Master motion mechanics, unit conversions ($1 km/h = 518 m/s$), inverse speed-time proportionality, harmonic average speed ($2aba+b$), railroad pole interval counting, and relative speed models across trains and streams.
Introductory Problem (CBSE Support Material)
Problem Statement:
An insect moves from point $A$ to point $C$ and returns to point $B$ taking a total time of $6 seconds$.
A • ————— 5 cm —————> B • —— 2 cm ——> C •
a) What is the total distance travelled by the insect?
Distance from $A$ to $C = 5 cm + 2 cm = 7 cm$.
Distance returning from $C$ to $B = 2 cm$.
$$Total Distance = AC + CB = 7 cm + 2 cm = \mathbf{9 cm}$$
b) Calculate the speed of the insect:
$$Speed = Total DistanceTotal Time = 9 cm6 s = \mathbf{1.5 cm/s}$$
Relation of Speed, Distance and Time
The fundamental definition connecting motion parameters is:
$$\mathbf{Speed = DistanceTime \iff Distance = Speed × Time \iff Time = DistanceSpeed}$$
Inverse Proportionality Principle:
Speed is inversely proportional to time when distance is kept constant ($Speed ∝ 1Time$). That is, when speed increases, time taken decreases, and vice versa.
Ratio of Speeds and Times for Two Vehicles:
Suppose speeds of two cars, A and B, are in the ratio $x : y$:
$$Speed of car ASpeed of car B = xy$$
For the same distance $d$:
$$Time taken by car A = dx, \quad Time taken by car B = dy$$
$$Time taken by car ATime taken by car B = d/xd/y = yx$$
$$\mathbf{Ratio of Time Taken (Car A : Car B) = y : x}$$
Key Insight: The ratio of times taken to cover a constant distance is strictly the reciprocal of their ratio of speeds!
Average Speed: Definition vs Misconceptions
A very common error is calculating average speed as the arithmetic mean of speeds $a+b2$. This is mathematically incorrect!
Universal Definition:
$$\mathbf{Average Speed = Total Distance TravelledTotal Time Taken}$$
Case of Equal Distances (Harmonic Mean):
If an object travels a distance $d$ at speed $a$, and returns over the same distance $d$ at speed $b$:
Total distance $= d + d = 2d$.
Total time $= da + db = d(a + bab
ight)$.
$$Average Speed = 2dd(a + bab
ight) = \mathbf{2aba + b}$$
The average speed for equal distance segments is always the Harmonic Mean of the two speeds.
Unit Conversions, Relative Speed & Trains
1. Unit Conversion Multipliers:
$$1 km/h = 1000 m3600 s = \mathbf{518 m/s}$$
$$1 m/s = 1/1000 km1/3600 h = \mathbf{185 km/h}$$
Same Direction (overtaking/chasing): Speeds subtract $\implies \mathbf{Srel = |S_1 - S_2|}$.
3. Train Crossing Equations:
Crossing a pole/standing person of negligible length: $Distance = \mathbf{Ltrain}$.
Crossing a platform/bridge/tunnel of length $L_p$: $Distance = \mathbf{Ltrain + L_p}$.
4. Boats and Streams:
Let speed of boat in still water $= u$, and speed of river stream $= v$.
Downstream speed (with current): $\mathbf{D = u + v}$.
Upstream speed (against current): $\mathbf{U = u - v}$.
Speed of boat in still water: $\mathbf{u = D + U2}$.
Speed of stream: $\mathbf{v = D - U2}$.
Check Your Progress 2.4
Official CBSE Applied Mathematics Support Material Questions with Full Step-by-Step Solutions
Question 1Speed Increase & Time Saved
If a car increases its speed from $40 km/hr$ to $60 km/hr$, how much time will it save on a $120-km$ journey?
Step-by-step Solution: 1. Given parameters:
Total distance of journey ($D$) $= 120 km$.
Initial speed ($S_1$) $= 40 km/hr$.
Increased speed ($S_2$) $= 60 km/hr$.
2. Compute time taken at initial speed:
$$T_1 = DistanceS_1 = 120 km40 km/hr = 3 hours$$
3. Compute time taken at increased speed:
$$T_2 = DistanceS_2 = 120 km60 km/hr = 2 hours$$
4. Calculate time saved:
$$Time Saved = T_1 - T_2 = 3 hours - 2 hours = \mathbf{1 hour} = \mathbf{60 minutes}$$
Final Answer: The car will save 1 hour (60 minutes).
Question 2Round Trip & Total Time
A person travels from one place to another at $30 km/hr$ and returns at $120 km/hr$. If the total time taken is $5 hours$, then find the Distance.
Step-by-step Solution: 1. Define variables:
Let the one-way distance between the two places be $D km$.
Speed going outward ($S_1$) $= 30 km/hr$.
Speed returning ($S_2$) $= 120 km/hr$.
2. Express time taken for each journey:
$$Time outward (T_1) = D30 hours, \quad Time return (T_2) = D120 hours$$
3. Total time equation:
$$Total Time = T_1 + T_2 = 5 hours$$
$$D30 + D120 = 5$$
Find common denominator (120):
$$4D + D120 = 5 \implies 5D120 = 5 \implies D24 = 5$$
$$D = 5 × 24 = \mathbf{120 km}$$
Verification: Time outward $= 120/30 = 4 hrs$. Time return $= 120/120 = 1 hr$. Total $= 4 + 1 = 5 hrs$. Perfect! Final Answer: The one-way distance is 120 km (Total round trip distance is $240 km$).
Question 3Fractional Speed in Fog
A train is running at $711$ of its own speed due to fog and reached a place in $44 hours$. What was the original time taken by the train if it runs at its own speed?
Step-by-step Solution: 1. Proportionality principle:
For a constant distance, speed and time are inversely proportional:
$$Speed × Time = Distance (constant)$$
Let original speed be $S$ and original time be $T$.
2. Relate reduced speed and time:
Reduced speed ($S'$) $= 711S$.
Time taken at reduced speed ($T'$) $= 44 hours$.
$$S × T = S' × T'$$
$$S × T = (711S
ight) × 44$$
Dividing both sides by $S$:
$$T = 711 × 44 = 7 × 4 = \mathbf{28 hours}$$
Additional Note: The time delayed by the fog was $44 - 28 = 16 hours$. Final Answer: The original time taken by the train is 28 hours.
Question 4 (Two Parts)Railroad Signal Poles
(a) The signal poles on a railroad are placed $100 m$ apart. How many poles will be passed by a train in $8 hours$ if the speed of the train is $45 km/h$?
(b) If $7201 poles$ are to be installed within two stations at equal distance covering a distance of $360 km$, find out the distance between two consecutive poles.
Step-by-step Solution:
Part (a):
1. Calculate total distance covered by the train in 8 hours:
$$Distance = Speed × Time = 45 km/h × 8 h = 360 km$$
Converting to metres:
$$360 km = 360 × 1000 m = 360,000 metres$$
2. Distance between adjacent poles $= 100 metres$.
Number of $100-metre$ intervals covered:
$$Number of Intervals = 360,000 m100 m = 3600 intervals$$
3. Since a pole is stationed at the initial position ($0 m$), the total number of poles passed is:
$$Total Poles = Number of Intervals + 1 = 3600 + 1 = \mathbf{3601 poles}$$
(Note: If only poles encountered after the starting point are counted, the answer is 3600 poles).
Part (b):
1. Total distance between the two stations $= 360 km = 360,000 metres$.
2. Total number of installed poles $= 7201 poles$.
3. Number of equal consecutive intervals created by $N$ poles $= N - 1$:
$$Number of Intervals = 7201 - 1 = 7200 intervals$$
4. Distance between two consecutive poles:
$$Distance = Total DistanceNumber of Intervals = 360,000 m7200 = 360072 = \mathbf{50 metres}$$
Final Answer: (a) 3601 poles (or 3600 poles excluding origin); (b) 50 metres between consecutive poles.
Targeted Practice Worksheet
Comprehensive multi-section drill problems structured according to CBSE board exam patterns.
Section A: Conversions & Speed-Time Inverses
WS.1: Late and Early Arrival Problem
Walking at $4 km/hr$, a student reaches school $10 minutes$ late. Next day, walking at $5 km/hr$, the student reaches $5 minutes$ early. Find the distance from home to school.
Solution:
Difference in time $= 10 mins (late) - (-5 mins (early)) = 15 minutes = 1560 = 14 hour$.
Let distance be $D km$.
$$D4 - D5 = 14 \implies 5D - 4D20 = 14 \implies D20 = 14 \implies D = 204 = \mathbf{5 km}$$
WS.2: Unit Conversion & Crossing
A car travels at a speed of $72 km/hr$. Convert this speed into $m/s$ and calculate how many metres it covers in $25 seconds$.
Solution:
$Speed in m/s = 72 × 518 = 4 × 5 = \mathbf{20 m/s}$.
$Distance in 25 s = 20 m/s × 25 s = \mathbf{500 metres}$.
Section B: Average Speed & Non-Uniform Trips
WS.3: Three Equal Distance Segments
An express bus travels three equal distances of a $300-km$ route at speeds of $20 km/hr$, $30 km/hr$, and $60 km/hr$. Find the average speed for the whole journey.
Solution:
Each segment $= 300 / 3 = 100 km$.
Time for 1st part $= 100 / 20 = 5 hrs$.
Time for 2nd part $= 100 / 30 = 313 hrs$.
Time for 3rd part $= 100 / 60 = 123 hrs$.
Total Time $= 5 + 313 + 123 = 5 + 5 = 10 hours$.
$$Average Speed = Total DistanceTotal Time = 300 km10 hrs = \mathbf{30 km/hr}$$
(Using 3-speed harmonic formula: $3120 + 130 + 160 = 3660 = 30 km/hr$).
Section C: Relative Speed & Train Crossings
WS.4: Train Crossing a Platform
A train $150 m$ long is running at $54 km/hr$. How much time will it take to pass a railway station platform of length $250 m$?
A motorboat can travel $36 km$ downstream in $3 hours$ and the same distance upstream in $6 hours$. Find the speed of the boat in still water and the speed of the river stream.
Solution:
Downstream speed ($D$) $= 36 / 3 = 12 km/hr$.
Upstream speed ($U$) $= 36 / 6 = 6 km/hr$.
Speed of boat in still water ($u$) $= D + U2 = 12 + 62 = \mathbf{9 km/hr}$.
Speed of stream ($v$) $= D - U2 = 12 - 62 = \mathbf{3 km/hr}$.
High-Yield Revision Sheet & Formula Matrix
Essential formulas, conversions, relative motion laws, and common board exam pitfalls:
Concept / Scenario
Formula
Operational Insight
Fundamental Motion
$$Speed = DistanceTime$$
Time and Speed are mutually inverse for fixed distance.
Applies only when equal distances are covered at speeds $a$ and $b$.
Opposite Relative Speed
$$Srel = S_1 + S_2$$
Two objects heading towards or moving away from each other.
Same Direction Relative Speed
$$Srel = |S_1 - S_2|$$
One object overtaking another in the same direction.
Railroad Poles Intervals
$$Distance = (N - 1) × d$$
$N$ consecutive poles create exactly $N - 1$ spatial intervals.
Boat in Still Water
$$u = D + U2$$
Half of sum of downstream and upstream speeds.
River Stream Current
$$v = D - U2$$
Half of difference between downstream and upstream speeds.
Examination Traps & Common Errors:
Unit Mismatch: Never divide $km$ by $seconds$ or $metres$ by $hours$ without converting first!
Arithmetic Mean Fallacy: If you travel to school at $20 km/h$ and return at $30 km/h$, your average speed is $2 × 20 × 3020 + 30 = 24 km/h$, NOT $25 km/h$!
Poles vs Intervals: For 100 poles, there are 99 intervals. Always check whether the question asks for poles passed or distance between stations.
CBSE Board-Level 10-Question Mock Test
Section 2.4 Speed, Distance & Time • Maximum Marks: 25 • Standard Time: 45 Minutes
CBSE Blueprint Aligned
Q1 • MCQ (1 Mark)
A car travels from City A to City B at $60 km/hr$ and returns from City B to City A at $40 km/hr$. The average speed of the car for the entire journey is:
A train $280 m$ long is running at $63 km/hr$. The time taken by the train to cross a person standing on the platform is:
A. 14 seconds
B. 15 seconds
C. 16 seconds
D. 18 seconds
Correct Answer: C (16 seconds)
Speed in m/s $= 63 × 518 = 3.5 × 5 = 17.5 m/s$.
$$Time = 28017.5 = 16 seconds$$
Q3 • MCQ (1 Mark)
A man can row $18 km/hr$ in still water. If the river current flows at $3 km/hr$, the ratio of time taken to row a certain distance upstream to that downstream is:
A. 7 : 5
B. 5 : 7
C. 4 : 3
D. 3 : 2
Correct Answer: A (7 : 5)
Downstream speed $D = 18 + 3 = 21 km/hr$.
Upstream speed $U = 18 - 3 = 15 km/hr$.
Ratio of time (Upstream : Downstream) $= DU = 2115 = 75 = 7 : 5$.
Q4 • Assertion & Reason (1 Mark)
Assertion (A): When two trains move in the same direction at speeds $u$ and $v$ ($u > v$), their relative speed is $u - v$. Reason (R): The relative speed of two objects moving in the opposite direction is the difference of their individual speeds.
A. Both (A) and (R) are true and (R) is the correct explanation of (A).
B. Both (A) and (R) are true but (R) is not the correct explanation of (A).
C. (A) is true but (R) is false.
D. (A) is false but (R) is true.
Correct Answer: C ((A) is true but (R) is false)
When two objects move in opposite directions, their relative speed is the sum $(u + v)$, not the difference. Therefore Reason (R) is false while Assertion (A) is true.
Q5 • Very Short Answer (2 Marks)
A cyclist covers a distance of $750 metres$ in $2 minutes 30 seconds$. What is the speed in $km/hr$ of the cyclist?
Click to View Model Solution
Time in seconds $= (2 × 60) + 30 = 150 seconds$.
$$Speed in m/s = 750 m150 s = 5 m/s$$
$$Speed in km/hr = 5 × 185 = \mathbf{18 km/hr}$$
Q6 • Very Short Answer (2 Marks)
Excluding stoppages, the speed of a bus is $54 km/hr$ and including stoppages, it is $45 km/hr$. For how many minutes does the bus stop per hour?
Click to View Model Solution
Distance lost per hour due to stoppages $= 54 - 45 = 9 km$.
Time taken to cover $9 km$ at non-stop speed ($54 km/h$):
$$Stoppage time = 954 hour = 16 hour = 16 × 60 mins = \mathbf{10 minutes per hour}$$
Q7 • Short Answer (3 Marks)
Two trains $140 m$ and $160 m$ long run at the speeds of $60 km/hr$ and $48 km/hr$ respectively in opposite directions on parallel tracks. Find the time taken by them to cross each other completely.
Click to View Model Solution
Total distance to be covered $= L_1 + L_2 = 140 + 160 = 300 metres$.
Relative speed in opposite directions $= 60 + 48 = 108 km/hr$.
Converting to m/s:
$$Srel = 108 × 518 = 6 × 5 = 30 m/s$$
$$Time taken = 300 m30 m/s = \mathbf{10 seconds}$$
Q8 • Short Answer (3 Marks)
A thief is spotted by a policeman from a distance of $200 metres$. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief is $10 km/hr$ and that of the policeman is $12 km/hr$, how far will the thief have run before he is overtaken?
Click to View Model Solution
Initial gap $= 200 m = 0.2 km$.
Relative speed (same direction) $= 12 - 10 = 2 km/hr$.
Time to overtake:
$$T = 0.2 km2 km/hr = 0.1 hour = 6 minutes$$
Distance run by the thief in $0.1 hour$:
$$Distance = 10 km/hr × 0.1 hr = 1 km = \mathbf{1000 metres}$$
(Distance run by policeman $= 12 × 0.1 = 1.2 km = 1200 m$).
Q9 • Long Answer (4 Marks)
A train leaves station P at $6:00 AM$ and reaches station Q at $10:00 AM$. Another train leaves station Q at $8:00 AM$ and reaches station P at $11:30 AM$. At what time do the two trains cross each other?
Click to View Model Solution
Time taken by 1st train ($P \to Q$) $= 10 - 6 = 4 hours$.
Time taken by 2nd train ($Q \to P$) $= 11:30 - 8:00 = 3.5 = 72 hours$.
Let distance between P and Q be $28 km$ (LCM of 4 and 7).
Speed of 1st train $= 28 / 4 = 7 km/hr$.
Speed of 2nd train $= 28 / (7/2) = 8 km/hr$.
By $8:00 AM$, the 1st train has run for 2 hours ($6 AM$ to $8 AM$):
Distance covered $= 7 × 2 = 14 km$.
Remaining distance between trains at $8:00 AM = 28 - 14 = 14 km$.
Both trains now move towards each other at relative speed $= 7 + 8 = 15 km/hr$.
Time taken to meet $= 1415 hour = 1415 × 60 mins = 56 minutes$.
Crossing time $= 8:00 AM + 56 minutes = \mathbf{8:56 AM}$.
Q10 • Case-Based Integrated Problem (4 Marks)
Case Study: High-Speed Rail Corridor Project
Indian Railways is testing a semi-high-speed train on a $450-km$ dedicated freight and passenger corridor between Station Alpha and Station Beta. During trial run 1, the train completes the outward journey at a planned uniform speed of $90 km/hr$. During the return journey (trial run 2), due to optimal weather and signaling priority, the train increases its speed to $150 km/hr$.
What was the total time taken for the round trip journey? (1 Mark)
Calculate the overall average speed for the round trip journey. (2 Marks)
By what percentage did the return trip travel time decrease compared to the outward trip? (1 Mark)
Click to View Model Solution
Solution:
1. Time outward $= 45090 = 5 hours$.
Time return $= 450150 = 3 hours$.
Total round-trip time $= 5 + 3 = \mathbf{8 hours}$.