Official Syllabus Item: 22, 24
Arithmetic & Applications
Speed, Time & Distance (Trains, Relative Motion & Races)
Master speed-distance-time relations, km/h to m/s conversions, train crossing dynamics (poles, platforms, tunnels), relative speed, early/late time shifts, and race lead problems for RIMC Entrance Exam.
Overview & Fundamental Concept
Speed, Time, and Distance constitutes one of the most prominent applied mathematics modules in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Serial Numbers 22 ('Speed and Time') and 24 ('Simple problems of time and distance') of Chapter 8 in the official RIMC brochure, questions in this section feature consistently across Part A (5 Marks) and Part B (10 Marks). The syllabus tests candidates on dimensional velocity conversions ($\text{km/h} \leftrightarrow \text{m/s}$), train passage dynamics across stationary point objects (poles, milestones) and extended structures (platforms, bridges, tunnels), relative speed vectors for bodies moving in parallel and opposing directions, multi-segment journey averages, early/late arrival time shifts, and linear pursuit races. Descriptive step marking requires candidates to explicitly derive unit conversion multipliers ($\frac{5}{18}$ or $\frac{18}{5}$), declare relative speed formulas before substitution, define variable distances, and box final answers with standard metric units.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 22 | Speed and Time: Formulas, units conversions, uniform motion, and kinematic relationships | Tests fundamental physical arithmetic, temporal conversions, rate equations, and proportional scaling. |
| Item 24 | Simple problems of time and distance: Train crossings, relative speed, and applied transit scenarios | Assesses extended object kinematics (trains passing platforms/tunnels), vector relative speed, and pursuit dynamics. |
1. Core Kinematic Equations & The 5/18 Conversion Bridge
The fundamental relationship governing uniform rectilinear motion is:
$$\text{Speed } (v) = \frac{\text{Distance } (d)}{\text{Time } (t)}, \quad \text{Distance } (d) = \text{Speed } (v) \times \text{Time } (t), \quad \text{Time } (t) = \frac{\text{Distance } (d)}{\text{Speed } (v)}$$
The 5/18 Metric Bridge: In RIMC descriptive questions, speed is commonly given in $\text{km/h}$ while distances are measured in metres and time in seconds. Candidates must never omit the formal conversion step:
$$1\text{ km/h} = \frac{1\text{ kilometre}}{1\text{ hour}} = \frac{1000\text{ metres}}{3600\text{ seconds}} = \frac{10}{36}\text{ m/s} = \mathbf{\frac{5}{18}\text{ m/s}}$$
$$1\text{ m/s} = \frac{1/1000\text{ km}}{1/3600\text{ h}} = \frac{3600}{1000}\text{ km/h} = \mathbf{\frac{18}{5}\text{ km/h}}$$
• To convert from $\text{km/h}$ to $\text{m/s}$: Multiply by $\frac{5}{18}$.
• To convert from $\text{m/s}$ to $\text{km/h}$: Multiply by $\frac{18}{5}$.
Speed Conversion Quick Lookup Table
- $18\text{ km/h} = 5\text{ m/s}$, $\quad 36\text{ km/h} = 10\text{ m/s}$, $\quad 54\text{ km/h} = 15\text{ m/s}$
- $72\text{ km/h} = 20\text{ m/s}$, $\quad 90\text{ km/h} = 25\text{ m/s}$, $\quad 108\text{ km/h} = 30\text{ m/s}$
- $126\text{ km/h} = 35\text{ m/s}$, $\quad 144\text{ km/h} = 40\text{ m/s}$.
2. The Mathematical Mechanics of Train Crossings
In problems involving trains, the physical length of the train and the object it traverses dictates the total effective distance:
| Encounter Scenario | Effective Distance Covered $(d)$ | Effective Speed $(v_{\text{eff}})$ | Time Formula $(t = d / v)$ |
|---|---|---|---|
| Train Passing a Stationary Point Object (Telegraph pole, tree, milestone, standing person) | $$d = L_{\text{train}}$$ (Object length is negligible) | $$v = v_{\text{train}}$$ | $$t = \frac{L_{\text{train}}}{v_{\text{train}}}$$ |
| Train Passing a Stationary Extended Object (Platform, railway bridge, tunnel, stationary train) | $$d = L_{\text{train}} + L_{\text{object}}$$ | $$v = v_{\text{train}}$$ | $$t = \frac{L_{\text{train}} + L_{\text{object}}}{v_{\text{train}}}$$ |
| Two Trains Moving in OPPOSITE Directions | $$d = L_1 + L_2$$ | $$v_{\text{rel}} = v_1 + v_2$$ (Relative speed adds) | $$t = \frac{L_1 + L_2}{v_1 + v_2}$$ |
| Two Trains Moving in the SAME Direction (Faster train overtaking slower train) | $$d = L_1 + L_2$$ | $$v_{\text{rel}} = v_{\text{fast}} - v_{\text{slow}}$$ (Relative speed subtracts) | $$t = \frac{L_1 + L_2}{v_{\text{fast}} - v_{\text{slow}}}$$ |
3. Relative Velocity Principles for Co-Linear Motion
Relative Speed is the rate at which the distance between two moving bodies is either shrinking or expanding:
• Opposite Directions: When two objects move towards each other (or away from each other in opposite directions), their relative speed is the sum of their individual speeds: $v_{\text{rel}} = v_1 + v_2$.
• Same Direction: When two objects move in the same direction, their relative speed is the difference between their speeds: $v_{\text{rel}} = |v_1 - v_2|$.
4. Classic RIMC Journey Archetypes
Examiners frequently test two specialized real-world travel models:
Archetype A: The Early / Late School Journey Formula
A student travels to school at speed $s_1$ and arrives $t_1$ minutes late. Next day, he travels at speed $s_2$ and arrives $t_2$ minutes early:
- Let true scheduled time be $T\text{ minutes}$ and distance be $D\text{ km}$.
- Time Difference: $\Delta t = (t_1 + t_2)\text{ minutes} = \frac{t_1 + t_2}{60}\text{ hours}$.
- Since $\frac{D}{s_1} - \frac{D}{s_2} = \Delta t$:
- $$\mathbf{D = \frac{s_1 \times s_2}{s_2 - s_1} \times \left( \frac{\text{Total Time Shift}}{60} \right)}$$
Archetype B: Races and Head Starts
- Head Start in Distance: 'A can give B a start of $20\text{ m}$ in a $100\text{ m}$ race' means when A runs $100\text{ m}$, B only needs to run $80\text{ m}$ in the same time.
- Head Start in Time: 'A beats B by $5\text{ seconds}$' means A finishes the race $5\text{ seconds}$ before B crosses the line.
Illustrated Conceptual Diagrams & Visual Schematics
Relative Speed & Boat Upstream/Downstream Dynamics
Key Principles: (1) Opposite Direction: $S_{\text{rel}} = u + v$ (Speeds add). (2) Same Direction: $S_{\text{rel}} = u - v$ (Speeds subtract). (3) Downstream Flow: $v_{\text{down}} = u + v$ (Boat + Stream). (4) Upstream Flow: $v_{\text{up}} = u - v$ (Boat - Stream).
Topper Strategy & Descriptive Step Breakdown
Mastering RIMC Mathematics Standard
In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.
Official Descriptive Marking Scheme & Step Allocation
In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:
Step 1: Metric Parameter Harmonization & Conversion
1 - 2 Marks
State given values, calculate total distance (Length of train + platform), and convert km/h to m/s explicitly showing the 5/18 factor.
Booklet Layout:
Given:\nLength of train $L_1 = 240\text{ m}$, Length of platform $L_2 = 260\text{ m}$\n$$\text{Total Distance Covered } d = 240 + 260 = 500\text{ m}$$\nTime taken $t = 25\text{ seconds}$.
Given:\nLength of train $L_1 = 240\text{ m}$, Length of platform $L_2 = 260\text{ m}$\n$$\text{Total Distance Covered } d = 240 + 260 = 500\text{ m}$$\nTime taken $t = 25\text{ seconds}$.
Step 2: Kinematic Formula Statement
1 - 2 Marks
State the velocity formula $v = d / t$ or $t = d / v_{\text{rel}}$ symbolically.
Booklet Layout:
$$\text{Speed in m/s} = \frac{\text{Total Distance}}{\text{Time Taken}} = \frac{d}{t}$$
$$\text{Speed in m/s} = \frac{\text{Total Distance}}{\text{Time Taken}} = \frac{d}{t}$$
Step 3: Stepwise Arithmetic Reduction & Unit Conversion
2 - 4 Marks
Evaluate speed in m/s, then convert back to km/h by multiplying by 18/5 with step-by-step cancellations.
Booklet Layout:
$$\text{Speed in m/s} = \frac{500}{25} = 20\text{ m/s}$$
$$\text{Converting to km/h: } 20 \times \frac{18}{5} = 4 \times 18 = 72\text{ km/h}$$
$$\text{Speed in m/s} = \frac{500}{25} = 20\text{ m/s}$$
$$\text{Converting to km/h: } 20 \times \frac{18}{5} = 4 \times 18 = 72\text{ km/h}$$
Step 4: Concluding Statement with Proper Units & Answer Box
1 Mark
State the final velocity or time clearly in words and enclose in an answer box.
Booklet Layout:
$$\mathbf{\text{Answer: The speed of the train is } 72\text{ km/h.}}$$
$$\mathbf{\text{Answer: The speed of the train is } 72\text{ km/h.}}$$
Examiner Traps & Common Blunders (Marks Penalties)
Omitting the Length of the Train When Passing Extended Objects -3 Marks
Common Mistake: Using only the platform length ($d = 260\text{ m}$) as the total distance covered.
Cadet Antidote: A train completely clears a platform only when its rear wagon exits the platform edge: $d = L_{\text{train}} + L_{\text{platform}}$.
Inverting the 5/18 and 18/5 Conversion Multipliers -2 Marks
Common Mistake: Multiplying by $\frac{18}{5}$ when converting from $\text{km/h}$ to $\text{m/s}$.
Cadet Antidote: Memory anchor: Going to a smaller unit (m/s) requires the smaller numerator ($ rac{5}{18}$). Going to a larger unit (km/h) requires the larger numerator ($ rac{18}{5}$).
Relative Speed Direction Confusion -3 Marks
Common Mistake: Subtracting speeds ($v_1 - v_2$) when two trains move towards each other in opposite directions.
Cadet Antidote: Opposite directions SHRINK distance faster $\implies$ speeds ADD ($v_1 + v_2$). Same direction $\implies$ speeds SUBTRACT ($v_1 - v_2$).
Forgetting to Convert Early/Late Minutes to Hours -2 Marks
Common Mistake: Multiplying speed difference directly by minutes without dividing by 60.
Cadet Antidote: If speed is in km/h, time differences MUST be converted to hours by dividing by 60.
Cadet Topper Mental Verification Shortcuts
Multiples of 18 Shortcut for km/h
Every $18\text{ km/h}$ equals exactly $5\text{ m/s}$. Divide the km/h by 18 and multiply by 5 to find m/s in your head.
$54\text{ km/h} = (54 \div 18) \times 5 = 3 \times 5 = 15\text{ m/s}$ in 1 second.
1-Line Early/Late Distance Formula
To find distance when given two speeds and early/late arrivals: $D = \frac{s_1 s_2}{s_2 - s_1} \times \left(\frac{\Delta t_{\text{total}}}{60}\right)$.
Speeds 4 km/h (10 mins late) and 5 km/h (5 mins early): $D = \frac{4 \times 5}{5 - 4} \times \frac{10 + 5}{60} = 20 \times \frac{15}{60} = 20 \times \frac{1}{4} = 5\text{ km}$ instantly.
Overtaking Trains Relative Distance Balance
When a faster train overtakes a slower train, the distance covered is ALWAYS the sum of lengths $(L_1 + L_2)$, while the speed is $(v_1 - v_2)$.
Crossing time $t = \frac{L_1 + L_2}{(v_1 - v_2) \times \frac{5}{18}}$.
Key Formulas & Conversion Cheat Sheet
Speed-Distance-Time Core Relation
$$v = \frac{d}{t}, \quad d = v \times t, \quad t = \frac{d}{v}$$
Fundamental uniform rectilinear kinematics.
km/h to m/s Conversion
$$v_{\text{m/s}} = v_{\text{km/h}} \times \frac{5}{18}$$
Converts kilometres per hour to metres per second.
m/s to km/h Conversion
$$v_{\text{km/h}} = v_{\text{m/s}} \times \frac{18}{5}$$
Converts metres per second to kilometres per hour.
Train Passing Stationary Extended Object
$$t = \frac{L_{\text{train}} + L_{\text{platform}}}{v_{\text{train}}}$$
Time taken to cross a platform, bridge, or tunnel.
Relative Speed (Opposite Directions)
$$v_{\text{rel}} = v_1 + v_2$$
Effective velocity when two objects move towards each other.
Relative Speed (Same Direction)
$$v_{\text{rel}} = |v_1 - v_2|$$
Effective velocity when two objects move in the same direction.
Harmonic Average Speed Formula
$$v_{\text{avg}} = \frac{2uv}{u + v}$$
Average speed for equal forward and return journeys.
Early / Late Journey Distance
$$D = \frac{s_1 \times s_2}{s_2 - s_1} \times \left( \frac{t_{\text{late}} + t_{\text{early}}}{60} \right)$$
Direct distance calculation from dual-speed schedule shifts.
Solved Model Examples (Descriptive Format with Step Mark Distribution)
Model 1: Train Crossing an Extended Platform
5 Marks
A train $240\text{ metres}$ long passes a railway platform $260\text{ metres}$ long in $25\text{ seconds}$. Calculate the speed of the train in kilometres per hour ($\text{km/h}$).
Step-by-Step Model Solution:
Step 1: Compute Total Effective Distance
1.5 Marks
Length of train $L_1 = 240\text{ m}$.
Length of platform $L_2 = 260\text{ m}$.
$$\text{Total Distance Covered } d = L_1 + L_2 = 240 + 260 = 500\text{ m}$$
Length of platform $L_2 = 260\text{ m}$.
$$\text{Total Distance Covered } d = L_1 + L_2 = 240 + 260 = 500\text{ m}$$
Step 2: Calculate Speed in Metres per Second (m/s)
1.5 Marks
Time taken $t = 25\text{ seconds}$.
$$\text{Speed in m/s} = \frac{d}{t} = \frac{500\text{ m}}{25\text{ s}} = 20\text{ m/s}$$
$$\text{Speed in m/s} = \frac{d}{t} = \frac{500\text{ m}}{25\text{ s}} = 20\text{ m/s}$$
Step 3: Convert Speed to km/h using 18/5 Multiplier
2 Marks
$$\text{Speed in km/h} = 20 \times \frac{18}{5} = (20 \div 5) \times 18 = 4 \times 18 = \mathbf{72\text{ km/h}}$$
$$\mathbf{\text{Answer: The speed of the train is } 72\text{ km/h.}}$$
$$\mathbf{\text{Answer: The speed of the train is } 72\text{ km/h.}}$$
Examiner Scoring Key: Adding 240 + 260 = 500 m carries 1.5 marks. Computing 20 m/s carries 1.5 marks. Correct conversion to 72 km/h earns full 5 marks.
Model 2: Two Trains Crossing in Opposite Directions
5 Marks
Two trains of lengths $120\text{ metres}$ and $80\text{ metres}$ 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 completely pass each other?
Step-by-Step Model Solution:
Step 1: Compute Total Combined Distance
1.5 Marks
$$d = L_1 + L_2 = 120 + 80 = 200\text{ metres}$$
Step 2: Compute Relative Speed in m/s
2 Marks
Since the trains move in opposite directions, their relative speed is the sum of their speeds:
$$v_{\text{rel}} = 42 + 30 = 72\text{ km/h}$$
Convert to $\text{m/s}$ by multiplying by $\frac{5}{18}$:
$$v_{\text{rel}} = 72 \times \frac{5}{18} = 4 \times 5 = 20\text{ m/s}$$
$$v_{\text{rel}} = 42 + 30 = 72\text{ km/h}$$
Convert to $\text{m/s}$ by multiplying by $\frac{5}{18}$:
$$v_{\text{rel}} = 72 \times \frac{5}{18} = 4 \times 5 = 20\text{ m/s}$$
Step 3: Calculate Crossing Time
1.5 Marks
$$t = \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{200\text{ m}}{20\text{ m/s}} = \mathbf{10\text{ seconds}}$$
$$\mathbf{\text{Answer: The trains will pass each other in } 10\text{ seconds.}}$$
$$\mathbf{\text{Answer: The trains will pass each other in } 10\text{ seconds.}}$$
Examiner Scoring Key: Stating relative speed = 42 + 30 = 72 km/h and converting to 20 m/s carries 2 marks. Concluding with 10 seconds gets full marks.
Model 3: Train Passing a Moving Man (Same Direction)
7 Marks
A train $150\text{ metres}$ long passes a man walking at $6\text{ km/h}$ in the same direction in which the train is going, in $10\text{ seconds}$. Find the speed of the train in kilometres per hour.
Step-by-Step Model Solution:
Step 1: Declare Variable & Formulate Relative Speed
2 Marks
Let the speed of the train be $v\text{ km/h}$.
Speed of the man $= 6\text{ km/h}$.
Since both move in the same direction:
$$\text{Relative Speed } v_{\text{rel}} = (v - 6)\text{ km/h}$$
Speed of the man $= 6\text{ km/h}$.
Since both move in the same direction:
$$\text{Relative Speed } v_{\text{rel}} = (v - 6)\text{ km/h}$$
Step 2: Relate Distance and Time to Relative Speed in m/s
2.5 Marks
Distance covered passing the man $= \text{Length of train} = 150\text{ metres}$.
Time taken $= 10\text{ seconds}$.
$$\text{Relative Speed in m/s} = \frac{150\text{ m}}{10\text{ s}} = 15\text{ m/s}$$
Time taken $= 10\text{ seconds}$.
$$\text{Relative Speed in m/s} = \frac{150\text{ m}}{10\text{ s}} = 15\text{ m/s}$$
Step 3: Convert Relative Speed to km/h and Solve for v
2.5 Marks
$$\text{Relative Speed in km/h} = 15 \times \frac{18}{5} = 3 \times 18 = 54\text{ km/h}$$
Now equate with our expression from Step 1:
$$v - 6 = 54 \implies v = 54 + 6 = \mathbf{60\text{ km/h}}$$
$$\mathbf{\text{Answer: The speed of the train is } 60\text{ km/h.}}$$
Now equate with our expression from Step 1:
$$v - 6 = 54 \implies v = 54 + 6 = \mathbf{60\text{ km/h}}$$
$$\mathbf{\text{Answer: The speed of the train is } 60\text{ km/h.}}$$
Examiner Scoring Key: Setting relative speed as (v - 6) carries 2 marks. Converting 15 m/s to 54 km/h carries 2.5 marks. Concluding with v = 60 km/h earns 7 marks.
Model 4: The Classic Early / Late Arrival Problem
7 Marks
If a cadet walks from his quarters to the drill ground at $4\text{ km/h}$, he arrives $10\text{ minutes}$ late for morning muster. If he walks at $5\text{ km/h}$, he arrives $5\text{ minutes}$ before muster time. Find the distance between his quarters and the drill ground.
Step-by-Step Model Solution:
Step 1: Declare Distance Variable & Time Difference
2 Marks
Let the distance to the drill ground be $D\text{ km}$.
Let the scheduled muster time be $T\text{ hours}$.
At $4\text{ km/h}$, time taken $t_1 = T + \frac{10}{60}\text{ h}$.
At $5\text{ km/h}$, time taken $t_2 = T - \frac{5}{60}\text{ h}$.
Let the scheduled muster time be $T\text{ hours}$.
At $4\text{ km/h}$, time taken $t_1 = T + \frac{10}{60}\text{ h}$.
At $5\text{ km/h}$, time taken $t_2 = T - \frac{5}{60}\text{ h}$.
Step 2: Formulate Time Difference Equation
2.5 Marks
The difference between the two travel times is:
$$t_1 - t_2 = \left(T + \frac{10}{60}\right) - \left(T - \frac{5}{60}\right) = \frac{10 + 5}{60} = \frac{15}{60} = \frac{1}{4}\text{ hour}$$
We also know that $t = \frac{D}{s}$:
$$\frac{D}{4} - \frac{D}{5} = \frac{1}{4}$$
$$t_1 - t_2 = \left(T + \frac{10}{60}\right) - \left(T - \frac{5}{60}\right) = \frac{10 + 5}{60} = \frac{15}{60} = \frac{1}{4}\text{ hour}$$
We also know that $t = \frac{D}{s}$:
$$\frac{D}{4} - \frac{D}{5} = \frac{1}{4}$$
Step 3: Solve for Distance D & Verify
2.5 Marks
$$\text{LCM}(4, 5) = 20 \implies \frac{5D - 4D}{20} = \frac{1}{4}$$
$$\frac{D}{20} = \frac{1}{4} \implies D = \frac{20}{4} = \mathbf{5\text{ km}}$$
Check: At $4\text{ km/h}$, time $= 5/4\text{ h} = 75\text{ mins}$ ($10$ mins late $\implies$ schedule is $65$ mins). At $5\text{ km/h}$, time $= 5/5\text{ h} = 60\text{ mins}$ ($5$ mins early $\implies$ schedule is $65$ mins). Verified!
$$\mathbf{\text{Answer: The distance is } 5\text{ km.}}$$
$$\frac{D}{20} = \frac{1}{4} \implies D = \frac{20}{4} = \mathbf{5\text{ km}}$$
Check: At $4\text{ km/h}$, time $= 5/4\text{ h} = 75\text{ mins}$ ($10$ mins late $\implies$ schedule is $65$ mins). At $5\text{ km/h}$, time $= 5/5\text{ h} = 60\text{ mins}$ ($5$ mins early $\implies$ schedule is $65$ mins). Verified!
$$\mathbf{\text{Answer: The distance is } 5\text{ km.}}$$
Examiner Scoring Key: Setting up D/4 - D/5 = 15/60 carries 4.5 marks. Solving for D = 5 km carries 2.5 marks.
Model 5: Overtaking Two Parallel Trains
10 Marks
A fast express train $180\text{ metres}$ long is travelling at $72\text{ km/h}$. A freight train $120\text{ metres}$ long is travelling ahead of it on a parallel track in the same direction at $36\text{ km/h}$.
(a) How long will it take for the express train to completely pass the freight train?
(b) How far does the express train travel during the overtaking maneuver?
(a) How long will it take for the express train to completely pass the freight train?
(b) How far does the express train travel during the overtaking maneuver?
Step-by-Step Model Solution:
Step 1: Compute Total Overtaking Distance
2.5 Marks
Length of express train $L_1 = 180\text{ m}$.
Length of freight train $L_2 = 120\text{ m}$.
$$\text{Total Distance to completely overtake } d = L_1 + L_2 = 180 + 120 = 300\text{ metres}$$
Length of freight train $L_2 = 120\text{ m}$.
$$\text{Total Distance to completely overtake } d = L_1 + L_2 = 180 + 120 = 300\text{ metres}$$
Step 2: Compute Relative Speed in m/s
2.5 Marks
Both trains travel in the same direction:
$$v_{\text{rel}} = 72 - 36 = 36\text{ km/h}$$
Convert to $\text{m/s}$:
$$v_{\text{rel}} = 36 \times \frac{5}{18} = 2 \times 5 = 10\text{ m/s}$$
$$v_{\text{rel}} = 72 - 36 = 36\text{ km/h}$$
Convert to $\text{m/s}$:
$$v_{\text{rel}} = 36 \times \frac{5}{18} = 2 \times 5 = 10\text{ m/s}$$
Step 3: Calculate Overtaking Time (Part a)
2.5 Marks
$$t = \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{300\text{ m}}{10\text{ m/s}} = \mathbf{30\text{ seconds}}$$
Step 4: Compute Actual Distance Travelled by Express Train (Part b)
2.5 Marks
Speed of express train $= 72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}$.
Distance travelled in $30\text{ seconds}$:
$$D_{\text{express}} = \text{Speed} \times \text{Time} = 20\text{ m/s} \times 30\text{ s} = \mathbf{600\text{ metres}}$$
$$\mathbf{\text{Answer: (a) Time = } 30\text{ seconds}; \quad \text{(b) Distance = } 600\text{ metres}}$$
Distance travelled in $30\text{ seconds}$:
$$D_{\text{express}} = \text{Speed} \times \text{Time} = 20\text{ m/s} \times 30\text{ s} = \mathbf{600\text{ metres}}$$
$$\mathbf{\text{Answer: (a) Time = } 30\text{ seconds}; \quad \text{(b) Distance = } 600\text{ metres}}$$
Examiner Scoring Key: In part (b), using the true ground speed of the express train (20 m/s * 30 s = 600 m) rather than relative speed is where top students shine.
Model 6: Multi-Stage Journey & Stoppage Delay
10 Marks
Excluding stoppages, the average speed of a long-distance bus is $54\text{ km/h}$, and including stoppages, its average speed drops to $45\text{ km/h}$. For how many minutes does the bus stop per hour? Furthermore, if the total journey is $270\text{ km}$, find the total time spent in transit including all stops.
Step-by-Step Model Solution:
Step 1: Analyze Loss of Distance Due to Stoppages per Hour
3 Marks
In $1\text{ hour}$, without stoppages, the bus covers $54\text{ km}$.
In $1\text{ hour}$, with stoppages, the bus covers $45\text{ km}$.
$$\text{Loss of distance in 1 hour due to stops} = 54 - 45 = 9\text{ km}$$
In $1\text{ hour}$, with stoppages, the bus covers $45\text{ km}$.
$$\text{Loss of distance in 1 hour due to stops} = 54 - 45 = 9\text{ km}$$
Step 2: Calculate Stoppage Time per Hour
3 Marks
The time required to cover this lost $9\text{ km}$ at the normal running speed of $54\text{ km/h}$ is the stoppage time:
$$\text{Stoppage Time} = \frac{9\text{ km}}{54\text{ km/h}} = \frac{1}{6}\text{ hour}$$
Convert to minutes:
$$\text{Stoppage Time per hour} = \frac{1}{6} \times 60 = \mathbf{10\text{ minutes}}$$.
$$\text{Stoppage Time} = \frac{9\text{ km}}{54\text{ km/h}} = \frac{1}{6}\text{ hour}$$
Convert to minutes:
$$\text{Stoppage Time per hour} = \frac{1}{6} \times 60 = \mathbf{10\text{ minutes}}$$.
Step 3: Calculate Total Journey Duration for 270 km
4 Marks
Total journey distance $= 270\text{ km}$.
Overall effective speed including stoppages $= 45\text{ km/h}$.
$$\text{Total Duration} = \frac{\text{Total Distance}}{\text{Speed with Stoppages}} = \frac{270}{45} = \mathbf{6\text{ hours}}$$
Breakdown: Running time $= 270 / 54 = 5\text{ hours}$; Total stoppage time $= 6 \times 10\text{ mins} = 60\text{ mins} = 1\text{ hour}$. $5 + 1 = 6\text{ hours}$ (Matches!).
$$\mathbf{\text{Answer: The bus stops for } 10\text{ minutes per hour; Total transit time = } 6\text{ hours.}}$$
Overall effective speed including stoppages $= 45\text{ km/h}$.
$$\text{Total Duration} = \frac{\text{Total Distance}}{\text{Speed with Stoppages}} = \frac{270}{45} = \mathbf{6\text{ hours}}$$
Breakdown: Running time $= 270 / 54 = 5\text{ hours}$; Total stoppage time $= 6 \times 10\text{ mins} = 60\text{ mins} = 1\text{ hour}$. $5 + 1 = 6\text{ hours}$ (Matches!).
$$\mathbf{\text{Answer: The bus stops for } 10\text{ minutes per hour; Total transit time = } 6\text{ hours.}}$$
Examiner Scoring Key: The formula (Speed without stops - Speed with stops) / Speed without stops * 60 mins carries 3 marks. Duration calculation carries 4 marks.
RIMC Practice & Drill Exercises
Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.
Question 1
5 Marks
Convert $90\text{ km/h}$ into metres per second ($\text{m/s}$), and $25\text{ m/s}$ into kilometres per hour ($\text{km/h}$).
Model Descriptive Solution:
Step 1: Convert 90 km/h to m/s
2.5 Marks
$$90 \times \frac{5}{18} = 5 \times 5 = 25\text{ m/s}$$
Step 2: Convert 25 m/s to km/h
2.5 Marks
$$25 \times \frac{18}{5} = 5 \times 18 = 90\text{ km/h}$$
Final Answer: $90\text{ km/h} = 25\text{ m/s}$; $25\text{ m/s} = 90\text{ km/h}$
Question 2
5 Marks
A boy walks to school at $4\text{ km/h}$ and returns home along the same route at $6\text{ km/h}$. Find his average speed for the whole round trip.
Model Descriptive Solution:
Step 1: State Harmonic Average Speed Formula
2 Marks
$$v_{\text{avg}} = \frac{2uv}{u + v}$$
Step 2: Substitute and Solve
3 Marks
$$v_{\text{avg}} = \frac{2 \times 4 \times 6}{4 + 6} = \frac{48}{10} = 4.8\text{ km/h}$$
Final Answer: $4.8\text{ km/h}$
Question 3
5 Marks
A train $160\text{ metres}$ long is running at $54\text{ km/h}$. In how many seconds will it pass a telegraph post?
Model Descriptive Solution:
Step 1: Convert Speed to m/s
2 Marks
$$54 \times \frac{5}{18} = 3 \times 5 = 15\text{ m/s}$$
Step 2: Compute Crossing Time
3 Marks
$$t = \frac{L_{\text{train}}}{\text{Speed}} = \frac{160}{15} = \frac{32}{3} = 10\frac{2}{3}\text{ seconds} \approx 10.67\text{ s}$$
Final Answer: $10\frac{2}{3}\text{ seconds}$ (or $10.67\text{ s}$)
Question 4
7 Marks
A train $300\text{ m}$ long crosses a platform of length $200\text{ m}$ in $25\text{ seconds}$. Find the speed of the train in km/h.
Model Descriptive Solution:
Step 1: Total Distance
2 Marks
$$d = 300 + 200 = 500\text{ metres}$$
Step 2: Speed in m/s and km/h
5 Marks
$$\text{Speed} = \frac{500}{25} = 20\text{ m/s}$$
$$\text{In km/h} = 20 \times \frac{18}{5} = 72\text{ km/h}$$
$$\text{In km/h} = 20 \times \frac{18}{5} = 72\text{ km/h}$$
Final Answer: $72\text{ km/h}$
Question 5
7 Marks
Two cyclists start from the same point at the same time in opposite directions. If their speeds are $15\text{ km/h}$ and $18\text{ km/h}$, after how many hours will they be $99\text{ km}$ apart?
Model Descriptive Solution:
Step 1: Compute Relative Speed
3 Marks
Moving in opposite directions $\implies v_{\text{rel}} = 15 + 18 = 33\text{ km/h}$.
Step 2: Calculate Time Needed
4 Marks
$$t = \frac{\text{Distance}}{v_{\text{rel}}} = \frac{99}{33} = 3\text{ hours}$$
Final Answer: $3\text{ hours}$
Question 6
7 Marks
A thief is spotted by a policeman from a distance of $200\text{ metres}$. When the policeman starts the chase, the thief also starts running. 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 overtaken?
Model Descriptive Solution:
Step 1: Compute Relative Speed
3 Marks
Same direction $\implies v_{\text{rel}} = 12 - 10 = 2\text{ km/h} = 2 \times \frac{5}{18} = \frac{5}{9}\text{ m/s}$.
Step 2: Time to Catch Thief
2 Marks
$$t = \frac{200\text{ m}}{5/9\text{ m/s}} = 200 \times \frac{9}{5} = 40 \times 9 = 360\text{ seconds} = 6\text{ minutes}$$
Step 3: Distance Run by Thief
2 Marks
$$D_{\text{thief}} = 10\text{ km/h} \times \frac{6}{60}\text{ h} = 10 \times \frac{1}{10} = 1\text{ km} = 1000\text{ metres}$$
Final Answer: $1\text{ km}$ ($1,000\text{ metres}$)
Question 7
7 Marks
A man covers a distance of $60\text{ km}$ in $6\text{ hours}$, partly on foot at $4\text{ km/h}$ and partly on bicycle at $12\text{ km/h}$. Find the distance traveled on foot.
Model Descriptive Solution:
Step 1: Formulate Algebraic Equation
3.5 Marks
Let distance walked on foot be $x\text{ km}$. Distance on cycle $= (60 - x)\text{ km}$.
$$\frac{x}{4} + \frac{60 - x}{12} = 6$$
$$\frac{x}{4} + \frac{60 - x}{12} = 6$$
Step 2: Solve for x
3.5 Marks
$$\text{LCM}(4, 12) = 12 \implies \frac{3x + 60 - x}{12} = 6$$
$$2x + 60 = 72 \implies 2x = 12 \implies x = 6\text{ km}$$
$$2x + 60 = 72 \implies 2x = 12 \implies x = 6\text{ km}$$
Final Answer: Distance on foot $= 6\text{ km}$
Question 8
10 Marks
Two trains $130\text{ m}$ and $110\text{ m}$ long are going in the same direction. The faster train completely overtakes the slower train in $12\text{ seconds}$. If the same trains run in opposite directions, they pass each other in $3\text{ seconds}$. Find the speed of each train in km/h.
Model Descriptive Solution:
Step 1: Set Up Total Distance
2 Marks
$$d = 130 + 110 = 240\text{ metres}$$
Step 2: Same Direction (Overtaking)
3 Marks
$$v_1 - v_2 = \frac{240\text{ m}}{12\text{ s}} = 20\text{ m/s}$$
Step 3: Opposite Direction
3 Marks
$$v_1 + v_2 = \frac{240\text{ m}}{3\text{ s}} = 80\text{ m/s}$$
Step 4: Solve Simultaneous Equations and Convert to km/h
2 Marks
Add equations: $2v_1 = 100 \implies v_1 = 50\text{ m/s}$.
$v_2 = 80 - 50 = 30\text{ m/s}$.
In km/h:
$$v_1 = 50 \times \frac{18}{5} = 180\text{ km/h}$$
$$v_2 = 30 \times \frac{18}{5} = 108\text{ km/h}$$
$v_2 = 80 - 50 = 30\text{ m/s}$.
In km/h:
$$v_1 = 50 \times \frac{18}{5} = 180\text{ km/h}$$
$$v_2 = 30 \times \frac{18}{5} = 108\text{ km/h}$$
Final Answer: Faster train: $180\text{ km/h}$; Slower train: $108\text{ km/h}$
Question 9
10 Marks
In a $1000\text{ m}$ race, Cadet A beats Cadet B by $50\text{ m}$ or $10\text{ seconds}$. Find the time taken by Cadet A to complete the race and his speed in km/h.
Model Descriptive Solution:
Step 1: Find Speed of Cadet B
3 Marks
Cadet B covers the remaining $50\text{ m}$ in $10\text{ seconds}$.
$$\text{Speed of B} = \frac{50}{10} = 5\text{ m/s}$$
$$\text{Speed of B} = \frac{50}{10} = 5\text{ m/s}$$
Step 2: Total Time Taken by Cadet B
3 Marks
$$\text{Time for B to run } 1000\text{ m} = \frac{1000}{5} = 200\text{ seconds}$$
Step 3: Calculate Time & Speed of Cadet A
4 Marks
Cadet A beats B by $10\text{ s}$, so:
$$\text{Time taken by A} = 200 - 10 = 190\text{ seconds} = 3\text{ mins } 10\text{ s}$$
$$\text{Speed of A} = \frac{1000\text{ m}}{190\text{ s}} = \frac{100}{19}\text{ m/s} = \frac{100}{19} \times \frac{18}{5} = \frac{360}{19} = 18\frac{18}{19}\text{ km/h}$$
$$\text{Time taken by A} = 200 - 10 = 190\text{ seconds} = 3\text{ mins } 10\text{ s}$$
$$\text{Speed of A} = \frac{1000\text{ m}}{190\text{ s}} = \frac{100}{19}\text{ m/s} = \frac{100}{19} \times \frac{18}{5} = \frac{360}{19} = 18\frac{18}{19}\text{ km/h}$$
Final Answer: Time for A: $190\text{ seconds}$; Speed of A: $18\frac{18}{19}\text{ km/h}$
Question 10
10 Marks
A motorboat goes downstream in a river and covers the distance between two towns in $6\text{ hours}$. Returning upstream, it covers the same distance in $8\text{ hours}$. If the speed of the stream is $3\text{ km/h}$, find the speed of the boat in still water and the distance between the two towns.
Model Descriptive Solution:
Step 1: Declare Variable & Relative Speeds
3 Marks
Let still water speed be $u\text{ km/h}$. Stream speed $= 3\text{ km/h}$.
Downstream speed $= (u + 3)\text{ km/h}$. Upstream speed $= (u - 3)\text{ km/h}$.
Downstream speed $= (u + 3)\text{ km/h}$. Upstream speed $= (u - 3)\text{ km/h}$.
Step 2: Equate Distance Expressions
4 Marks
$$\text{Distance} = 6(u + 3) = 8(u - 3)$$
$$6u + 18 = 8u - 24 \implies 8u - 6u = 18 + 24 \implies 2u = 42 \implies u = 21\text{ km/h}$$
$$6u + 18 = 8u - 24 \implies 8u - 6u = 18 + 24 \implies 2u = 42 \implies u = 21\text{ km/h}$$
Step 3: Compute Distance
3 Marks
$$\text{Distance} = 6(21 + 3) = 6 \times 24 = 144\text{ km}$$
Check upstream: $8(21 - 3) = 8 \times 18 = 144\text{ km}$.
Check upstream: $8(21 - 3) = 8 \times 18 = 144\text{ km}$.
Final Answer: Speed in still water $= 21\text{ km/h}$; Distance $= 144\text{ km}$
High-Yield FAQs & Revision Notes
Why do we multiply by 5/18 to convert km/h to m/s?
Because $1\text{ km} = 1000\text{ m}$ and $1\text{ hour} = 3600\text{ s}$. The ratio $\frac{1000}{3600}$ reduces to $\frac{5}{18}$.
What is the distance covered when a train crosses a platform?
The total distance equals the length of the train PLUS the length of the platform ($L_{\text{train}} + L_{\text{platform}}$).
What is the distance covered when a train crosses a pole?
Just the length of the train ($L_{\text{train}}$), since the width of a pole or standing person is negligible compared to the train's length.
Why do relative speeds add in opposite directions?
Because the gap between the two objects closes at the combined rate of both speeds ($v_1 + v_2$).
Why do relative speeds subtract in the same direction?
Because the faster object gains on the slower object only by the difference in their velocities ($v_1 - v_2$).
What is the average speed formula for equal distances?
The harmonic mean: $\frac{2uv}{u + v}$, where $u$ and $v$ are the two speeds.
How do you calculate stoppage time per hour?
$\text{Stoppage Time} = \frac{\text{Speed without stops} - \text{Speed with stops}}{\text{Speed without stops}} \times 60\text{ minutes}$.
What does 'A beats B by 10 metres' mean in a 100m race?
When A crosses the 100m finish line, B is 10 metres behind (at the 90m mark).
What does 'A gives B a start of 5 seconds' mean?
B starts running 5 seconds before A starts.
How does this module map to the official RIMC syllabus?
Directly covers Serial No. 22 ('Speed and Time') and Serial No. 24 ('Simple problems of time and distance') of Chapter 8 in the RIMC brochure.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 22, 24 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
| Topic / Concept | High-Yield Exam Rules & Memory Anchor |
|---|---|
| 5/18 Multiplier | $\text{km/h} \to \text{m/s}$: multiply by $\frac{5}{18}$. $\text{m/s} \to \text{km/h}$: multiply by $\frac{18}{5}$. |
| Train on Pole | $d = L_{\text{train}}$. Length of pole is zero. |
| Train on Platform | $d = L_{\text{train}} + L_{\text{platform}}$. Always add both lengths. |
| Relative Speed | Opposite: $v_1 + v_2$; Same direction: $|v_1 - v_2|$. |
| Harmonic Speed | Equal distances $\implies v_{\text{avg}} = \frac{2uv}{u+v}$. Never use arithmetic mean. |
| Early / Late Shift | $D = \frac{s_1 s_2}{s_2 - s_1} \times \left(\frac{t_{\text{late}} + t_{\text{early}}}{60}\right)$. |