Special case of Time, Speed and Distance - Train, Boat and Stream

Trains

Trains are special case in the concept of speed, time and distance. The basic relation is same as,
Speed × time = distance.
These points need to be kept in mind while solving a train related questions.

1. When a train is crossing a moving object, the speed has to be taken as relative speed of the train with respect to the object.

2. The distance to be covered while crossing an object whenever a train crosses an object will be equal to
Length of train + Length of the object

3. Time taken by the train to cross a pole, man etc. by a train,
= $$\frac{{{\text{Length}}\,{\text{of}}\,{\text{train}}}}{{{\text{Speed}}\,{\text{of}}\,{\text{train}}}}$$

Example:
A train crosses a pole in 8 seconds. If the length of the train is 200 meters, find the speed of train.

Solution:
Time taken to cross the pole = $$\frac{{{\text{Length}}\,{\text{of}}\,{\text{train}}}}{{{\text{Speed}}\,{\text{of}}\,{\text{train}}}}.$$
Speed of train = $$\frac{{{\text{Length}}\,{\text{of}}\,{\text{train}}}}{{{\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{cross}}\,{\text{the}}\,{\text{pole}}}}$$
Speed of train = $$\frac{{200}}{8}$$ = 25 m/s.
= $$\frac{{25 \times 18}}{5}$$  = 90 km/h.

1. Time taken to cross another train or bridge or platform etc. by a train is given by,
= $$\frac{{{\text{Length}}\,{\text{of}}\,\left( {{\text{train}} + {\text{platform}}\,{\text{etc}}{\text{.}}} \right)}}{{{\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}}}$$

Example:
A 150 m long train passes another train 100 m long traveling in opposite direction. If the speeds of the trains are 39 km/h and 21 km/h respectively, how long will it take to cross each other?

Solution:
Total distance = length of 1^st train + length of 2^nd train
= 150 + 100 = 250 m.
Since, trains are moving in opposite directions, their relative speed= 39 + 21 = 60 km/h.
$$ = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3}\,{\text{m/sec}}{\text{.}}$$
Hence, time taken to cross = $$\frac{{250}}{{\left( {\frac{{50}}{3}} \right)}}$$
= $$\frac{{250 \times 3}}{{50}}$$ = 15 seconds.

Key Facts:

1) If two persons start at the same time from two different positions towards each other and they complete their journey in 'a' and 'b' hours respectively after meeting each other then-
$$\frac{{{\text{Speed}}\,{\text{of}}\,{\text{A}}}}{{{\text{Speed}}\,{\text{of}}\,{\text{B}}}} = {\text{sqrt}}\,{\text{of}}\,\frac{{\text{b}}}{{\text{a}}}$$

Example:
Two trains are approaching each other from two different points at the same time. They complete their journey in $$3\frac{1}{3}$$ and $$4\frac{4}{5}$$ hours after they met each other. Speed of A is 8 km/h. what is that of speed of B.

Solution:
$$\eqalign{ & \frac{{{\text{Speed}}\,{\text{of}}\,{\text{A}}}}{{{\text{Speed}}\,{\text{of}}\,{\text{B}}}} = \frac{{{\text{sqrt}}\,{\text{of}}\,4\frac{4}{5}}}{{{\text{sqrt}}\,{\text{of}}\,3\frac{1}{3}}} \cr & \frac{8}{{\text{B}}} = \frac{{{\text{sqrt}}\,{\text{of}}\,\frac{{24}}{5}}}{{{\text{sqrt}}\,{\text{of}}\,\frac{{10}}{3}}} \cr & {\text{On}}\,{\text{squaring}}\,{\text{both}}\,{\text{sides,}}\,{\text{we}}\,{\text{get,}} \cr & \frac{{64}}{{{{\text{B}}^2}}} = \frac{{24 \times 3}}{{10 \times 5}} \cr & {{\text{B}}^2} = \frac{{64 \times 10 \times 5}}{{24 \times 3}} \cr & {{\text{B}}^2} = \frac{{400}}{9} \cr & {\text{B}} = \frac{{20}}{3} \cr & {\text{B}} = 6\frac{2}{3} \cr} $$

1. A monkey or man tries to climb a pole 'x' m high. In 't' time he climbs 'y' m but slips down 'z' m. How much time will be required by him to reach the top?

$$\eqalign{ & \left[ {\frac{{{\text{ht}}{\text{.}}\,{\text{of}}\,{\text{pole}} - {\text{slipped}}\,{\text{dist}}{\text{.}}}}{{{\text{climbed}}\,{\text{dist}}{\text{.}} - {\text{slipped}}\,{\text{dist}}{\text{.}}}}} \right] \times {\text{t}} \cr & {\text{ = }}\left[ {\frac{{x - z}}{{y - z}}} \right] \times t \cr} $$

Example:
A monkey tries to climb a pole 12 m high. In each minute he climbs 2 m but slips down 1 m. how much time will be required by him to reach the top?

Solution:
$$\eqalign{ & \left[ {\frac{{{\text{ht}}{\text{.}}\,{\text{of}}\,{\text{pole}} - {\text{slipped}}\,{\text{dist}}{\text{.}}}}{{{\text{climbed}}\,{\text{dist}}{\text{.}} - {\text{slipped}}\,{\text{dist}}{\text{.}}}}} \right] \times {\text{t}} \cr & {\text{ = }}\left[ {\frac{{x - z}}{{y - z}}} \right] \times t \cr & = \left[ {12 - \frac{1}{2} - 1} \right] \times 1 \cr & = 11\,{\text{minute}} \cr} $$

Boat and Stream

The problem of boats and streams are also based on formula,
Speed × time = distance
Some points need to be kept in mind while solving problem related with Boat and stream.

1. Speed of boat moving downstream or in direction of flow of stream = speed of boat in still water+ speed of stream.
2. Speed of boat moving upstream or in opposite direction of the flow of stream = Speed of boat in still water - speed of stream.
3. Speed of boat in still water = speed of boat

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