 # Chapter 1

• 2 May 1994

I can't wait until school is over. Only two more years and I can finally begin a new chapter of my life. Today, dad helped me out with my homework again. This maths exercise was really confusing, but he helped me figure out how to solve it. He's really good with numbers. No wonder he's so successful with his shop. Yesterday, he asked me again if I wanted to take over his shop. I know it'd make him happy, but I have my own plans and dreams. Once I graduate from school, I'm going to become a top manager and create my own business. I can't wait! Time to sleep now. But first I'll review this maths exercise once again, just to make sure I can present the right answers in class tomorrow. Due to a technical error, two trains were accidently assigned to the same rail track and are now approaching each other. The communication system of both trains is broken and the conductors don't know about the imminent danger they're in.

By the time a supervisor at one of the offices notices the problem, the trains are only 140 km away from each other. The speed of train A is 160 km/h; the speed of train B is 120 km/h.

There's a manual switch that could stop both trains from colliding, but the supervisor is 6,300 metres away from it and he needs to get there first. If he runs at an average speed of 14 km/h, will he be in time to save the trains?

• 3 points for the correct answer
• 7 points for the correct explanation • Let us first look at the trains: using non-relativistic mechanics we see that from one train's perspective the other train is moving in with a speed of 160+120=280km/h. Hence the distance to the other train (140km) with be covered in exactly 30min. The supervisor needs to run 6.3km at a speed of 14km/h, thus takes 6.3/14=0.45h=27min. Since 27<30, the supervisor can pull the switch before he sees the trains collide.

We only need to hope the switch is in the right place...

Addendum: when considering the relativistic case, in the frame of the switch, the switch will be pulled in 27min (same calculation).

In the frame of each train, the speed of the other train moving in, is slightly smaller than 280km/h (it is (120+160)/(1+120*160/c^2), where c^2 is the speed of light).

In the frame of a train the distance to be covered will be shortened by a factor γ (which is different for both trains), so for the train the collision happens a factor γ earlier than for the switch. However the clock in each trains is running slower than the clock in the switch, by the same factor γ. So a train sees the switch being pulled earlier by a factor γ than the switch sees it.

We conclude that both trains see the switch being pulled earlier than they "see" collision happen, in fact relativity will help avoid the collision.

PS: γ=1/sqrt(1-v^2/c^2), where v is the speed of the train and c the speed of light

• Problem:

Due to a technical error, two trains were accidently assigned to the same rail track and are now approaching each other. The communication system of both trains is broken and the conductors don't know about the imminent danger they're in.

By the time a supervisor at one of the offices notices the problem, the trains are only 140 km away from each other. The speed of train A is 160 km/h; the speed of train B is 120 km/h.

There's a manual switch that could stop both trains from colliding, but the supervisor is 6,300 metres away from it and he needs to get there first. If he runs at an average speed of 14 km/h, will he be in time to save the trains?

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### Correct explanation:

1 -The speed of train A is 160 km/h = 160.000 m/h = 2 667 metres / minute

2 -The speed of train B is 120 km/h = 120.000 m/h = 2 000 metres / minute

3 -They are 140 km away from each other and will be in touch in 30 minutes:

Train A - 30 min x 2 667 m = 80 000 mt = 80 Km

Train B - 30 min x 2 000 m = 60 000 mt = 60 Km

80 Km + 60 Km = 140 Km in 30 minutes

4 - The supervisor is 6,300 metres away, at an average speed of 14 km/h:

14 Km/h = 14 000 m/h = 233 metres / minute

6,300 / 233 = 27 minutes

The supervisor arrived early less 3 minutes than the pervasively trains contact.

LYNA

LYNA • The supervisor will make it in time.

This is how it goes:

The supervisor runs at 14 km/h, that is the same as 3,88 meters per second. He will reach the switch after 1620 seconds.

Train A travels at 160 km/h, which is the same as 44,44 meters per second. So this train will travel 72000 meters in 1620 seconds. This is 72 km.

Train B travels at 120 km/h which is the same as 33,33 meters per second. So this train will travel 54000 meters in 1620 seconds. This is 54 km.

The trains will in total have covered 128 km before the supervisor reaches the switch. They will then still be 12 km apart.

• if the supervisor runs to the manual switch that is 6.300 metres away with a speed of 14 km/h he will be there at 27 mins.

train A in 27 mins will be managed to cover 72 km with a speed of 160 km/h.

train B in 27 mins will be managed to cover 54 km with a speed of 120 km/h.

that means that trains will be already covered the distance of 126 km between them and the wiill still having 14km remaing distance to each other from the 140 km that they had in first place.

So yes the supervisor will manage to stop the accident and save the trains!!!

Explanation:

As train A covers 160 km in an hour, it will cover 80 km in half an hour.

As train B covers 120 km in an hour, it will cover 60 km in half an hour.

After 30 minutes both trains will have approached by 140 kms (80km + 60km).

As the supervisor notices the problem when the trains are 140 km away from each other, he will have exactly 30 minutes to reach the switch.

As the supervisor can run at 14km/h, it will cover 700 meters in 3 minutes.

Then, it will take him 27 minutes to run 6300 meters (3 * 6300 / 700).

Consequently, he will reach the switch 3 minutes before the trains crash.

I hope the trains have a good braking system.

yes, he will be in the time to save the trains

2) explanation:

speed train 1 -> v1 = 160km/h; distance train 1 -> s1 = ?

speed train 2 -> v2 = 120km/h; distance train 2 -> s2 = ?

distance -> s = 140km

v = s/t -> t = s/v

t1 = s1 / v1 and t2 = s2 / v2

As the trains go against each other, they will go for exact the same time to the collision -> t1 = t2

And they go together the distance of s => s = s1 + s2

t1 = t2 = t

s1 / v1 = s2 / v2 // s1 = s - s2

(s - s2) / v1 = s2 / v2

- - - - - - - - - - - - - - -

s2 = (v2s) / (v1 + v2)

s2 = (120 x 140) / (160 + 120)

s2 = 60km

t = s2 / v2

t = 60 / 120

t = 0,5h -> time to collision

speed supervizor -> v3 = 14km/h

distance to switch -> s3 = 6,3km

t3 = s3 / v3

t3 = 6,3 / 14

t3 = 0,45h -> time to switch

If the time to switch is shorter as the time to collision, the trains will be stopped before the collision • This is the solution and explanation The supervisor makes it on time ## Images

• Time until the 2 trains will meet/crash:

t1=d/(VA+VB) > t1=140/(160+120)=0.5h=30 min

Time the supervisor reaches the switch:

t2=d/Vc > t2=6.3/14=0.45h=27 min

t2<t1 > the supervisor has time to operate the manual switch.

• :большой палец вверх:

• Yes he will be able to save the trains ,

explanation: the trains will meet in 30 minutes , train A needs 30 min for 80KM and train B needs 30 min fpr 60KM , together 140KM

the superviser needs 27 minutes for the 6300 meters , so he will be in time