## Trains Part 5

Important formulas:

Time is taken to cross a moving object in the direction of the train
Given Length of the object as $L_{1}$
Length of the train as $L_{2}$
The speed of moving the object as $S_{1}$ and
The speed of the train as $S_{2}$
Time taken to cross a moving object in the direction of train = $T=\frac{L_{1}+L_{2}}{S_{1}-S_{2}}$

Time taken to cross a moving object in the opposite direction of train = $T=\frac{L_{1}+L_{2}}{S_{1}-S_{2}}$

Time taken by the train to cross a stationary object like a tree, electric pole, a standing man

$T=&space;\frac{L}{S}$

Time taken by the train to cross a stationary object for e.g. a bridge or a platform.

$T=\frac{L_{1}+L_{2}}{S}$

Time taken by the train to cross a moving object like a walking man of negligible length

When an object moves in the same direction as the train, $T=\frac{L}{S_{1}-S_{2}}$
When an object moves in the opposite direction of the train, $T=\frac{L}{S_{1}+S_{2}}$

Time taken by the train to cross a moving object like another moving train of length Lo

When an object moves in the same direction of the train, $T=\frac{L_{1}+L_{2}}{S_{1}-S_{2}}$
When an object moves in the opposite direction of the f train, $T=\frac{L_{1}+L_{2}}{S_{1}+S_{2}}$
Let two trains of lengths $L_{1}$ and $L_{2}$ takes t1s time to cross each other if they travel in the opposite direction and t2s is they travel in the same direction. Let $S_{f}$ and $S_{s}$ be the speed of the faster and slower train respectively, then
$(S_{f}+S_{s})=\frac{L_{1}+L_{2}}{T_{1}}$
$(S_{f}-S_{s})=\frac{L_{1}+L_{2}}{T_{2}}$
Speed of faster train, $S_{f}=\frac{L_{1}+L_{2}}{2}\left&space;(&space;\frac{1}{T_{1}}+\frac{1}{T_{2}}&space;\right&space;)$
Speed of faster train, $S_{s}=\frac{L_{1}+L_{2}}{2}\left&space;(&space;\frac{1}{T_{1}}-\frac{1}{T_{2}}&space;\right&space;)$

## Trains Part 4

Problem 1: An express train traveled at an avg speed of 100km/h stopping for 3 min after 75km. A local train travel at a speed of 50km/h stopping for 1 min after every 25km. If the train began traveling at the same time, how many km did the local train travel in the time it took the express train to travel 600 km?

Solution:

Express train——->600km——->100m/s

stopping for 3 min after 75km

$\frac{600}{75}-1=7$

7*3=21

Total time =6hours 21min

Local train —–>50km/h

stopping for 1 min after 25km

In 6hours it travels 300km i.e 6*50=300

$\frac{300}{25}=12$

including stopages in 6h 12min it covers 300km

21-12=9

now we have to find in 9 min how many km it will travel

60min———>50km

9 min———->?$\frac{50\times&space;9}{60}=7.5km$

total distance traveled by local train is 300+7.5=307.5km

Problem 2: A, B, C start from Delhi at 10 AM, 11 AM, 12 AM towards Goa and their speed are 3 km/h, 4 km/h, 5 km/h and after meeting on the way, B sends back A to C with a message at what time C will get the message?

Solution:

At 2 PM both A and B will meet

L.C/M OF 3 and 4 is 12

C travels 10 km

12-10=2km

$\frac{2}{8}\times&space;60=15min$

Total time=2:15pm

Problem 3: A train left station A form station B at a certain speed after traveling for 100km the train meets with an accident and could travel at $\frac{4}{5}$th of the original speed and reaches 45 min late at the station B. Had the accident taken place 50km further on it would have reached 30min late at station B. What is the distance between station A and station B?

Solution:

usual             new

S               5            :          4

T              4             :           5

4-5=1

After traveling 100km train meet with an accident and reaches 45min late

1——–>$\frac{45}{60}=\frac{3}{4}$

4———->?$\frac{3\times&space;4}{4}=3$h

Had the accident taken place 50km further on it would have reached 30min late

1———>$\frac{30}{60}=\frac{1}{2}$

4———->?$\frac{1\times&space;4}{2}=2$h

Time gap=3h-2h=1hr

in one hour it travels 50km

1 h———>50km

3*1———->50*3 km

3h————>150km

total distance=100+150=250

Shortcut:

45min-30min=15min

3*50km=15min*3

150km——->45min

100+150=250km

## Trains Part 3

Problem 1: Two trains one from A to B, another from B to A start simultaneously after they meet the train reach their destination after 9h and 16h respectively the ratio of speeds is?

Solution:

$\sqrt{\frac{T_{1}}{T_{2}}}=\frac{S_{2}}{S_{1}}$

$\sqrt{\frac{9}{16}}=\frac{S_{2}}{S_{1}}$

$\frac{3}{4}=\frac{S_{2}}{S_{1}}$

$S_{1}:S_{2}=3:4$

Problem 2: The distance between Delhi and Amritsar is 450km. A train starts from Delhi at 4 pm and travels towards Amritsar at 60km/h and B train starts from Amritsar and travels towards Delhi at 3:20 PM at the speed of 80km/hr. At what time they will meet?

Solution:

40+450=490

$\frac{490}{140}=\frac{7}{2}=3\tfrac{1}{2}h$

3:20+3:30=6:30pm

Problem 3: A travel 600km to his home partly by train and partly by car he takes 8 hours. If he travels 120km by train and rest by car it takes 20min more. When he travel 200km by train and rest by car. Find the speed of the car?

Solution:

$\frac{120}{T}+\frac{480}{C}=8$————->EQU1

$\frac{200}{T}+\frac{400}{C}=8\frac{1}{3}$————->EQU2

$5\times&space;\left&space;[&space;\frac{120}{T}+\frac{480}{C}&space;\right&space;]=8\times&space;5$

$3\times&space;\left&space;[&space;\frac{200}{T}+\frac{400}{C}&space;\right&space;]=8\tfrac{1}{3}\times&space;5$

$\frac{2400}{C}-\frac{1200}{C}=40-25$

$\frac{1200}{C}=15$

C=80

## Trains Part 2

Problem 1: Two trains start at the same time from two stations and proceed towards each other at the speeds oh 15km/h and 20km/h respectively. When the train meets it is found that one train has travel 50km more than the other. Find the distance between two stations?

Solution:

Train A can travel 15km/h and train B can travel 20km/h

1 hour ———>   5km

10*1 hour————> 5km*10

10h———->50km

10*15=150

10*20=200

150+200=350km

Problem 2: Two trains running in opposite direction crosses a man standing on a platform in 27sec and 17sec respectively and they cross each other in 23 sec the ratio of their speed is?

Solution:

$S_{1}=\frac{L_{1}}{27},S_{2}=\frac{L_{2}}{17}$

$L_{1}=27S_{1},L_{2}=17S_{2}$

$S_{1}+S_{2}=\frac{L_{1}+L_{2}}{t}$

$S_{1}+S_{2}=\frac{27S_{1}+17S_{2}}{23}$

$23S_{1}+23S_{2}=27S_{1}+17S_{2}$

$S_{1}:S_{2}=3:2$

Method 2:

27                     17

23

6                          4

3           :              2

Problem 3: A train overtakes two persons walking along the railway track the first person walks at 4.5km/h and the other walks at 5.4km/hr the train needs 8.4 and 8.5 seconds respectively to overtake them. what is the speed of train if both the persons are walking in the same direction?

Solution:

$\frac{(S_{1}-4.5)\times&space;5}{18}=\frac{L_{1}}{8.4}$

$\frac{(S_{1}-5.4)\times&space;5}{18}=\frac{L_{1}}{8.5}$

$\frac{S_{1}-4.5}{S_{1}-5.4}=\frac{8.5}{8.4}$

$S_{1}=81m/s$

## Trains Part 1

Problem 1: A train travels at the rate of 45km/h. How many seconds will it take to cover a dist of  $\frac{4}{5}$ km?

Solution:

$Speed=\frac{Distance}{Time}$

$45=\frac{4}{5\times&space;t}$

$t=\frac{4}{5\times&space;45}\times&space;60\times&space;60=64sec$

Problem 2: A train passes two bridges of length 800m and 400m in 100sec and 60sec respectively. Find the length of the train?

Solution:

$\frac{800+x}{100}=\frac{400+x}{60}$

$2400+3x=2000+5x$

$2x=400$

$x=200$

Problem 3: A train 150m long is moving at a speed of 85km/hr. Will it cross a  man coming from opposite direction at a speed of 5kms?

Solution:

85+5=90

$90\times&space;\frac{5}{18}=25$

$Speed=\frac{Distance}{Time}$

$25=\frac{150}{T}$

$T=\frac{150}{25}=6$

Problem 4: Two trains of equal length are running on parallel lines in the same direction at the rate of 46km/h and 36km/hr. The faster train passes the slower train in 36 sec. The length of the train is?

Solution:

$S=\frac{L_{1}+L_{2}}{T}$

$\frac{10\times&space;5}{18}=\frac{2L}{6}$

L=50m

Problem 5: A goods train 150m long moving at the speed of 54km/hr. A man is sitting on a passenger train which is moving with 18km/hr in the same direction, the length of the passenger train is 100m. In how many seconds the goods train crosses the man sitting in the passenger train?

Solution:

$S_{1}-S_{2}=\frac{L}{T}$

54-18=36

$36\times&space;\frac{5}{18}=\frac{150}{T}$

$10=\frac{150}{T}$

T=15 sec