Problems on Trains

When a question talks about a train going from one city to another, over hundreds of kilometres and hours of travel, the length of the train (a few hundred metres) does not matter. Here the train is a moving point, and the whole TSD chapter applies directly. The question says "train", but the working is plain D = S × T.

1. The basic relation. Distance = Speed × Time. A train running at 50 km/h for 3 hours covers 150 km. Watch the units: if the time is in seconds and the answer is wanted in metres, convert the speed to m/s first.

2. Same speed: distance and time go together. If the speed does not change, distance ∝ time. A train that covers 90 km in 2 hours will cover 225 km in 5 hours. You don't even need the speed: 52 × 90 = 225.

3. A journey in several parts. Find the time (or distance) of each part separately and add. Average speed is always Total distanceTotal time, never the simple average of the speeds. For two equal distances at a and b, average speed = 2aba + b; for three equal distances at a, b and c it is 3abcab + bc + ca.

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4. Two speeds, late and early. If at speed S₁ a train is late and at S₂ it is early, the two journey times differ by "late + early". So DS₁ − DS₂ = total time difference, which gives D = S₁ × S₂S₂ − S₁ × (time difference).

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5. Speed reduced, same distance. For a fixed distance, speed ∝ 1time. If the time goes from 3 hours to 4 hours, the speed has become 34 of the original; it has fallen by 14.

For example, Radhika's train leaves at 7:30 a.m. and reaches at 11:00 a.m., running at 64 km/h. Time = 312 hours, so distance = 64 × 72 = 224 km. Marks are usually lost in counting the clock time across a.m. and p.m., not in the multiplication.

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This is the first real train type, and the simplest. A pole, a signal post, a tree, a milestone or a man standing on the platform: all of these are point objects. Their width is so small compared with the train that we take it as zero.

So, as we saw in the foundation, while crossing a point object the train covers only its own length:

Length of train = Speed × Time taken to cross the pole

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There are only three quantities, length L, speed S and time t, and the question hides one of them.

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1. Find the speed. A 280 m train passes a signal in 14 seconds. S = 28014 = 20 m/s = 20 × 185 = 72 km/h.

2. Find the length. A train at 63 km/h passes a tree in 8 seconds. 63 km/h = 63 × 518 = 17.5 m/s, so L = 17.5 × 8 = 140 m.

3. Find the time. A 330 m train runs at 99 km/h = 27.5 m/s. It passes a standing man in 33027.5 = 12 seconds.

Read the sentence carefully. "A man standing on the platform" is still a point object, even though the platform is mentioned. The train only has to get past him, not past the platform. Always ask: what exactly is being crossed?

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Radhika is waiting at a level crossing, and a 160 m goods train takes 16 seconds to pass her. Its speed is 16016 = 10 m/s, that is 36 km/h. The width of the crossing did not matter, because she herself is the point being crossed.

Now the object has a length of its own. With a platform, a bridge or a tunnel, the train has to go over the whole of it and then get its last coach off it. Look at the figure: when the engine is at the start of the platform, the last coach is still L metres behind. The crossing is over only when that last coach reaches the far end of the platform, so the last coach travels L + P.

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Length of train + Length of platform = Speed × Time

The same unknowns appear again, but now the length side has two parts.

1. Find the speed. A 220 m train crosses a 330 m bridge in 25 seconds. Distance = 220 + 330 = 550 m. Speed = 55025 = 22 m/s = 22 × 3.6 = 79.2 km/h.

2. Find the time. A 260 m train at 81 km/h (= 22.5 m/s) enters a 640 m tunnel. Time = 260 + 64022.5 = 90022.5 = 40 seconds.

3. Find a length. A train at 45 km/h (= 12.5 m/s) crosses a 175 m platform in 30 seconds. Total distance = 12.5 × 30 = 375 m, so train length = 375 − 175 = 200 m. If the train length is given instead, the same subtraction gives the platform length.

4. Lengths linked to each other. Sometimes the platform is twice the train or *50% longer than the train*. Then write everything in terms of L. If the platform is twice the train, total distance = L + 2L = 3L. A train at 20 m/s crossing such a platform in 24 seconds covers 480 m, so 3L = 480 and L = 160 m.

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Radhika's train, 180 m long, runs at 54 km/h = 15 m/s through a tunnel 420 m long. From the moment the engine enters till the last coach comes out, it takes 180 + 42015 = 40 seconds. But the coach in which Radhika sits is in darkness for only 42015 = 28 seconds, because for one coach (a point) only the tunnel's length counts.

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Here the question shows the same train crossing two different things, usually a pole (or a standing man) and a platform, or two bridges of different lengths. Often the length of the train is not given. Don't panic. One thing does not change between the two situations, the speed of the train, and everything is worked out from it.

Idea 1: the extra time is for the extra length. The train takes t₁ seconds for a pole and t₂ seconds for a platform of length P. In both cases it covers its own length; for the platform it covers P more. So the extra time (t₂ − t₁) was spent only on those extra P metres:

Speed = Pt₂ − t₁, and then L = Speed × t₁

A train passes a signal in 9 seconds and a 270 m platform in 24 seconds. The extra 15 seconds were for the 270 m of platform, so speed = 27015 = 18 m/s and length = 18 × 9 = 162 m.

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Idea 2: two platforms or two bridges. If the same train crosses P₁ in t₁ and P₂ in t₂, the difference of the lengths is covered in the difference of the times: Speed = P₂ − P₁t₂ − t₁.

Idea 3: constant speed ⇒ distance ∝ time. Since the speed is fixed, the distances are in the same ratio as the times. Pole : platform crossing = L : (L + P) = t₁ : t₂. This gives the length at once: L = P × t₁t₂ − t₁. And when the length is known and one crossing is given, a new crossing time is t_new = t × L + P_newL + P.

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Idea 4: find the speed first, then everything else. If the length of the train is given along with one crossing, the speed follows directly; any other crossing is then (new distance) ÷ speed.

The same idea also comes as data sufficiency: is statement I alone sufficient, is statement II alone sufficient? Ask yourself one thing only. Does this statement let me find the speed? If yes, the time for any platform follows.

Two trains on parallel tracks are coming towards each other. Watch the figure: the crossing begins when the two engines are side by side and ends when the two last coaches separate. In that time the two trains together have moved L₁ + L₂. And since they are coming towards each other, the gap closes at S₁ + S₂.

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Time to cross each other = L₁ + L₂S₁ + S₂

So in opposite directions lengths add and speeds add. Every question is this one equation with one unknown.

1. Find the time. Trains of 150 m and 175 m run at 50 km/h and 40 km/h. Relative speed = 90 km/h = 25 m/s (add first, convert once). Time = 32525 = 13 s.

2. Find the other speed. A 210 m train at 36 km/h (10 m/s) crosses a 240 m train coming the other way in 9 s. Relative speed = 4509 = 50 m/s, so the other train = 50 − 10 = 40 m/s = 144 km/h.

3. Equal speeds. If both trains have the same speed v, relative speed = 2v. Two 200 m trains crossing in 10 s ⇒ 2v = 40010 = 40 ⇒ v = 20 m/s each.

4. Two-step questions. Often you first get a speed or a length from a pole crossing (Type 2) and then use it here. Do the first crossing on its own, write down L and S, and only then go to the two-train equation.

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5. A passenger sitting in one train. Suppose Radhika sits in a train at 40 km/h and a 250 m train passes her from the opposite side at 50 km/h. She is a point, so the distance is only the other train's length, 250 m, and the speed is 90 km/h = 25 m/s. It passes her in 10 s. Her own train's length does not come in.

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Now both trains run the same way on parallel tracks and the faster one overtakes the slower. The overtaking starts when the engine of the faster train comes level with the last coach of the slower one. It ends when the last coach of the faster train goes past the engine of the slower one. So the faster train has to gain both lengths, L₁ + L₂, on the slower train.

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But it gains only at the rate of the difference of speeds, because the slower train is also moving forward:

Time to overtake = L₁ + L₂S₁ − S₂

So in the same direction lengths still add, but speeds subtract. That is why overtaking takes much longer than crossing a train that comes from the front.

1. Find the time. A 180 m train at 70 km/h overtakes a 220 m train at 34 km/h. Relative speed = 36 km/h = 10 m/s. Time = 40010 = 40 s.

2. Find the slower speed. A 260 m train at 90 km/h overtakes a 190 m train in 45 s. Relative speed = 45045 = 10 m/s = 36 km/h, so the slower train = 90 − 36 = 54 km/h.

3. Same pair, both directions. If the same two trains take t_o to cross in opposite directions and t_s to overtake, then the sum of speeds = L₁ + L₂t_o and the difference = L₁ + L₂t_s: two equations, two speeds. With a total length of 360 m, 12 s in opposite directions and 36 s in the same direction, S₁ + S₂ = 30 and S₁ − S₂ = 10, so the speeds are 20 m/s and 10 m/s.

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4. Two trains of equal length that pass a pole (or a man) in t₁ and t₂. Here L = S₁t₁ = S₂t₂. Putting this into the formulas gives two neat results (take t₁ < t₂, so train 1 is the faster): in the same direction they cross in 2t₁t₂t₂ − t₁, and in opposite directions in 2t₁t₂t₁ + t₂.

Radhika's train, 160 m long, is running at 72 km/h when a 200 m express at 108 km/h overtakes it. The express gains 36 km/h = 10 m/s, so it needs 36010 = 36 seconds to get fully ahead.

A walking man is still a point object, so his length is ignored, but now he is moving. The distance is only the train's length, while the speed becomes relative:

• Man walking towards the train (opposite direction): t = LS + u
• Man walking in the same direction as the train: t = LS − u

Take a 200 m train running at 85 km/h, with Radhika walking beside the track at 5 km/h. If she walks towards the train, relative speed = 90 km/h = 25 m/s, and the train passes her in 20025 = 8 s. If she walks in the train's direction, relative speed = 80 km/h = 2009 m/s, and it takes 200 ÷ 2009 = 9 s. The one-second difference comes only from the direction in which she walks.

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If the moving thing has a length, such as a tractor, a bus or a car whose length is given, it is a long object, exactly like a second train. Distance = L + (its length), speed = S ± u.

Two men at different speeds. A train passes one cyclist at 3 m/s in 10 s and another at 5 m/s in 12 s, both going the train's way. The length is the same both times: (S − 3) × 10 = (S − 5) × 12 ⇒ 10S − 30 = 12S − 60 ⇒ S = 15 m/s, and L = 12 × 10 = 120 m.

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Speeds given as a ratio. If the train is k times as fast as the man, take the man's speed as x and the train's as kx. Towards each other the relative speed is (k + 1)x; in the same direction it is (k − 1)x. Find x from the crossing, then multiply by k.

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In this last type the trains are again points, because the distance between the stations is hundreds of kilometres. The question is when and where the two trains meet. This is the relative speed method of the TSD chapter, now with trains.

1. Starting together, towards each other. The gap D closes at a + b, so they meet after Da + b hours. If one train is a certain percent faster, first write the speeds as a ratio: 25% faster ⇒ 5 : 4.

2. Starting together in the same direction (the faster one behind). The gap closes at a − b, so the faster one catches up after Da − b hours. If a question gives both, the meeting time when moving towards each other and the catching-up time when moving the same way, you get a + b and a − b, and the two speeds follow at once.

3. One starts late: move the early one first. Train A starts at 7 a.m. at 50 km/h, and train B starts from the other end at 9 a.m. at 70 km/h, with 460 km between them. By 9 a.m. A has covered 100 km. The remaining 360 km closes at 120 km/h, so they meet 3 hours after 9 a.m., at 12 noon. Always measure the meeting time from the moment both are moving, then add it to that clock time.

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4. Converting decimal hours to a clock time. 1.7 h is not 1 h 70 min. 0.7 h = 0.7 × 60 = 42 min, so 1.7 h = 1 h 42 min. Be careful with a.m. and p.m. while adding.

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Two more results from the TSD chapter also come with trains. If after meeting train A takes t₁ hours and train B takes t₂ hours to reach its destination, then S_AS_B = √t₂t₁. And if at the meeting one has run d km more than the other, then D = a + ba − b × d.

Here is a quick one. Radhika's train leaves station P at 60 km/h, and at the same time another train leaves Q towards P at 40 km/h. When they meet, Radhika's train has run 50 km more than the other. So D = 10020 × 50 = 250 km.

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A railway line often runs along a river, so a question can put a train and a boat in the same picture. Nothing new is needed. The boat is a moving object whose speed over the ground you must find first, using the boat chapter. After that it is an ordinary train-and-moving-object question.

• Boat sailing downstream with still-water speed b and stream s: ground speed = b + s.
• Boat sailing upstream: ground speed = b − s.

Then compare the directions. If the train and the boat move the same way, the relative speed is the difference of the two. If they move in opposite ways, it is the sum.

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Take Radhika's boat with b = 14 km/h in a stream of 4 km/h. Downstream she moves at 18 km/h over the ground. A 200 m train running the same way at 90 km/h passes her with relative speed 90 − 18 = 72 km/h = 20 m/s, so it takes 20020 = 10 s. If she were sailing upstream at 10 km/h, the train running with the flow would meet her head on, with relative speed 90 + 10 = 100 km/h.

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A boat or swimmer with a length. If the length of the boat is given, it is a long object: distance = length of train + length of boat. A swimmer, or a boat whose length is ignored, is a point.

Two crossings, same boat speed. Two crossing times (one with a downstream boat, one with an upstream boat) give two ground speeds d and u. Their average is the still-water speed and half their difference is the stream: b = d + u2 and s = d − u2.