Time, Speed & Distance

Study Materials › Arithmetic › Time, Speed & Distance

This is the largest single topic in the arithmetic section — the 2022 SI paper alone carried questions on trains, boats, races and relative speed. All of them run off one relation; the difficulty is only ever in deciding which speed is relative to what.

Core Formulas

Speed = Distance / Time

km/hr → m/s: multiply by 5/18 · m/s → km/hr: multiply by 18/5

Same distance: speed and time are inversely proportional. If speeds are in the ratio a : b, the times are in the ratio b : a.

Average speed over two equal distances at x and y = 2xy / (x + y)

The inverse-ratio trick. Questions of the form "he travelled at 5/4 of his usual speed and saved 10 minutes" never need a distance. Speeds 5 : 4 mean times 4 : 5; the one-part difference is the time saved. This single idea solves a whole family of questions in seconds.

Trains

Situation	Distance covered
Crossing a pole, man or point	Length of the train
Crossing a platform or bridge	Train length + platform length
Two trains, opposite directions	Sum of lengths, at speed v₁ + v₂
Two trains, same direction	Sum of lengths, at speed v₁ − v₂
Passenger inside a train sees another pass	Only the other train's length

Boats & Streams

Downstream speed = b + s · Upstream speed = b − s

Boat in still water b = (downstream + upstream) / 2

Stream s = (downstream − upstream) / 2

Races

"A beats B by x metres" in a race of d metres means: when A covers d, B covers d − x.

Chain the ratios — if B/A and C/B are known, then C/A = (B/A) × (C/B).

Worked Examples from Real Papers

TS Police SI 2022 — Q17

A man started 10 minutes late, travelled at 1¼ times his usual speed and reached the office on time. Find the time he usually takes.

New speed : usual speed = 5 : 4, so new time : usual time = 4 : 5.
The difference of one part is the 10 minutes saved.
Usual time = 5 parts = 50 minutes.

TS Police Constable 2022 — Q44

A train crosses a 200 m platform in 45 seconds and a man standing on the platform in 25 seconds. If the man walks in the direction of the train at 2 m/s, how long does the train take to cross him?

Crossing the man gives the train's length: L = 25v.
Crossing the platform: (L + 200) = 45v → 25v + 200 = 45v → v = 10 m/s, so L = 250 m.
With the man walking the same way, relative speed = 10 − 2 = 8 m/s.
Time = 250 / 8 = 31¼ seconds.

TS Police SI 2022 — Q31

Two trains run at 126 km/hr and 108 km/hr. Running in opposite directions they cross each other completely in 5 seconds. Running in the same direction, a passenger sitting in the faster train sees the other train pass completely in 30 seconds. Find the length of the faster train.

Opposite directions: relative speed = 234 km/hr = 65 m/s. So L₁ + L₂ = 65 × 5 = 325 m.
Same direction, and the observer is inside a train — so only the other train's length counts. Relative speed = 18 km/hr = 5 m/s, giving L₂ = 5 × 30 = 150 m.
Faster train: L₁ = 325 − 150 = 175 m.
This is the question the "passenger inside" rule exists for — treating it as train-crossing-train would wrongly use the sum of lengths in the second condition.

TS Police SI 2022 — Q36

A motor boat travels from A to B and back. The stream flows at 3 km/hr, the distance AB is 2 km, and the round trip takes 30 minutes. Find the boat's speed in still water.

Let the still-water speed be b.
2/(b + 3) + 2/(b − 3) = 1/2
→ 4b / (b² − 9) = 1/2 → 8b = b² − 9 → b² − 8b − 9 = 0
→ (b − 9)(b + 1) = 0, so b = 9 km/hr (a negative speed is rejected).

TS Police SI 2022 — Q37

In a 1,200 m race A beats B by 120 m. In an 800 m race B beats C by 40 m. In a 400 m race, by how many metres does A beat C?

When A runs 1200, B runs 1080 → B/A = 0.9.
When B runs 800, C runs 760 → C/B = 0.95.
Chaining: C/A = 0.9 × 0.95 = 0.855.
In a 400 m race, C covers 400 × 0.855 = 342 m, so A wins by 400 − 342 = 58 m.

Convert units before anything else. Mixing km/hr with metres and seconds is the most frequent source of lost marks here. Pick one system at the start of the question and stay in it.

Also see: Ratio & Proportion · Averages · Time & Work