Speed, Distance & Time

Speed, distance, and time problems are a staple of aptitude tests. They involve trains, boats, relative motion, and average speed calculations.

Core Formula

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

Unit Conversions

Quick conversion:

72 km/h = 72 × 5/18 = 20 m/s
15 m/s = 15 × 18/5 = 54 km/h

Average Speed

Equal Distances, Different Speeds

If a person covers distance d at speed s₁ and the same distance d at speed s₂:

Average speed = 2s₁s₂/(s₁+s₂) [Harmonic Mean]

Example: Travel 100 km at 50 km/h, return 100 km at 50 km/h:

Average = 2×50×50/(50+50) = 5000/100 = 50 km/h

Example: Travel 100 km at 40 km/h, return at 60 km/h:

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

Important: Average speed is NOT the arithmetic mean (which would be 50 here).

Equal Times, Different Speeds

If a person travels at speed s₁ for time t, then at speed s₂ for time t:

Average speed = (s₁+s₂)/2 [Arithmetic Mean]

Three Segments (Equal Distances)

Average speed = 3s₁s₂s₃/(s₁s₂+s₂s₃+s₃s₁)

Relative Speed

Same Direction (Chasing)

Relative speed = |s₁ - s₂|

Opposite Direction (Approaching)

Relative speed = s₁ + s₂

Example: Two trains at 60 km/h and 40 km/h in the same direction:

Relative speed = 60 - 40 = 20 km/h

Same trains approaching each other:

Relative speed = 60 + 40 = 100 km/h

Train Problems

Train Crossing a Pole/Person

Time = Length of train / Speed of train

Example: A 200m train at 72 km/h crosses a pole.

Speed = 72 × 5/18 = 20 m/s
Time = 200/20 = 10 seconds

Train Crossing a Platform/Bridge

Time = (Length of train + Length of platform) / Speed

Example: A 300m train at 36 km/h crosses a 150m platform.

Speed = 36 × 5/18 = 10 m/s
Time = (300+150)/10 = 45 seconds

Two Trains Crossing Each Other

Opposite direction:

Time = (L₁ + L₂) / (s₁ + s₂)

Same direction:

Time = (L₁ + L₂) / |s₁ - s₂|

Example: Two trains of 200m and 300m at 72 km/h and 54 km/h cross each other (opposite direction):

Relative speed = (72+54) × 5/18 = 126 × 5/18 = 35 m/s
Time = (200+300)/35 = 500/35 = 14.29 seconds

Train Passing Through Another Train (Same Direction)

Example: A 200m train at 72 km/h overtakes a 150m train at 36 km/h.

Relative speed = (72-36) × 5/18 = 10 m/s
Time = (200+150)/10 = 35 seconds

Boats and Streams

Key Terms

Formulas

Speed in still water = (Downstream + Upstream) / 2
Speed of stream = (Downstream - Upstream) / 2

Example: A boat goes 24 km downstream in 3 hours and 24 km upstream in 6 hours.

Downstream speed = 24/3 = 8 km/h
Upstream speed = 24/6 = 4 km/h
Still water speed = (8+4)/2 = 6 km/h
Stream speed = (8-4)/2 = 2 km/h

Time Ratio

Time upstream / Time downstream = (u+v)/(u-v)

For the same distance:

T_up/T_down = (u+v)/(u-v)

Example: If downstream speed = 10, upstream = 6, same distance:

T_up/T_down = 10/6 = 5/3

Finding Distance (Boat Problem)

Problem: A boat takes 4 hours to go 20 km downstream and return. Speed in still water is 8 km/h. Find stream speed.

Solution:

20/(8+v) + 20/(8-v) = 4
20(8-v) + 20(8+v) = 4(64-v²)
160 + 160 = 256 - 4v²
320 = 256 - 4v²
4v² = -64 → No real solution (check: this means the problem setup is inconsistent)

Let me use a better example:

Problem: A boat goes 16 km upstream in 4 hours and 16 km downstream in 2 hours. Find the speed of the stream.

Solution:

Upstream speed = 16/4 = 4 km/h
Downstream speed = 16/2 = 8 km/h
Stream speed = (8-4)/2 = 2 km/h
Still water speed = (8+4)/2 = 6 km/h

Circular Motion

Meeting on a Circular Track

Same direction:

Time to meet = Track length / |s₁ - s₂|

Opposite direction:

Time to meet = Track length / (s₁ + s₂)

Number of Meetings

Same direction: In time T, they meet T × |s₁-s₂|/L times (where L = track length)

Opposite direction: In time T, they meet T × (s₁+s₂)/L times

Example: Track = 400m, A at 5 m/s, B at 3 m/s (same direction):

Time to first meet = 400/(5-3) = 200 seconds
In 1 hour (3600s): 3600/200 = 18 meetings

Escalator Problems

Concept

Escalator problems are similar to boats and streams:

• Walking speed = person’s own speed
• Escalator speed = stream speed
• Same direction (walking up on up-escalator) = speeds add
• Opposite direction = speeds subtract

Problem: A person takes 20 steps on a stationary escalator and 10 steps on a moving escalator (both times reaching the top). How many visible steps on the escalator?

Solution:

Let escalator speed = e steps/second, person speed = p steps/second
Stationary: 20 steps, time = 20/p
Moving: 10 steps by person, escalator adds (e/p)×10 steps
Total steps = 10 + 10e/p
From stationary case: total = 20
20 = 10 + 10e/p → e/p = 1 → escalator speed = person speed
Total visible steps = 20

Speed & Time Ratio Problems

If Speed Ratio is a:b

For the same distance:

Time ratio = b:a (inverse)

Example: If A’s speed is 3/2 of B’s speed, and they travel the same distance:

Time_A : Time_B = 2:3

If Time Ratio is a:b

For the same distance:

Speed ratio = b:a (inverse)

Tricks & Shortcuts

Trick 1: Quick km/h to m/s

Multiply by 5/18:

36 km/h = 36 × 5/18 = 10 m/s
54 km/h = 54 × 5/18 = 15 m/s
72 km/h = 72 × 5/18 = 20 m/s
90 km/h = 90 × 5/18 = 25 m/s
108 km/h = 108 × 5/18 = 30 m/s

Trick 2: Average Speed for Three Equal Distances

If three equal distances at s₁, s₂, s₃:

Avg = 3/(1/s₁ + 1/s₂ + 1/s₃)

Trick 3: If A Reaches x Minutes Early and y Minutes Late

If at speed s₁, A is x minutes late, and at speed s₂, A is y minutes early:

Distance = s₁ × s₂ × (x+y) / (s₂-s₁)  [times in same units]

Trick 4: Train Problem Shortcut

If a train of length L crosses a pole in t₁ seconds and a platform of length P in t₂ seconds:

L/t₁ = (L+P)/t₂
P = L(t₂-t₁)/t₁

Trick 5: Boat Round Trip

Time for round trip of distance d:

T = d/(u+v) + d/(u-v) = 2du/(u²-v²)

Practice Questions

Q1: Average Speed

A car travels from A to B at 40 km/h and returns at 60 km/h. Find average speed.

Solution:

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

Q2: Train Crossing

A 150m train at 54 km/h crosses a 250m platform. Find time.

Solution:

Speed = 54 × 5/18 = 15 m/s
Time = (150+250)/15 = 400/15 = 26.67 seconds

Q3: Two Trains

Two trains of 120m and 80m travel at 60 km/h and 40 km/h in opposite directions. Find crossing time.

Solution:

Relative speed = (60+40) × 5/18 = 100 × 5/18 = 27.78 m/s
Time = (120+80)/27.78 = 200/27.78 = 7.2 seconds

Q4: Boat Problem

A boat goes 30 km downstream in 3 hours and returns in 5 hours. Find speed in still water.

Solution:

Downstream = 30/3 = 10 km/h
Upstream = 30/5 = 6 km/h
Still water = (10+6)/2 = 8 km/h
Stream = (10-6)/2 = 2 km/h

Q5: Relative Speed

A 100m train at 36 km/h overtakes a man walking at 4 km/h in the same direction. Find time.

Solution:

Relative speed = (36-4) × 5/18 = 32 × 5/18 = 8.89 m/s
Time = 100/8.89 = 11.25 seconds

Q6: Meeting Problem

A and B start from two places 100 km apart. A at 20 km/h, B at 30 km/h (towards each other). When do they meet?

Solution:

Relative speed = 20 + 30 = 50 km/h
Time = 100/50 = 2 hours

Q7: Mixed Journey

A person travels 200 km: first 100 km at 50 km/h, next 50 km at 25 km/h, last 50 km at 10 km/h. Find average speed.

Solution:

Time = 100/50 + 50/25 + 50/10 = 2 + 2 + 5 = 9 hours
Average speed = 200/9 = 22.22 km/h

Q8: Escalator

Walking up a moving escalator, a person takes 30 steps. Walking at twice the speed, takes 20 steps. How many visible steps?

Solution:

Let escalator moves e steps while person takes 30 steps (at speed p)
Total steps N = 30 + 30e/p
At 2p speed: N = 20 + 20e/(2p) = 20 + 10e/p
30 + 30e/p = 20 + 10e/p
10 = -20e/p → e/p = -0.5 → This means escalator helps
N = 30 + 30(-0.5) = 30 - 15 = 15
Or: N = 20 + 10(-0.5) = 20 - 5 = 15 ✓
Visible steps = 15... but wait, this seems small.

Now consider the opposite case: escalator moves down while person walks up:

N = 30 - 30e/p (net progress per step is 1-e/p)
At 2p: N = 20 - 20e/(2p) = 20 - 10e/p
30 - 30e/p = 20 - 10e/p → 10 = 20e/p → e/p = 0.5
N = 30 - 15 = 15 ✓
Total visible steps = 15

The above was for an escalator going down (opposing the person). Now redo for an escalator going up (assisting):

If the person walks up and the escalator also goes up:

At speed p: person takes 30 steps, escalator moves e×(30/p) steps
Total = 30 + 30e/p
At speed 2p: person takes 20 steps, escalator moves e×(20/2p) = 10e/p steps
Total = 20 + 10e/p
30 + 30e/p = 20 + 10e/p → 10 = -20e/p → e/p = -0.5 (negative = going down)

The escalator is going down. Total visible = 15 steps. This is actually a valid problem — the escalator is going down and the person is walking up.

Let me use a cleaner example:

Revised Q8: A person walking up a stationary escalator counts 50 steps. On a moving escalator going down, they count 30 steps. How many steps are visible?

Solution:

Let escalator adds/removes e steps while person takes 30
50 = 30 + 30e/p → 20 = 30e/p → e/p = 2/3
Or: 50 = 30 + 30(2/3) = 30 + 20 = 50 ✓
Visible steps = 50

Summary Table