solutions above are complex and so we would like to get our hands on a couple of solutions (“nice enough” of course…) that are real.

To do this we’ll need Euler’s Formula.

eiθ=cosθ+isinθeiθ=cos⁡θ+isin⁡θ
A nice variant of Euler’s Formula that we’ll need is.

e−iθ=cos(−θ)+isin(−θ)=cosθ−isinθe−iθ=cos⁡(−θ)+isin⁡(−θ)=cos⁡θ−isin⁡θ
Now, split up our two solutions into exponentials that only have real exponents and exponentials that only have imaginary exponents. Then use Euler’s formula, or its variant, to rewrite the second exponential.

y1(t)=eλteiμt=eλt(cos(μt)+isin(μt))y2(t)=eλte−iμt=eλt(cos(μt)−isin(μt))y1(t)=eλteiμt=eλt(cos⁡(μt)+isin⁡(μt))y2(t)=eλte−iμt=eλt(cos⁡(μt)−isin⁡(μt))
This doesn’t eliminate the complex nature of the solutions, but it does put the two solutions into a form that we can eliminate the complex parts.

Recall from the basics section that if two solutions are “nice enough” then any solution can be written as a combination of the two solutions. In other words,

y(t)=c1y1(t)+c2y2(t)y(t)=c1y1(t)+c2y2(t)
will also be a solution.

Using this let’s notice that if we add the two solutions together we will arrive at.

y1(t)+y2(t)=2eλtcos(μt)y1(t)+y2(t)=2eλtcos⁡(μt)
This is a real solution and just to eliminate the extraneous 2 let’s divide everything by a 2. This gives the first real solution that we’re after.

u(t)=12y1(t)+12y2(t)=eλtcos(μt)u(t)=12y1(t)+12y2(t)=eλtcos⁡(μt)
Note that this is just equivalent to taking

c1=c2=12c1=c2=12
Now, we can arrive at a second solution in a similar manner. This time let’s subtract the two original solutions to arrive at.

y1(t)−y2(t)=2ieλtsin(μt)y1(t)−y2(t)=2ieλtsin⁡(μt)
On the surface this doesn’t appear to fix the problem as the solution is still complex. However, upon learning that the two constants, c1c1 and c2c2 can be complex numbers we can arrive at a real solution by dividing this by 2i2i. This is equivalent to taking

c1=12iandc2=−12ic1=12iandc2=−12i
Our second solution will then be

v(t)=12iy1(t)−12iy2(t)=eλtsin(μt)v(t)=12iy1(t)−12iy2(t)=eλtsin⁡(μt)
We now have two solutions (we’ll leave it to you to check that they are in fact solutions) to the differential equation.

u(t)=eλtcos(μt)andv(t)=eλtsin(μt)u(t)=eλtcos⁡(μt)andv(t)=eλtsin⁡(μt)
It also turns out that these two solutions are “nice enough” to form a general solution.

So, if the roots of the characteristic equation happen to be r1,2=λ±μir1,2=λ±μi the general solution to the differential equation is.

y(t)=c1eλtcos(μt)+c2eλtsin(μt)y(t)=c1eλtcos⁡(μt)+c2eλtsin⁡(μt)

 

[Least Concern]

Section 3-3 : Complex Roots​

In this section we will be looking at solutions to the differential equation

ay′′+by′+cy=0ay″+by′+cy=0
in which roots of the characteristic equation,

ar2+br+c=0ar2+br+c=0
are complex roots in the form r1,2=λ±μir1,2=λ±μi.

Now, recall that we arrived at the characteristic equation by assuming that all solutions to the differential equation will be of the form

y(t)=erty(t)=ert
Plugging our two roots into the general form of the solution gives the following solutions to the differential equation.

y1(t)=e(λ+μi)tandy2(t)=e(λ−μi)ty1(t)=e(λ+μi)tandy2(t)=e(λ−μi)t
Now, these two functions are “nice enough” (there’s those words again… we’ll get around to defining them eventually) to form the general solution. We do have a problem however. Since we started with only real numbers in our differential equation we would like our solution to only involve real numbers. The two solutions above are complex and so we would like to get our hands on a couple of solutions (“nice enough” of course…) that are real.

To do this we’ll need Euler’s Formula.

eiθ=cosθ+isinθeiθ=cos⁡θ+isin⁡θ
A nice variant of Euler’s Formula that we’ll need is.

e−iθ=cos(−θ)+isin(−θ)=cosθ−isinθe−iθ=cos⁡(−θ)+isin⁡(−θ)=cos⁡θ−isin⁡θ
Now, split up our two solutions into exponentials that only have real exponents and exponentials that only have imaginary exponents. Then use Euler’s formula, or its variant, to rewrite the second exponential.

y1(t)=eλteiμt=eλt(cos(μt)+isin(μt))y2(t)=eλte−iμt=eλt(cos(μt)−isin(μt))y1(t)=eλteiμt=eλt(cos⁡(μt)+isin⁡(μt))y2(t)=eλte−iμt=eλt(cos⁡(μt)−isin⁡(μt))
This doesn’t eliminate the complex nature of the solutions, but it does put the two solutions into a form that we can eliminate the complex parts.

Recall from the basics section that if two solutions are “nice enough” then any solution can be written as a combination of the two solutions. In other words,

y(t)=c1y1(t)+c2y2(t)y(t)=c1y1(t)+c2y2(t)
will also be a solution.

Using this let’s notice that if we add the two solutions together we will arrive at.

y1(t)+y2(t)=2eλtcos(μt)y1(t)+y2(t)=2eλtcos⁡(μt)
This is a real solution and just to eliminate the extraneous 2 let’s divide everything by a 2. This gives the first real solution that we’re after.

u(t)=12y1(t)+12y2(t)=eλtcos(μt)u(t)=12y1(t)+12y2(t)=eλtcos⁡(μt)
Note that this is just equivalent to taking

c1=c2=12c1=c2=12
Now, we can arrive at a second solution in a similar manner. This time let’s subtract the two original solutions to arrive at.

y1(t)−y2(t)=2ieλtsin(μt)y1(t)−y2(t)=2ieλtsin⁡(μt)
On the surface this doesn’t appear to fix the problem as the solution is still complex. However, upon learning that the two constants, c1c1 and c2c2 can be complex numbers we can arrive at a real solution by dividing this by 2i2i. This is equivalent to taking

c1=12iandc2=−12ic1=12iandc2=−12i
Our second solution will then be

v(t)=12iy1(t)−12iy2(t)=eλtsin(μt)v(t)=12iy1(t)−12iy2(t)=eλtsin⁡(μt)
We now have two solutions (we’ll leave it to you to check that they are in fact solutions) to the differential equation.

u(t)=eλtcos(μt)andv(t)=eλtsin(μt)u(t)=eλtcos⁡(μt)andv(t)=eλtsin⁡(μt)
It also turns out that these two solutions are “nice enough” to form a general solution.

So, if the roots of the characteristic equation happen to be r1,2=λ±μir1,2=λ±μi the general solution to the differential equation is.

y(t)=c1eλtcos(μt)+c2eλtsin(μt)y(t)=c1eλtcos⁡(μt)+c2eλtsin⁡(μt)

i'll take your word for it

 

[Slimy Time]

Is this your way of telling us you saw your mom get double teamed?

 

[Absolute Brainlet]

Since I feel like it, might as well provide an actual answer.

V¹ = 70 mph = 31.293 meters/s
V² = 60 mph = 26.822 m/s
S = 260 miles = 418.429 km
t¹ (train 1 reaches Eastford) = x¹
t² (train 2 reaches Westford) = x²
t³ (trains meeting) = x³
S¹ (from Westford) = x⁴
S² (from Eastford) = x⁵
Solution:
t¹ = S/V¹ = 260/70 ≈ 3.7 hours = 222 minutes = 13320 seconds
t² = S/V² = 260/60 ≈ 4.3 hours = 258 minutes = 15480 seconds
t³ = a² - a¹ = (t²/V²) - (t¹/V¹) ≈ 577.138 - 425.654 = 151.484 minutes = 9089.04 seconds
Therefore:
Train 1 will meet Train 2 at: t¹ - t³ = 70.516 minutes = 4230.96 seconds into its journey.
Train 2 will meet Train 1 at: t² - t³ = 106.516 minutes = 6390.96 seconds into its journey.
S¹ = S/(t³/V¹) ≈ 53.71 miles = 86.438 km from Westford
S² = S/(t³/V²) ≈ 46.09 miles = 74.175 km from Eastford

 

[Spl00gies]

instructions unclear, boarded a plane to Northford

 

[Product Placement]

When will they meet, you mean crash into each other?

 

[Gigantic Faggot]

70t + 60t = 260
t = 2
70 x 2 = 140 from Westford
60 x 2 = 120 from Eastford

 

[HensKenKline]

Depends, how often does the train from the East stop to pick up refugees?

 

[Rekkington]

How about my cock is leaving from West Turdshire at midnight and your ass is stationary in San Francisco, how long does it take me to fuck you if I'm going 100k an hour in a straight line, bitch?

 

[Gigantic Faggot]

Since I feel like it, might as well provide an actual answer.

V¹ = 70 mph = 31.293 meters/s
V² = 60 mph = 26.822 m/s
S = 260 miles = 418.429 km
t¹ (train 1 reaches Eastford) = x¹
t² (train 2 reaches Westford) = x²
t³ (trains meeting) = x³
S¹ (from Westford) = x⁴
S² (from Eastford) = x⁵
Solution:
t¹ = S/V¹ = 260/70 ≈ 3.7 hours = 222 minutes = 13320 seconds
t² = S/V² = 260/60 ≈ 4.3 hours = 258 minutes = 15480 seconds
t³ = a² - a¹ = (t²/V²) - (t¹/V¹) ≈ 577.138 - 425.654 = 151.484 minutes = 9089.04 seconds
Therefore:
Train 1 will meet Train 2 at: t¹ - t³ = 70.516 minutes = 4230.96 seconds into its journey.
Train 2 will meet Train 1 at: t² - t³ = 106.516 minutes = 6390.96 seconds into its journey.
S¹ = S/(t³/V¹) ≈ 53.71 miles = 86.438 km from Westford
S² = S/(t³/V²) ≈ 46.09 miles = 74.175 km from Eastford

Not exactly sure what's going on here but the red flag is the trains not meeting at the same time.

 

[nippleonbonerfart]

Train A is full of somali muslim immigrants and it blows up shortly after departure.

 

[Happy Fish]

They both arrive at their scheduled stop in Nagoya at separate times

 

[Feline Supremacist]

I'm not doing your homework.

 

[General Tug Boat]

Does the amount of Jews on the inbound train have any impact on the problem? Asking for a friend.

 

[Gigantic Faggot]

Does the amount of Jews on the inbound train have any impact on the problem? Asking for a friend.

Adolf Eichmann was once asked this question. He just smiled and said "what difference does it make?".