PROBLEMS ON TRAIN
Before, entering into the topic one must know the common relation which is
S=D/T
S- speed of the train in (km/hr) or (m/s)
D- distance of the train or object in (km) or (m)
T- time it takes in (hr) or (sec)
CONVERSION:
Km/hr to m/s multiply by 5/18
m/s to km/hr multiply by 18/5
what is this 5/18???
It is 1 km equals to 1000m & 1 hr equals to 3600secs
So 1km/1hr = 1000m/3600secs on further simplifying 5/18 m/s.
NOTE- always convert speed rather than time & distance.
Some factors to be known before solving the problem based on this, the problem classified into several types…
LENGTH OBJECTS- Platforms, bridges, tunnels, trains
NO LENGTH OBJECTS- Tree, man, pole, post, car, bus.
WHEN THE TRAIN CROSSES NO LENGTH OBJECTS | |||
IF, STANDING | DIRECTION | SPEED | DISTANCE |
| - | Train speed S | Train distance D |
IF, MOVING | DIRECTION | SPEED | DISTANCE |
| SAME | S1-S2 S1-speed of train S2-speed of man/car etc | Train distance D |
OPPOSITE | S1+S2 S1-speed of train S2-speed of car/man etc | Train distance D | |
WHEN THE TRAIN CROSSES LENGTH OBJECTS | |||
IF, STANDING | DIRECTION | SPEED | DISTANCE |
| - | Train speed S | D1+D2 D1- distance of train D2- Distance of length objects |
IF, MOVING | DIRECTION | SPEED | DISTANCE |
s | SAME | S1-S2 S1-speed of train P S2-speed of train Q | D1+D2 D1-distance of train P D2-distance of train Q |
OPPOSITE | S1+S2 S1-speed of train P S2-speed of train Q | D1+D2 D1-distance of train P D2-distance of train Q | |