Speed, Distance and Time Problems: The Zero-Confusion Formula Guide

Speed, distance and time is one of those topics where students consistently lose marks — not because the concepts are hard, but because the problem types are diverse and the unit conversions are easy to get wrong under pressure. A train problem, a boat problem, a meeting point problem, and a circular track problem all use the same three variables — but each has a different structural setup that confuses candidates who have not systematically categorized the problem types.

This guide eliminates that confusion entirely. Every major problem type is presented with its exact formula, the reasoning behind it, and at least two fully worked examples. By the end, you will have a complete mental template for every speed-distance-time variation that appears in SSC, RRB, IBPS, and school-level examinations.

The Foundation: Three Forms of One Formula

Everything in this topic derives from one relationship:

Distance = Speed × Time

Rearranged:

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

These three forms must be instantaneously accessible — you should not need to think about which form to use. The rule is simple: the unknown variable is isolated; the other two are multiplied or divided accordingly.

Unit Conversion — The Most Common Source of Errors

Unit mismatches cause more wrong answers in this topic than any conceptual error.

Essential conversions:

Memory trick for km/h ↔ m/s:

• To go from km/h to m/s: multiply by 5/18 (think: 5 is smaller, you are going to smaller units)
• To go from m/s to km/h: multiply by 18/5 (think: 18 is larger, you are going to larger units)

Worked Example:
Convert 72 km/h to m/s:

• 72 × 5/18 = 360/18 = 20 m/s

Convert 15 m/s to km/h:

• 15 × 18/5 = 270/5 = 54 km/h

Problem Type 1: Basic Speed-Distance-Time

The simplest application — find one variable given the other two.

Worked Example 1:
A car travels 360 km in 4 hours. Find its speed.

• Speed = 360/4 = 90 km/h

Worked Example 2:
A train moves at 80 km/h for 2 hours 30 minutes. Distance covered?

• Time = 2.5 hours
• Distance = 80×2.5 = 200 km

Worked Example 3:
A cyclist covers 48 km at 16 km/h. Time taken?

• Time = 48/16 = 3 hours

Worked Example 4:
A person walks at 5 km/h and runs at 15 km/h. He covers a total of 50 km in 5 hours. How much did he run?

• Let running time = t hours, walking time = (5−t) hours
• 15t + 5(5−t) = 50
• 15t + 25 − 5t = 50
• 10t = 25 → t = 2.5 hours
• Running distance = 15×2.5 = 37.5 km

Problem Type 2: Average Speed

Average speed is the total distance divided by total time — never the simple average of speeds.

Formula:
Average Speed = Total Distance / Total Time

Special Case: Same Distance at Two Different Speeds

When equal distances are covered at speeds S₁ and S₂:

Average Speed = 2S₁S₂ / (S₁ + S₂)

This is the harmonic mean formula — memorize it and never use simple average for speed.

Worked Example 1:
A car goes from A to B at 60 km/h and returns at 40 km/h. Average speed?

• = 2×60×40 / (60+40) = 4800/100 = 48 km/h

Worked Example 2:
A person covers first half of journey at 30 km/h and second half at 20 km/h. Average speed?

• = 2×30×20 / (30+20) = 1200/50 = 24 km/h

Worked Example 3:
A train covers first 100 km at 50 km/h, next 100 km at 100 km/h, last 100 km at 25 km/h. Average speed?

• Total distance = 300 km
• Time 1 = 100/50 = 2 hrs
• Time 2 = 100/100 = 1 hr
• Time 3 = 100/25 = 4 hrs
• Total time = 7 hrs
• Average speed = 300/7 = 42.86 km/h

Common trap: When three different speeds are involved, you cannot use the harmonic mean formula — you must calculate each time separately and use total distance / total time.

Problem Type 3: Relative Speed

Relative speed is used when two objects are moving simultaneously — toward each other, away from each other, or in the same direction.

Rule 1: Opposite Directions (Moving Toward Each Other)

Relative speed = S₁ + S₂ (speeds add)

Rule 2: Same Direction

Relative speed = |S₁ − S₂| (speeds subtract — take absolute value)

Meeting Point Problems

Worked Example 1 — Toward Each Other:
A and B are 240 km apart. A moves at 60 km/h and B at 80 km/h toward each other. When do they meet?

• Relative speed = 60+80 = 140 km/h
• Time = 240/140 = 12/7 hours = 1 hour 42.86 minutes

Worked Example 2 — Same Direction:
A starts at 8 AM at 40 km/h. B starts at 10 AM at 60 km/h from the same point in the same direction. When does B catch A?

• Head start of A = 2 hours × 40 = 80 km
• Relative speed = 60−40 = 20 km/h
• Time for B to catch A = 80/20 = 4 hours after B starts = 2 PM

Worked Example 3 — Meeting Point Location:
A and B start simultaneously from X and Y (300 km apart) toward each other at 70 and 80 km/h. Where do they meet from X?

• Time to meet = 300/(70+80) = 300/150 = 2 hours
• Distance from X = 70×2 = 140 km

Problem Type 4: Train Problems

Train problems are relative speed problems with one added element — the length of the train (and sometimes the platform) must be included in the distance.

The Four Train Templates

Template 1: Train Crosses a Pole or Person

• Distance = Length of train
• Time = Length of train / Speed of train

Template 2: Train Crosses a Platform or Bridge

• Distance = Length of train + Length of platform
• Time = (Train length + Platform length) / Speed

Template 3: Two Trains Opposite Directions

• Distance = Sum of both train lengths
• Relative speed = Sum of both speeds
• Time = Sum of lengths / Sum of speeds

Template 4: Two Trains Same Direction

• Distance = Sum of both train lengths
• Relative speed = Difference of speeds
• Time = Sum of lengths / Difference of speeds

Worked Examples

Example 1 — Template 1:
A 180m train moves at 72 km/h. Time to cross a signal post?

• Convert: 72 km/h = 72×5/18 = 20 m/s
• Time = 180/20 = 9 seconds

Example 2 — Template 2:
A 250m train at 90 km/h crosses a 350m platform. Time?

• Convert: 90×5/18 = 25 m/s
• Total distance = 250+350 = 600m
• Time = 600/25 = 24 seconds

Example 3 — Template 3:
Two trains of 200m and 300m approach each other at 60 km/h and 90 km/h. Time to cross?

• Relative speed = 150 km/h = 150×5/18 = 125/3 m/s
• Total distance = 500m
• Time = 500/(125/3) = 500×3/125 = 12 seconds

Example 4 — Template 4:
A 400m train at 100 km/h overtakes a 300m train at 64 km/h. Time?

• Relative speed = 36 km/h = 36×5/18 = 10 m/s
• Total distance = 700m
• Time = 700/10 = 70 seconds

Example 5 — Finding Train Length:
A train crosses a 200m bridge in 20 seconds and a 400m bridge in 30 seconds. Find train length and speed.

• Let length = L, speed = S
• L+200 = 20S ... (1)
• L+400 = 30S ... (2)
• Subtract: 200 = 10S → S = 20 m/s
• L = 20×20−200 = 400−200 = 200m
• Speed = 20 m/s = 72 km/h

Problem Type 5: Boats and Streams

Boats and streams problems follow the same relative speed principle — the stream speed adds to or subtracts from the boat's speed.

Core Formulas

Worked Examples

Example 1 — Finding Boat Speed:
A boat goes 36 km downstream in 3 hours and 24 km upstream in 4 hours. Find boat speed and stream speed.

• Downstream speed = 36/3 = 12 km/h
• Upstream speed = 24/4 = 6 km/h
• Boat speed = (12+6)/2 = 9 km/h
• Stream speed = (12−6)/2 = 3 km/h

Example 2 — Time for Round Trip:
A boat whose speed in still water is 15 km/h travels in a stream of 3 km/h. Time to go 36 km downstream and return?

• Downstream speed = 15+3 = 18 km/h → Time = 36/18 = 2 hours
• Upstream speed = 15−3 = 12 km/h → Time = 36/12 = 3 hours
• Total time = 5 hours

Example 3 — Finding Distance:
A man can row at 8 km/h in still water. Stream speed is 2 km/h. He rows upstream for 3 hours. How far did he go?

• Upstream speed = 8−2 = 6 km/h
• Distance = 6×3 = 18 km

Example 4 — Speed of Stream Unknown:
A boat covers 40 km downstream in 2 hours. It takes 4 hours to return. Find stream speed.

• Downstream speed = 40/2 = 20 km/h
• Upstream speed = 40/4 = 10 km/h
• Stream speed = (20−10)/2 = 5 km/h

Problem Type 6: Circular Track Problems

When two people run around a circular track, meeting point and meeting time problems arise.

Rules

Same direction:
Time to first meet = Track length / |S₁ − S₂|

Opposite directions:
Time to first meet = Track length / (S₁ + S₂)

All three meet (three runners):
Find when all are at the starting point simultaneously → LCM of individual lap times

Worked Examples

Example 1 — Opposite Directions:
A and B run around a 400m track at 5 m/s and 3 m/s in opposite directions. When do they first meet?

• Relative speed = 5+3 = 8 m/s
• Time = 400/8 = 50 seconds

Example 2 — Same Direction:
A and B start from the same point at 8 m/s and 6 m/s in the same direction on a 600m track. When does A lap B?

• Relative speed = 8−6 = 2 m/s
• Time = 600/2 = 300 seconds = 5 minutes

Example 3 — Three Runners:
A, B, C complete one lap in 6, 8, and 12 minutes. When do all three meet at the start?

• LCM(6, 8, 12) = 24 minutes

Problem Type 7: Time Gained/Lost Problems

These problems involve a person changing speed mid-journey and calculating the effect on arrival time.

Formula:
Extra time if slowed = (Required time at reduced speed) − (Original time)

Worked Example 1:
A man walks at 4 km/h and reaches 30 minutes late. At 5 km/h he reaches 12 minutes early. Find the distance.

• Let distance = D
• D/4 − D/5 = (30+12)/60 hours = 42/60 = 7/10
• D(5−4)/20 = 7/10
• D/20 = 7/10
• D = 14 km

Worked Example 2:
A train leaves 2 hours late. To reach on time, it increases speed by 25 km/h. Normal speed = 75 km/h. Find distance.

• New speed = 100 km/h
• Let normal time = T hours
• Distance: 75T = 100(T−2)
• 75T = 100T − 200
• 25T = 200 → T = 8 hours
• Distance = 75×8 = 600 km

Quick Formula Reference Card