Word Problems

Speed, Distance & Time

Solve word problems involving speed, distance and time with unit conversions

Explanation & Worked Examples

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The Formula Triangle

All speed, distance and time problems use one of three related formulas:

Speed

S = D ÷ T

Distance

D = S × T

Time

T = D ÷ S

The SDT Triangle

Cover the value you want to find — the remaining two show you what to do:

Cover D → S × T
Cover S → D ÷ T
Cover T → D ÷ S

Key rule: Speed, distance and time must all use consistent units. If speed is in km/h, distance must be in km and time in hours.

Unit Conversions

Many exam questions mix units on purpose. Always convert before applying the formula.

Time: Hours ↔ Minutes

To convert	Operation	Example
Hours → Minutes	× 60	1.5 h = 90 min
Minutes → Hours	÷ 60	45 min = 0.75 h

Speed: km/h ↔ m/s

To convert	Operation	Example
km/h → m/s	÷ 3.6	72 km/h = 20 m/s
m/s → km/h	× 3.6	25 m/s = 90 km/h

Why 3.6? Because 1 km/h = 1000 m ÷ 3600 s = 1/3.6 m/s.

Quick check: m/s values are always much smaller than km/h values (roughly 3.6× smaller). If your answer looks unexpectedly large or small, you may have converted the wrong way.

Easy Examples

These problems use the formula directly — no unit conversion needed.

Example 1 — Find Distance

A car travels at a constant speed of 60 km/h for 3 hours. How far does it travel?

1Identify what you know: S = 60 km/h, T = 3 h, find D

2Use D = S × T

3D = 60 × 3 = 180 km

Example 2 — Find Speed

A cyclist covers 40 km in 2 hours. What is their average speed?

1Identify: D = 40 km, T = 2 h, find S

2Use S = D ÷ T

3S = 40 ÷ 2 = 20 km/h

Example 3 — Find Time

A runner jogs at 8 km/h. How long does it take to run 12 km?

1Identify: S = 8 km/h, D = 12 km, find T

2Use T = D ÷ S

3T = 12 ÷ 8 = 1.5 hours = 1 hour 30 minutes

Medium Examples

These problems require a unit conversion step before or after applying the formula.

Example 1 — Answer in Minutes

A train travels at a constant speed of 90 km/h. How many minutes will it take to cover 60 km?

1Identify: S = 90 km/h, D = 60 km, find T in minutes

2T = D ÷ S = 60 ÷ 90 = ²/₃ hours

3Convert to minutes: ²/₃ × 60 = 40 minutes

Example 2 — Time Given in Minutes

A bus covers 30 km in 45 minutes. What is its average speed in km/h?

1Convert time to hours: 45 min ÷ 60 = 0.75 h

2S = D ÷ T = 30 ÷ 0.75

3S = 40 km/h

Example 3 — Speed in m/s

A sprinter runs 100 metres in 10 seconds. What is their speed in km/h?

1S = D ÷ T = 100 ÷ 10 = 10 m/s

2Convert to km/h: 10 × 3.6 = 36 km/h

Strategy: Always write down what you know and what units the answer needs to be in before calculating. That prevents unit errors.

Complex Examples

These problems combine multiple steps, involve two objects, or require the average speed formula.

Example 1 — Two Objects Moving Apart

Two cars start from the same point and travel in opposite directions. Car A travels at 60 km/h and Car B at 80 km/h. How far apart are they after 2.5 hours?

1Moving apart → add speeds. Combined speed = 60 + 80 = 140 km/h

2D = S × T = 140 × 2.5 = 350 km

Example 2 — Two Objects Moving Towards Each Other

Two trains are 450 km apart and travel towards each other. Train A travels at 90 km/h and Train B at 60 km/h. When do they meet?

1Moving towards each other → add speeds. Combined speed = 90 + 60 = 150 km/h

2T = D ÷ S = 450 ÷ 150 = 3 hours

Example 3 — Average Speed (Round Trip)

A person walks from A to B at 4 km/h and returns at 6 km/h. What is their average speed for the whole journey?

1Average speed ≠ average of the two speeds. Use: Avg S = 2 × S₁ × S₂ ÷ (S₁ + S₂)

2Avg S = 2 × 4 × 6 ÷ (4 + 6) = 48 ÷ 10 = 4.8 km/h

Why? Let distance A→B = d km. Time A→B = d/4 h, time B→A = d/6 h. Total distance = 2d. Total time = d/4 + d/6 = 5d/12. Avg S = 2d ÷ (5d/12) = 24/5 = 4.8 km/h.

Example 4 — Train Passing a Point

A train 400 m long passes a stationary signal post in 20 seconds. Find the train's speed in km/h.

1Distance covered = length of train = 400 m

2S = D ÷ T = 400 ÷ 20 = 20 m/s

3Convert: 20 × 3.6 = 72 km/h

Common trap: For average speed over a round trip, you cannot just average the two speeds — you must use the harmonic mean formula (or work from total distance ÷ total time).

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