3.3 Time, Speed & Distance

Why Time, Speed & Distance Matters

Distance, speed and time (TSD) problems are the most common arithmetic word problems in the RRB Mathematics section. They appear as train problems, boat-and-stream problems, and simple journey problems. The math is light — the marks go to candidates who handle units confidently.

Core Formula

• Distance = Speed × Time
• Speed = Distance / Time
• Time = Distance / Speed

Unit Conversion (must memorise)

• 1 km/h = 5/18 m/s
• 1 m/s = 18/5 km/h

Worked example 1 (unit conversion)

Convert 72 km/h to m/s.

• 72 × 5/18 = 4 × 5 = 20 m/s

Average Speed

When equal distances are travelled at speeds u and v:

• Average = 2uv / (u + v) (harmonic mean)

When equal times are spent at speeds u and v:

• Average = (u + v) / 2 (arithmetic mean)

Worked example 2 (round trip)

A man goes to office at 30 km/h and returns at 20 km/h by the same route. Average speed for the whole trip?

• Equal distances, so use harmonic mean: 2 × 30 × 20 / (30 + 20) = 1200 / 50 = 24 km/h

Note that (30 + 20)/2 = 25 km/h would be wrong — that applies only for equal times, not equal distances.

Relative Speed

• Same direction: relative speed = u − v (faster catches slower)
• Opposite directions: relative speed = u + v (they approach each other faster)

Trains

• Time for a train of length L metres to pass a pole or a stationary person = L / (speed in m/s)
• Time to pass a platform/bridge of length P = (L + P) / (speed in m/s)
• Two trains passing each other (opposite directions) = (L1 + L2) / (u + v)
• Same direction = (L1 + L2) / (u − v)

Worked example 3 (train + platform)

A 150 m long train runs at 54 km/h. How long does it take to cross a platform 250 m long?

• Speed = 54 × 5/18 = 15 m/s
• Total distance to cover = 150 + 250 = 400 m
• Time = 400 / 15 = 80/3 ≈ 26.67 seconds

Boats & Streams

• Speed of boat in still water = b
• Speed of stream/current = s
• Downstream speed = b + s (current helps)
• Upstream speed = b − s (current opposes)
• b = (downstream + upstream) / 2
• s = (downstream − upstream) / 2

Worked example 4 (find boat and stream)

A boat goes 30 km downstream in 2 hours and the same distance upstream in 5 hours. Find the speed of the boat in still water and the stream.

• Downstream = 30 / 2 = 15 km/h
• Upstream = 30 / 5 = 6 km/h
• b = (15 + 6) / 2 = 10.5 km/h
• s = (15 − 6) / 2 = 4.5 km/h

Diagram — Boat Speeds Relative to the Stream

graph LR
  B["Boat in still water: b"] --> D["Downstream: b + s"]
  B --> U["Upstream: b − s"]
  S["Stream: s"] --> D
  S --> U

Worked Examples — Extended Scenarios

Worked example 5 (two trains, opposite directions)

Two trains of length 120 m and 80 m are moving in opposite directions at 60 km/h and 40 km/h. How long do they take to completely cross each other?

• Relative speed (opposite) = 60 + 40 = 100 km/h = 100 × 5/18 = 250/9 m/s
• Total distance = 120 + 80 = 200 m
• Time = 200 / (250/9) = 200 × 9 / 250 = 1800 / 250 = 7.2 seconds

Worked example 6 (catch-up, same direction)

A thief running at 30 km/h is 200 m ahead of a policeman running at 40 km/h. How long will the policeman take to catch the thief?

• Relative speed (same direction) = 40 − 30 = 10 km/h = 10 × 5/18 = 50/18 = 25/9 m/s
• Gap to close = 200 m
• Time = 200 / (25/9) = 200 × 9 / 25 = 1800 / 25 = 72 seconds

Worked example 7 (average speed over three equal distances)

A car covers three equal stretches of 60 km each at 30 km/h, 40 km/h and 60 km/h. Find the average speed for the whole journey.

• For equal distances, the harmonic-mean rule extends: average = 3 / (1/u + 1/v + 1/w)
• 1/30 + 1/40 + 1/60 = 4/120 + 3/120 + 2/120 = 9/120 = 3/40
• Average = 3 / (3/40) = 40 km/h

This is the general pattern — the arithmetic mean (30 + 40 + 60)/3 = 43.33 km/h is wrong here because the car spends unequal times on each stretch.

Common Speed Conversion Table

Memorise a few anchor rows (18 km/h = 5 m/s is the most useful) — most RRB train problems use one of these speeds.

Common Exam Traps

• Always convert km/h to m/s (multiply by 5/18) when distance is given in metres.
• Two trains "crossing each other" require the SUM of their lengths — both trains must fully pass.
• A train passing a stationary man has only its own length to cover (the man is a point).
• "Against the stream" means upstream — subtract the stream speed.
• "Average speed for the whole journey" with equal distances is the harmonic mean, not the arithmetic mean.
• When two objects move towards each other, the closing speed is the sum; when one chases the other, it is the difference.
• For a train crossing a bridge, the bridge length is added to the train's length only because the train must clear the bridge; a man standing on the bridge is treated as a point.
• If a question gives speed in m/min, convert to m/s by dividing by 60 — never compare m/min with km/h directly.
• "Speed of the stream" is positive; "upstream speed" can be zero (boat just holds position) but is never negative in a valid RRB question.