Shortcut - Problems on Trains
Train crossing a stationary object with negligible width.
Time taken to cross =$\dfrac{L_{t}}{S_{t}}$
$L_{t}$ – Length of the train
$S_{t} $– Speed of the train
Question:
A train of length 450m is travelling at a speed of 54kmph. How long will it take to cross a pole?
Answer: Speed of train$ S_{T}$ in m/s = 54x(5/18)
= 15mps
Length of train = $L_{T} $= 450m
Time taken to cross = 450/15=30 seconds.
Shortcut - Problems on Trains
Train crossing a stationary object with considerable length.
Time taken to cross=$\dfrac{L_{t}+L_{p}}{S_{t]}}$
$L_{t} $– Length of the train;
$ L_{p}$ – Length of the platform
$S_{t}$ – Speed of the train
Question:
A train of length 200m crosses a platform of length 150m at a speed of 72kmph. How long will the train take to cross the platform?
Answer: $L_{t}$ = 200m
$L_{p}$ = 150m
$S_{t}$ = 72x(5/18) = 20mps
Time taken to cross the platform = (200+150)/20
= 17.5 seconds
Shortcut - Problems on Trains
Two trains crossing in the opposite directions.
Time taken to cross=$\dfrac{L_{1}+L_{2}}{S_{1}+S_{2}}$
$L_{1}$ – Length of the train 1;
$ L_{2}$ – Length of train 2
$S_{1}$– Speed of the train 1;
$S_{2}$ – Speed of train 2
Question:
Two trains A and B of length 200m and 300 are travelling at a speed of 54kmph and 36kmph respectively in opposite directions. How long will they take to cross each other?
Answer: $L_{1}$= 200;
$L_{2}$= 300
$S_{1}+S_{2}$= 66 + 24 = 90 kmph = 25 mps.
Substitute the values in the above equation, we get
T = (200+300)/(25)
= 20 seconds
Shortcut - Problems on Trains
Train crossing another train from the same direction.
Time taken to cross=$\dfrac{L_{1}+ L_{2}}{S_{1}- S_{2}} $
$L_{1} $ – Length of the train 1;
$ L_{2} $ – Length of train 2
$S_{1} $ – Speed of the train 1;
$ S_{2} $ – Speed of train 2
Question:
Two trains A and B of length 200m and 300 are travelling at a speed of 66 kmph and 30 kmph respectively in same directions. How long will they take to cross each other?
Answer: $L_{1}$ = 200;
$L_{2}$= 300
$S_{1}$ - $S_{2}$ = 66 – 30 = 36 kmph = 10 mps
Substitute the values in the above equation, we get
T = (200+300)/(10)
= 50 seconds
Shortcut - Problems on Trains
Train crossing a man sitting on another train travelling in opposite direction.
Time taken to cross=$\dfrac{L_{1}}{S_{1}+S_{2}}$
$L_{1}$– Length of the train which is crossing the man
$S_{1}$ – Speed of the train 1;
$ S_{2}$ – Speed of train 2
Question:
Two trains A and B of length 300m and 400 are travelling at a speed of 72kmph and 36kmph respectively in opposite directions. How long will train A take to cross a man travelling in train B?
Answer: L1 = 300, (Length of train crossing the man)
$S_{1}$ + $ S_{2} $ = 72 + 36 = 108 kmph = 30 mps
Substitute the values in the above equation, we get
T = (300)/(30)
= 10 seconds
Shortcut - Problems on Trains
Train crossing a man sitting on another train travelling in same direction.
Time taken to cross=$\dfrac{L_{1}}{S_{1} - S_{2}}$
$L_{1}$ – Length of the train which is crossing the man
$S_{1}$– Speed of the train 1;
$ S_{2}$– Speed of train 2
Question:
Two trains A and B of length 100m and 200 are travelling at a speed of 72kmph and 54kmph respectively in same directions. How long will train A take to cross a man travelling in train B?
Answer: $L_{T}$= 100, (Length of train crossing the man)
$S_{1}$= 72kmph = 72(5/18) mps = 20mps
$S_{2}$ = speed of the man travelling in train B
$S_{2}$ = 54kmph = 36(5/18) mps = 10mps
Substitute the values in the above formula, we get
T = (100)/(20-10)
= 10 seconds