Speed and Time
Pointwise AISSEE Class 6 Mathematics guide for Speed and Time (Topic 22). Master D = S × T, km/h to m/s conversion (5/18), average speed formula 2xy/(x+y), and train crossing poles and platforms.
Overview & Fundamental Concept
Pointwise AISSEE Class 6 Mathematics guide for Speed and Time (Topic 22). Master D = S × T, km/h to m/s conversion (5/18), average speed formula 2xy/(x+y), and train crossing poles and platforms.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 22 | Relationship between speed, distance and time, average speed, conversion between km/h and m/s, problems on trains and moving bodies. | Master kinematic distance-speed-time formulas, 5/18 and 18/5 unit conversion factors, harmonic average speed, and train platform/pole crossing models for AISSEE Class VI. |
1. The Fundamental Kinematic Formula Triangle
- Speed: The rate of distance covered per unit of time: $$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$
- Distance: Total path length traveled: $$\text{Distance} = \text{Speed} \times \text{Time}$$
- Time: Duration taken to cover the path: $$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$
- Standard Units: Distance in kilometers ($\text{km}$) or meters ($\text{m}$); Time in hours ($\text{h}$) or seconds ($\text{s}$); Speed in $\text{km/h}$ or $\text{m/s}$.
| Target Measure | Formula | Units Combination |
|---|---|---|
| Speed | $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$ | $\text{km/h}$ or $\text{m/s}$ |
| Distance | $\text{Distance} = \text{Speed} \times \text{Time}$ | $\text{km}$ or $\text{m}$ |
| Time | $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ | $\text{hours}$ or $\text{seconds}$ |
Cadet Mnemonic • D-S-T Triangle
Cover what you want to find:
- Distance on Top ($D = S \times T$)
- Speed bottom-left ($S = D / T$)
- Time bottom-right ($T = D / S$)
2. The Universal Conversion Factors: 5/18 & 18/5
- Derivation of Conversion Factor:
$$1\text{ km/h} = \frac{1,000\text{ meters}}{3,600\text{ seconds}} = \frac{10}{36} = \mathbf{\frac{5}{18}\text{ m/s}}$$ - Rule 1 (km/h to m/s): To convert speed from $\text{km/h}$ to $\text{m/s}$, multiply by $\frac{5}{18}$: $$\text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18}$$ Example: $72\text{ km/h} = 72 \times \frac{5}{18} = 4 \times 5 = 20\text{ m/s}$.
- Rule 2 (m/s to km/h): To convert speed from $\text{m/s}$ to $\text{km/h}$, multiply by $\frac{18}{5}$: $$\text{Speed in km/h} = \text{Speed in m/s} \times \frac{18}{5}$$ Example: $25\text{ m/s} = 25 \times \frac{18}{5} = 5 \times 18 = 90\text{ km/h}$.
| Speed in km/h (Multiples of 18) | Conversion Operation | Speed in m/s (Multiples of 5) |
|---|---|---|
| $18\text{ km/h}$ | $18 \times \frac{5}{18}$ | $5\text{ m/s}$ |
| $36\text{ km/h}$ | $36 \times \frac{5}{18}$ | $10\text{ m/s}$ |
| $54\text{ km/h}$ | $54 \times \frac{5}{18}$ | $15\text{ m/s}$ |
| $72\text{ km/h}$ | $72 \times \frac{5}{18}$ | $20\text{ m/s}$ |
| $90\text{ km/h}$ | $90 \times \frac{5}{18}$ | $25\text{ m/s}$ |
| $108\text{ km/h}$ | $108 \times \frac{5}{18}$ | $30\text{ m/s}$ |
Exam Trick • Multiples of 18 and 5 Anchor
Speed in km/h is usually a multiple of 18! Speed in m/s is the corresponding multiple of 5. For 54 km/h: 54 is 18 × 3 => Speed is 5 × 3 = 15 m/s instantly!
3. Average Speed & Train Crossing Scenarios
- Universal Average Speed Formula: $$\text{Average Speed} = \frac{\text{Total Distance Covered}}{\text{Total Time Taken}}$$
- Harmonic Mean Shortcut for Equal Distances: If a vehicle travels from $A$ to $B$ at speed $x$ and returns from $B$ to $A$ at speed $y$ over the same distance: $$\text{Average Speed} = \frac{2xy}{x + y}$$ Caution: Never take the simple arithmetic mean $\frac{x + y}{2}$!
- Train Crossing a Point Object (Pole, Tree, Person): The distance covered by the train equals its own length: $$\text{Distance} = \text{Length of Train} (L_T) \implies \text{Time} = \frac{L_T}{\text{Speed}}$$
- Train Crossing an Extended Object (Platform, Bridge, Tunnel): The total distance covered equals the length of the train PLUS the length of the platform: $$\text{Distance} = L_T + L_P \implies \text{Time} = \frac{L_T + L_P}{\text{Speed}}$$
AISSEE Trap Alert • The Train Crossing Unit Trap
Train length is given in METERS, but speed is given in KM/H! You MUST convert the speed to m/s (by multiplying by 5/18) before computing time in seconds. Never mix meters with km/h.
Visual Concept Blueprint • Vector Architecture
Speed Shortcuts & Mental Calculation Strategies
Master these high-yield time-saving heuristics curated by top AISSEE scorers to eliminate calculation bottlenecks and answer questions in under 45 seconds.
Shortcut Technique 1: Harmonic Average Speed Anchor
Going at 60 km/h and returning at 40 km/h: Average speed is $\frac{2 \times 60 \times 40}{60 + 40} = \frac{4800}{100} = 48\text{ km/h}$ (NOT 50 km/h!).
Shortcut Technique 2: Relative Speed in Same Direction
Two bodies moving in the SAME direction: $\text{Relative Speed} = S_1 - S_2$ (subtract speeds).
Shortcut Technique 3: Relative Speed in Opposite Direction
Two bodies moving in OPPOSITE directions towards each other: $\text{Relative Speed} = S_1 + S_2$ (add speeds).
Interactive 3D Revision Flashcards Deck
Click any card to flip and reveal the key exam rule or topper secret.
REVISION FLASHCARD
What is the core rule & definition for Speed and Time?
CADET MEMORY KEY
Speed Formula: \text{Speed} = \frac{\text{Distance}}{\text{Time}}
REVISION FLASHCARD
Cadet Shortcut: How to solve Speed and Time in under 30 seconds?
CADET MEMORY KEY
Fast MCQ Elimination: Check the last digit (unit digit) and estimate the order of magnitude to eliminate 2 distractors in under 10 seconds.
REVISION FLASHCARD
Exam Trap Alert: What pitfall do most candidates fall into?
CADET MEMORY KEY
Units Match: If distance is in km, time MUST be in hours; if distance is in meters, time MUST be in seconds.
REVISION FLASHCARD
Rapid Memory Anchor for AISSEE Hall Recall:
CADET MEMORY KEY
Average Speed: Average Speed = (Total Distance) ÷ (Total Time). NOT the average of speeds!
REVISION FLASHCARD
A car travels from A to B at 30 km/h and returns at 20 km/h. What is the average speed?
CADET MEMORY KEY
Harmonic mean: 2xy / (x + y) = (2 × 30 × 20) / (30 + 20) = 1200 / 50 = 24 km/h.
Common Traps & Examination Pitfalls
Avoid the Speed-Guess Trap: Under AISSEE pressure, candidates frequently misread units (e.g., mixing metres and centimetres, or per hour and per second) and confuse complementary ($90^\circ$) with supplementary ($180^\circ$) angles. Read every stem carefully and check units before selecting your OMR option.
Key Identities & Formulas
Conversion Factors
$$1\text{ km/h} = \frac{5}{18}\text{ m/s}, \quad 1\text{ m/s} = \frac{18}{5}\text{ km/h}$$
Standard velocity conversion multipliers.
Harmonic Average Speed
$$\text{Avg Speed} = \frac{2xy}{x + y}$$
Equal forward and return distance journey.
Train Crossing Platform
$$T = \frac{L_{\text{train}} + L_{\text{platform}}}{\text{Speed (in m/s)}}$$
Time taken to clear stationary extended structure.
Step-by-Step Solved Examination Models
Solved Example 1
Model Problem
A train running at a speed of $90\text{ km/h}$ crosses a pole in 10 seconds. Find the length of the train.
Detailed Solution:
Convert speed to m/s: $90 \times \frac{5}{18} = 25\text{ m/s}$.
Distance = $\text{Speed} \times \text{Time} = 25 \times 10 = 250$ m.
Length of the train is 250 m.
Solved Example 2
Model Problem
A bus travels 180 km in 4 hours. How long will it take to travel 270 km at the same speed?
Detailed Solution:
Speed = $180 / 4 = 45\text{ km/h}$.
Time required = $\text{Distance} / \text{Speed} = 270 / 45 = 6$ hours.
Solved Example 3
Model Problem
A train 150 meters long is running at a speed of 54 km/h. How much time will it take to cross a railway platform 250 meters long?
Detailed Solution:
1. Convert train speed from km/h to m/s: $54 \times \frac{5}{18} = 3 \times 5 = 15\text{ m/s}$.
2. Total distance to be covered $= \text{Length of Train} + \text{Length of Platform} = 150 + 250 = 400\text{ meters}$.
3. $\text{Time} = \frac{\text{Total Distance}}{\text{Speed}} = \frac{400}{15} = \frac{80}{3} = 26.67\text{ seconds}$ (or $26\frac{2}{3}\text{ s}$). Correct Option is (A).
2. Total distance to be covered $= \text{Length of Train} + \text{Length of Platform} = 150 + 250 = 400\text{ meters}$.
3. $\text{Time} = \frac{\text{Total Distance}}{\text{Speed}} = \frac{400}{15} = \frac{80}{3} = 26.67\text{ seconds}$ (or $26\frac{2}{3}\text{ s}$). Correct Option is (A).
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Question 1 of 10
Exam Standard
What is the speed of a car that covers 300 km in 5 hours?
Correct Answer: Option (A)
Speed = $\text{Distance} / \text{Time} = 300 / 5 = 60\text{ km/h}$.
Question 2 of 10
Exam Standard
Convert 54 km/h into metres per second.
Correct Answer: Option (A)
$54 \times \frac{5}{18} = 3 \times 5 = 15\text{ m/s}$.
Question 3 of 10
Exam Standard
How much distance will an aeroplane cover in 3 hours 30 minutes at a speed of $600\text{ km/h}$?
Correct Answer: Option (A)
Time = $3.5$ hours. Distance = $600 \times 3.5 = 2,100$ km.
Question 4 of 10
Exam Standard
A cyclist rides at $12\text{ km/h}$. How long will he take to cover 18 km?
Correct Answer: Option (A)
Time = $18 / 12 = 1.5$ hours = 1 hour 30 minutes.
Question 5 of 10
Exam Standard
A 150 m long train crosses a 250 m long bridge at $72\text{ km/h}$. Find the time taken.
Correct Answer: Option (A)
Total distance = $150 + 250 = 400$ m. Speed = $72 \times \frac{5}{18} = 20\text{ m/s}$. Time = $400 / 20 = 20$ seconds.
Question 6 of 10
Exam Standard
If a person walks at 4 km/h, he reaches school 10 minutes late. At 5 km/h, he reaches on time. What is the distance to school?
Correct Answer: Option (A)
Let distance be $d$. $\frac{d}{4} - \frac{d}{5} = \frac{10}{60} \implies \frac{d}{20} = \frac{1}{6} \implies d = \frac{20}{6} = 3\frac{1}{3}$ km.
Question 7 of 10
Exam Standard
Convert 20 m/s into km/h.
Correct Answer: Option (A)
$20 \times \frac{18}{5} = 4 \times 18 = 72\text{ km/h}$.
Question 8 of 10
Exam Standard
A boy runs 200 metres in 25 seconds. Find his speed in km/h.
Correct Answer: Option (A)
Speed in m/s = $200 / 25 = 8\text{ m/s}$. Convert to km/h: $8 \times \frac{18}{5} = 144 / 5 = 28.8\text{ km/h}$.
Question 9 of 10
Exam Standard
Two cars start from points A and B towards each other at 40 km/h and 60 km/h. If distance AB is 200 km, after how many hours will they meet?
Correct Answer: Option (A)
Relative speed = $40 + 60 = 100\text{ km/h}$. Time = $200 / 100 = 2$ hours.
Question 10 of 10
Exam Standard
A man covers half his journey at 20 km/h and the remaining half at 30 km/h. Find his average speed for the whole journey.
Correct Answer: Option (A)
Average speed for equal distances = $\frac{2xy}{x + y} = \frac{2 \times 20 \times 30}{20 + 30} = \frac{1200}{50} = 24\text{ km/h}$.
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Last-Minute Cadet Revision Matrix
Rapid recall checkpoints to review 24 hours before stepping into the examination hall.
| Concept Checkpoint | High-Yield Exam Rule & Memory Anchor |
|---|---|
| km/h to m/s | Multiply by 5/18 to convert km/h to m/s. |
| m/s to km/h | Multiply by 18/5 to convert m/s to km/h. |
| Average Speed | Average speed is Total Distance / Total Time. Never take the simple average. |
| Platform Crossing | Distance = Length of Train + Length of Platform. |
Frequently Asked Questions
How many questions from Speed and Time appear in the AISSEE Class VI exam?
Typically, 2 to 3 questions directly test Speed and Time in the 50-question Mathematics paper, contributing 6 to 9 marks towards the 150-mark section total.
Is there any negative marking for incorrect answers in AISSEE Class VI?
No. The National Testing Agency (NTA) does not apply negative marking in the AISSEE Class VI entrance examination. Candidates should attempt every single question.
What is the best way to revise Speed and Time before the exam?
Review the high-yield formulas, practice the 10 interactive drill MCQs on this page, and memorize the revision checkpoints matrix.