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ICSE • Class 7 • Mathematics • Ch 8

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Speed, Distance and Time

In ICSE Class 7 Mathematics, "Speed, Distance and Time" provides an authoritative, kinematical master study guide analyzing the mathematical relations governing uniform motion. This comprehensive chapter explores Fundamental Relations (Speed as distance traversed per unit time: $S = \frac{D}{T}$, $D = S \times T$, $T = \frac{D}{S}$), Unit Conversions (The critical scale factors: converting $\text{km/h} \to \text{m/s}$ by multiplying by $\frac{5}{18}$, and converting $\text{m/s} \to \text{km/h}$ by multiplying by $\frac{18}{5}$), Average Speed (Calculation over multi-leg journeys: $\text{Average Speed} = \frac{\text{Total Distance Traveled}}{\text{Total Time Taken}}$; The harmonic mean formula $\frac{2xy}{x+y}$ for equal outward and return distances), Motion of Trains (A train passing a stationary point object like a pole/person: distance traveled equals the length of the train; A train crossing a long stationary object like a platform/bridge/tunnel: distance traveled equals the length of the train plus the length of the platform), Relative Speed (Bodies moving in opposite directions: relative speed $= S_1 + S_2$; Bodies moving in the same direction: relative speed $= |S_1 - S_2|$), and Real-World Applied Problems (Meeting times, overtake problems, race tracking) aligned with the 2026–27 CISCE curriculum.

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How Did an 18-Wheeler Truck Moving at 72 km/h and a Sprinter Running at 20 m/s Discover They Were Moving at the EXACT Same Speed?

Imagine standing along a highway. A massive 18-wheeler truck roars past with its speedometer reading $72\text{ km/h}$. On a parallel track, an Olympic athlete sprints past at $20\text{ meters per second}$. To an untrained observer, 72 seems vastly greater than 20! But convert the units: $72\text{ km/h} = 72 \times \frac{5}{18} = \mathbf{20\text{ m/s}}$! They are moving side-by-side at the exact same physical velocity! Why is the conversion factor between kilometers per hour and meters per second precisely $\frac{5}{18}$? If you drive to your friend's house at $60\text{ km/h}$ and drive back along the same road at $40\text{ km/h}$, why is your average speed NOT $50\text{ km/h}$, but $48\text{ km/h}$? What happens to the distance when a $200\text{-meter}$ passenger train crosses a $300\text{-meter}$ railway bridge? Let's master the kinematics of speed, distance, and time.

Why This Chapter Matters

Speed, distance, and time problems are among the most tested topics in ICSE school examinations, NTSE, banking aptitude, and engineering physics. They provide the practical mathematical tools to calculate journey times, fuel efficiency, relative velocities, and GPS navigation metrics.

Before You Begin (Prerequisites)

Multiplication and division of fractions and decimals.
Ratio and proportion from Chapter 6.
Solving simple linear equations in one variable.

What You Will Learn (Core Objectives)

Apply the fundamental triad formulas: $S = D/T$, $D = S \times T$, and $T = D/S$.
Convert seamlessly between $\text{km/h}$ and $\text{m/s}$ using $\frac{5}{18}$ and $\frac{18}{5}$.
Calculate Average Speed over non-uniform and multi-stage journeys.
Solve train crossing problems for stationary point objects (poles) and extended objects (platforms).
Calculate relative speeds for objects moving in opposite and identical directions.
Solve applied word problems involving races, overtaking, and meeting times.

Chapter Roadmap & Progression

1 1. Fundamental Formulas & The 5/18...

2 2. Average Speed (Avoiding the Arit...

3 3. Train Motion: Crossing Poles vs...

4 4. Relative Speed: Opposite vs Same...

Complete Concept Guide (100% Curriculum Coverage)

1. Fundamental Formulas & The 5/18 Conversion Rule

2. Average Speed (Avoiding the Arithmetic Trap)

3. Train Motion: Crossing Poles vs Crossing Platforms

4. Relative Speed: Opposite vs Same Direction

Key Formulas, Identities & Theorems

Speed-Distance-Time Triad

$$S = \frac{D}{T}, \quad D = S \cdot T, \quad T = \frac{D}{S}$$

All units must be mutually consistent.

Harmonic Mean for Average Speed

$$S_{\text{avg}} = \frac{2xy}{x + y}$$

Applies when equal distances are traversed at speeds x and y.

Kinematics: Conversion Factors, Trains & Relative Speed

₁ + S₂ • Same Direction (→ →): Relative Speed = S₁ - S₂ • Average Speed Formula: Total Distance / Total Time Equal Distances: 2xy / (x+y) • Never average speeds directly! 1 km/h = 5/18 m/s • TRAIN PASSING PLATFORM: D = L + P • RELATIVE SPEED: ADD OR SUBTRACT

Chapter Summary & 10 Key Takeaways

Takeaway 1

Fundamental equations: Speed = Distance/Time, Distance = Speed * Time, Time = Distance/Speed.

Takeaway 2

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

Takeaway 3

Average speed is defined as Total Distance divided by Total Time.

Takeaway 4

For equal outward and return distances at speeds x and y, Average Speed = (2xy) / (x + y).

Takeaway 5

When a train passes a pole or person, distance covered equals the length of the train.

Takeaway 6

When a train crosses a platform, bridge, or tunnel, distance covered equals train length + platform length.

Takeaway 7

When two objects move towards each other (opposite direction), relative speed is S1 + S2.

Takeaway 8

When two objects move in the same direction, relative speed is |S1 - S2|.

Takeaway 9

Always ensure distance and speed units match (meters with m/s, kilometers with km/h).

Takeaway 10

Time taken for two trains to cross each other is (L1 + L2) divided by their relative speed.

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 train $280\text{ meters}$ long is running at a speed of $63\text{ km/h}$. In what time will it pass a telegraph post?

Reveal Answer & Explanation

Answer: Step 1: Convert speed from $\text{km/h}$ to $\text{m/s}$ by multiplying by $\frac{5}{18}$:
$$\text{Speed} = 63 \times \frac{5}{18} = \frac{7 \times 5}{2} = \frac{35}{2}\text{ m/s} = 17.5\text{ m/s}$$
Step 2: Distance covered in passing a telegraph post equals the length of the train ($D = 280\text{ m}$):
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{280}{\frac{35}{2}} = \frac{280 \times 2}{35}$$
Cancel by 7: $\frac{40 \times 2}{5} = 8 \times 2 = \mathbf{16\text{ seconds}}$.
Convert $63\text{ km/h}$ to $\text{m/s}$ by multiplying by $5/18$ ($35/2\text{ m/s}$), then divide $280$ by $35/2$.

A train $350\text{ meters}$ long takes $30\text{ seconds}$ to cross a railway platform $250\text{ meters}$ long. Find the speed of the train in $\text{km/h}$.

Reveal Answer & Explanation

Answer: Step 1: Total distance covered = Length of Train + Length of Platform:
$$\text{Total Distance } D = 350 + 250 = 600\text{ meters}$$
Step 2: Time taken $T = 30\text{ seconds}$.
$$\text{Speed in m/s} = \frac{D}{T} = \frac{600}{30} = 20\text{ m/s}$$
Step 3: Convert speed to $\text{km/h}$ by multiplying by $\frac{18}{5}$:
$$\text{Speed in km/h} = 20 \times \frac{18}{5} = 4 \times 18 = \mathbf{72\text{ km/h}}$$.
Total distance $= 350 + 250 = 600\text{ m}$. Speed $= 600 / 30 = 20\text{ m/s}$. Multiply by $18/5$ to get $\text{km/h}$.

A motorist travels from Town A to Town B at a speed of $60\text{ km/h}$, and returns along the same route at $40\text{ km/h}$. Find his average speed for the entire round trip.

Reveal Answer & Explanation

Answer: Let the one-way distance between Town A and Town B be $D\text{ km}$.
Time taken for onward journey: $T_1 = \frac{D}{60}\text{ hours}$.
Time taken for return journey: $T_2 = \frac{D}{40}\text{ hours}$.
Total distance $= D + D = 2D\text{ km}$.
Total time: $T_1 + T_2 = \frac{D}{60} + \frac{D}{40} = \frac{2D + 3D}{120} = \frac{5D}{120} = \frac{D}{24}\text{ hours}$.
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2D}{\frac{D}{24}} = 2 \times 24 = \mathbf{48\text{ km/h}}$$.
*(Harmonic Formula: $\frac{2xy}{x+y} = \frac{2 \times 60 \times 40}{60 + 40} = \frac{4,800}{100} = 48\text{ km/h}$)*.
Do not average 60 and 40! Use $\frac{2xy}{x+y} = \frac{2 \times 60 \times 40}{100} = 48\text{ km/h}$.

Two trains of lengths $160\text{ m}$ and $140\text{ m}$ are running in opposite directions on parallel tracks at speeds of $42\text{ km/h}$ and $30\text{ km/h}$ respectively. In what time will they cross each other?

Reveal Answer & Explanation

Answer:

Step 1: Total distance to be covered to clear each other:

$$\text{Total Distance } D = 160 + 140 = 300\text{ meters}$$

Step 2: Since they move in opposite directions, add speeds to find Relative Speed:

$$\text{Relative Speed} = 42 + 30 = 72\text{ km/h}$$

Convert to $\text{m/s}$:

$$72 \times \frac{5}{18} = 4 \times 5 = 20\text{ m/s}$$

Step 3: Time taken to cross each other:

$$\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{300}{20} = \mathbf{15\text{ seconds}}$$

Add lengths ($300\text{ m}$) and add speeds ($72\text{ km/h} = 20\text{ m/s}$). Time $= 300 / 20 = 15\text{ s}$.

A thief is spotted by a policeman from a distance of $200\text{ meters}$. When the policeman starts chasing, the thief also runs. If the speed of the thief is $10\text{ km/h}$ and that of the policeman is $12\text{ km/h}$, how far will the thief have run before he is caught?

Reveal Answer & Explanation

Answer:

Initial gap $= 200\text{ m}$.
Both run in the same direction, so subtract speeds to find Relative Speed:

$$\text{Relative Speed} = 12 - 10 = 2\text{ km/h} = 2 \times \frac{5}{18} = \frac{5}{9}\text{ m/s}$$

Time taken by policeman to catch the thief:

$$\text{Time } T = \frac{\text{Initial Distance}}{\text{Relative Speed}} = \frac{200}{\frac{5}{9}} = \frac{200 \times 9}{5} = 40 \times 9 = 360\text{ seconds}$$

Distance run by the thief in $360\text{ seconds}$:

$$\text{Speed of thief} = 10\text{ km/h} = 10 \times \frac{5}{18} = \frac{25}{9}\text{ m/s}$$

$$\text{Distance} = \text{Speed} \times \text{Time} = \frac{25}{9} \times 360 = 25 \times 40 = \mathbf{1,000\text{ meters = 1 km}}$$

Relative speed is $2\text{ km/h} = 5/9\text{ m/s}$. Time taken is $200 / (5/9) = 360\text{ s}$. Multiply by thief's speed.

Walking at $\frac{3}{4}$ of his usual speed, a man reaches his office $20\text{ minutes}$ late. Find his usual time to reach the office.

Reveal Answer & Explanation

Answer:

Let usual speed be $S$ and usual time be $T$. Distance $D = S \times T$.
New speed $= \frac{3}{4}S$.
Since distance is constant, new time is reciprocal of speed ratio: $T_{\text{new}} = \frac{4}{3}T$.
Given that new time is $20\text{ minutes}$ longer:

$$\frac{4}{3}T - T = 20$$

$$\frac{1}{3}T = 20 \implies T = 20 \times 3 = \mathbf{60\text{ minutes = 1 hour}}$$

.
His usual time to reach the office is $60\text{ minutes}$.

If speed becomes $3/4$, time becomes $4/3$. The delay is $4/3 T - T = 1/3 T = 20\text{ min} \implies T = 60\text{ min}$.

A car covers a distance of $450\text{ km}$ at a uniform speed. If the speed had been $15\text{ km/h}$ more, it would have taken $1\text{ hour}$ less for the journey. Find the original speed of the car.

Reveal Answer & Explanation

Answer: Let the original speed be $x\text{ km/h}$. Original time $= \frac{450}{x}\text{ hours}$.
New speed $= (x + 15)\text{ km/h}$. New time $= \frac{450}{x + 15}\text{ hours}$.
$$\frac{450}{x} - \frac{450}{x + 15} = 1$$
$$450 \left[ \frac{(x + 15) - x}{x(x + 15)} \right] = 1$$
$$450 \times 15 = x^2 + 15x$$
$$x^2 + 15x - 6750 = 0$$
Factor the quadratic: $(x + 90)(x - 75) = 0$.
Since speed must be positive: $x = \mathbf{75\text{ km/h}}$.
Set $\frac{450}{x} - \frac{450}{x+15} = 1$. Multiply out to get $x^2 + 15x - 6750 = 0$. Factors are $(x-75)(x+90)=0$.

Explain why the time taken by a train to cross a bridge is always greater than the time taken to cross a pole at the same speed.

Reveal Answer & Explanation

Answer:

• When crossing a pole, the pole is considered a point object with zero length; the train needs to cover only its own length ($L$).
• When crossing a bridge of length $P$, the train must clear both its own length and the bridge's length (Total Distance $= L + P$).
• Since $L + P > L$, the distance to be traversed is greater, requiring more time at the same speed ($T = D/S$).

Crossing a pole requires covering only length $L$; crossing a bridge requires covering $L + P$.

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