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WBB • Class 7 • Mathematics (গণিত প্রভা) • Ch 15

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Time and Distance

Welcome to Chapter 15: "Time and Distance" (সময় ও দূরত্ব / समय और दूरी) of the West Bengal Board (WBBSE) Class 7 Mathematics curriculum (Ganit Prabha). Formatted under the TargetExams Gold-Standard 5-Step Pedagogy System, this practical arithmetic chapter covers the fundamental distance-speed-time relationship ($D = S \times T$), instant unit conversions between $\text{km/h}$ and $\text{m/s}$ via factors of $5/18$ and $18/5$, direct and inverse proportionality (Rule of Three), problems on trains passing stationary objects versus platforms, relative speed of objects moving in opposite and identical directions, and average speed (harmonic mean).

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Overview Core Concepts Formulas & Identities Solved Examples Common Traps Summary Self Check Take MCQ Test

🚄 Speed Trials of the Rajdhani Express: When Does a Train Cross Its Own Length?

When the Howrah-New Delhi Rajdhani Express flashes past a telegraph post along the tracks, it clears it in a few quick seconds. Yet it takes significantly longer to clear a station platform. Why?

Because to clear a point-sized post or a stationary observer, the train needs to cover only its own length! But to clear a bridge or platform, it must travel a distance equal to its own length plus the length of the platform!

In this chapter, you will master the principles of train dynamics, the lightning unit conversion shortcut ($\times 5/18$), and avoid the classic average speed trap!

Why This Chapter Matters

Before You Begin (Prerequisites)

Basic units of time and distance (hours, minutes, seconds; kilometers, meters)
Unitary method and Rule of Three proportions
Multiplication and division of fractions
Understanding direct and inverse relationships

What You Will Learn (Core Objectives)

Apply the core formula $\text{Distance} = \text{Speed} \times \text{Time}$ to real-world scenarios
Convert fluently between $\text{km/h}$ and $\text{m/s}$ using $\times 5/18$ and $\times 18/5$
Solve train problems involving poles, bridges, and stationary observers
Determine relative speed and crossing times for two moving trains
Calculate average speed over round trips using the harmonic mean formula

Chapter Roadmap & Progression

1 Concept 1: Master Formulas & Conver...

2 Concept 2: Direct & Inverse Proport...

3 Concept 3: Train Problems—Poles, Br...

4 Concept 4: Relative Speed & Two Mov...

5 Concept 5: Average Speed & the Harm...

Complete Concept Guide (100% Curriculum Coverage)

Concept 1: Master Formulas & Conversion Factors ($\times \frac{5}{18}$ and $\times \frac{18}{5}$)

Step 1

Definitions & Equations

Speed is distance traversed per unit time: $\text{Distance} = \text{Speed} \times \text{Time}$.

Step 2

Derivation of 5/18 and 18/5

$1\text{ km/h} = \frac{1000\text{ m}}{3600\text{ s}} = \frac{5}{18}\text{ m/s}$. To convert $\text{m/s}$ to $\text{km/h}$, multiply by $\frac{18}{5}$.

Step 3

Worked Example

$90\text{ km/h} = 90 \times \frac{5}{18} = 25\text{ m/s}$. Conversely, $25\text{ m/s} = 25 \times \frac{18}{5} = 90\text{ km/h}$.

Step 4

Examiner Traps

Going from large ($\text{km/h}$) to small ($\text{m/s}$), put the smaller number on top ($5/18$).

Step 5

Real-World Application

Traffic Radar Guns: Police speed guns measure distance over microseconds and instantly output vehicle speed in $\text{km/h}$.

Concept 2: Direct & Inverse Proportionality (Rule of Three)

Step 1

Proportionality Relationships

For a constant distance, speed and time vary inversely: higher speed requires less time.

Step 2

Setting up Proportions

In inverse proportions, invert the ratio: $\text{Time}_2 = \text{Time}_1 \times \frac{\text{Speed}_1}{\text{Speed}_2}$.

Step 3

Worked Example

At $40\text{ km/h}$, a journey takes $3\text{ hours}$. At $60\text{ km/h}$: $3 \times \frac{40}{60} = 2\text{ hours}$.

Step 4

Examiner Traps

Never apply direct proportion to speed and time when distance is fixed.

Step 5

Real-World Application

Flight Dispatch: Headwinds reducing ground speed automatically trigger delay calculations using inverse variation.

Concept 3: Train Problems—Poles, Bridges & Platforms

Step 1

Distance Covered Geometry

Crossing point-object (pole, tree): Distance $= L$ (train length).

Crossing extended object (platform, bridge): Distance $= L + P$ (train length + platform length).

Step 2

Crossing Time Formula

$\text{Time} = \frac{L + P}{\text{Speed}}$.

Step 3

Worked Example

A $200\text{ m}$ train at $36\text{ km/h}$ ($10\text{ m/s}$) clears a post in $\frac{200}{10} = 20\text{ seconds}$.

Step 4

Examiner Traps

Never omit the train's own length when passing a bridge or platform.

Step 5

Real-World Application

Railway Safety Automation: Automatic level crossing gates compute clearance times based on train length and speed.

Concept 4: Relative Speed & Two Moving Trains

Step 1

Directional Rules

Opposite directions: $S_{\text{rel}} = S_1 + S_2$ (speeds add).

Same direction: $S_{\text{rel}} = S_1 - S_2$ (speeds subtract).

Step 2

Total Crossing Distance

In all cases, total distance to completely clear each other is $L_1 + L_2$.

Step 3

Worked Example

Trains of $120\text{ m}$ and $80\text{ m}$ at $42\text{ km/h}$ and $30\text{ km/h}$ in opposite directions: Distance $= 200\text{ m}$, $S_{\text{rel}} = 72\text{ km/h} = 20\text{ m/s}$. Time taken $= 10\text{ s}$.

Step 4

Examiner Traps

Do not subtract train lengths when moving in the same direction. Lengths are always added ($L_1 + L_2$).

Step 5

Real-World Application

Overtaking Maneuvers: Determining safe overtaking sight distance on two-lane highways relies entirely on relative speed.

Concept 5: Average Speed & the Harmonic Mean Formula ($2xy / (x+y)$)

Step 1

Universal Definition

Average Speed = Total Distance / Total Time.

Step 2

Equal Distance Formula

For identical forward and return distances: $\mathbf{\frac{2xy}{x+y}}$.

Step 3

Worked Example

Going at $20\text{ km/h}$ and returning at $30\text{ km/h}$: $\frac{2 \times 20 \times 30}{20 + 30} = 24\text{ km/h}$.

Step 4

Examiner Traps

Never write $\frac{20+30}{2} = 25\text{ km/h}$. More time is spent traveling at the lower speed.

Step 5

Real-World Application

Logistics Fleet Optimization: Delivery services compute true fuel economy and realistic ETAs using harmonic mean averages.

Key Formulas, Identities & Theorems

Speed-Distance-Time Relations

$$d = s \times t \iff s = \frac{d}{t}, \quad t = \frac{d}{s}$$

Where $d$ = distance, $s$ = speed, and $t$ = time taken.

Unit Conversion: km/h to m/s

$$1 \text{ km/h} = \frac{5}{18} \text{ m/s}$$

Multiply km/h by $\frac{5}{18}$ to get m/s ($\frac{1000\text{ m}}{3600\text{ s}}$).

Unit Conversion: m/s to km/h

$$1 \text{ m/s} = \frac{18}{5} \text{ km/h}$$

Multiply m/s by $\frac{18}{5}$ (or $3.6$) to get km/h.

Train Crossing a Platform / Bridge

$$t = \frac{L_{\text{train}} + L_{\text{platform}}}{s}$$

Total distance covered = train length ($L_{\text{train}}$) + platform length ($L_{\text{platform}}$).

Relative Speed

$$s_{\text{opp}} = s_1 + s_2, \quad s_{\text{same}} = s_1 - s_2$$

Speeds add when moving in opposite directions, and subtract in same direction ($s_1 > s_2$).

Average Speed (Equal Distances)

$$s_{\text{avg}} = \frac{2 s_1 s_2}{s_1 + s_2} = \frac{\text{Total Distance}}{\text{Total Time}}$$

Harmonic mean formula $\frac{2s_1s_2}{s_1+s_2}$ applies when equal distances are covered at speeds $s_1, s_2$.

Conceptual Solved Examples & Case Studies

Example 1

A train moves at $54\text{ km/h}$. Express its speed in $\text{m/s}$ and find the distance covered in 10 seconds.

Step-by-Step Solution:

Conversion: $54 \times \frac{5}{18} = 15\text{ m/s}$.
Distance in 10 s $= \text{Speed} \times \text{Time} = 15 \times 10 = 150\text{ m}$.

Example 2

A train $150\text{ m}$ long traveling at $72\text{ km/h}$ crosses a bridge of length $250\text{ m}$. Find the time taken.

Step-by-Step Solution:

Speed $= 72 \times \frac{5}{18} = 20\text{ m/s}$.
Total distance $= 150 + 250 = 400\text{ m}$.
Time taken $= \frac{400}{20} = 20\text{ seconds}$.

Example 3

A student cycles to school at $12\text{ km/h}$ and returns along the same route at $8\text{ km/h}$. Find the average speed.

Step-by-Step Solution:

Equal distances formula: $\text{Average Speed} = \frac{2 \times 12 \times 8}{12 + 8} = \frac{192}{20} = 9.6\text{ km/h}$.

Common Misconceptions & Examiner Traps

Common Misconception

Calculating average speed as the arithmetic mean $\frac{S_1 + S_2}{2}$.

Scientific Reality & Correction

This is mathematically incorrect! More time is spent traveling at the lower speed, pulling the average down. Use $\frac{\text{Total Distance}}{\text{Total Time}}$ or $\frac{2xy}{x+y}$.

Common Misconception

Mixing mismatched units (e.g. dividing meters by km/h).

Scientific Reality & Correction

Always convert speed to m/s when distances are given in meters, or convert distance to km.

Common Misconception

Forgetting the train's length when crossing a platform or bridge.

Scientific Reality & Correction

The total distance is always $\text{Length of Train} + \text{Length of Platform}$.

Train Crossing Model & Speed Unit Conversion Reference

Chapter Summary & 10 Key Takeaways

Takeaway 1

Fundamental formula: $\text{Distance} = \text{Speed} \times \text{Time}$.

Takeaway 2

Multiply by 5/18 to convert km/h to m/s.

Takeaway 3

Multiply by 18/5 to convert m/s to km/h.

Takeaway 4

For fixed distance, speed and time are inversely proportional.

Takeaway 5

Distance when crossing a pole = train's own length ($L$).

Takeaway 6

Distance when crossing a platform/bridge = train + platform ($L + P$).

Takeaway 7

Speeds add in opposite directions and subtract in identical directions.

Takeaway 8

Average speed for equal distances is $\frac{2xy}{x+y}$ (harmonic mean).

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

A car travels at $36\text{ km/h}$. How many meters does it cover in 1 second?

Reveal Answer & Explanation

Answer: $36 \times \frac{5}{18} = 10\text{ meters}$.
Multiply by 5/18.

A train $180\text{ m}$ long moving at $54\text{ km/h}$ clears a tree in how many seconds?

Reveal Answer & Explanation

Answer: $\frac{180}{15} = 12\text{ seconds}$.
$54\text{ km/h} = 15\text{ m/s}$.

A $100\text{ m}$ train at $60\text{ km/h}$ clears a $140\text{ m}$ platform in how many seconds?

Reveal Answer & Explanation

Answer: $\frac{240 \times 3}{50} = 14.4\text{ seconds}$.
Distance $= 240\text{ m}$, Speed $= 50/3\text{ m/s}$.

A vehicle travels forward at $60\text{ km/h}$ and returns at $40\text{ km/h}$. Find average speed.

Reveal Answer & Explanation

Answer: $48\text{ km/h}$.
$\frac{2 \times 60 \times 40}{60 + 40}$.

Two trains travel in opposite directions at $50\text{ km/h}$ and $40\text{ km/h}$. Find relative speed.

Reveal Answer & Explanation

Answer: $50 + 40 = 90\text{ km/h}$ ($25\text{ m/s}$).
Speeds add in opposite directions.

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All Class 7 Mathematics (গণিত প্রভা) Chapters

Ch 1: Revision of Previous Lessons Ch 2: Ratio Ch 3: Proportion Ch 4: Addition, Subtraction, Multiplication and Division of Integers Ch 5: Concept of Indices Ch 6: Algebraic Operations Ch 7: Construction of Specific Angles with Compass Ch 8: Construction of Triangles Ch 9: Concept of Congruence Ch 10: Approximation Ch 11: Square Root of Fractions Ch 12: Algebraic Formulae Ch 13: Parallel Lines and Transversals Ch 14: Properties of Triangles Ch 15: Time and Distance Ch 16: Double Bar Graph Ch 17: Area of Rectangle and Square Ch 18: Symmetry Ch 19: Factorization Ch 20: Classification of Quadrilaterals Ch 21: Construction of Quadrilaterals Ch 22: Formation and Solution of Equations

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