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Time Speed & Distance

Lines Angles & Triangles17 / 51Ratio Proportion & Variation
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Concepts (3)

Boats & Streams problems involve calculating relative speeds: downstream (boat + stream) and upstream (boat - stream) to find time, distance, or individual speeds. Master these for quick solutions.

Core Formulas

Boats & Streams problems are an application of Time, Speed & Distance, focusing on relative speeds. The key is understanding how the speed of the stream affects the boat's speed.

  • Speed of Boat in Still Water (B): The boat's actual speed without any current.
  • Speed of Stream/Current (S): The speed of the water flow.
  1. Downstream Speed (Ds): When the boat moves with the current. Ds = B + S
  2. Upstream Speed (Us): When the boat moves against the current. Us = B - S (Note: For upstream travel to be possible, B must be greater than S)
  3. Speed of Boat in Still Water (B): If downstream and upstream speeds are known. B = (Ds + Us) / 2
  4. Speed of Stream (S): If downstream and upstream speeds are known. S = (Ds - Us) / 2
  5. Time (T): T = Distance (D) / Speed

Worked Example 1

Q: A boat's speed in still water is 15 km/h and stream speed is 3 km/h. What is the boat's downstream speed and upstream speed? A:

  • Downstream Speed (Ds) = B + S = 15 km/h + 3 km/h = 18 km/h
  • Upstream Speed (Us) = B - S = 15 km/h - 3 km/h = 12 km/h

Worked Example 2

Q: A boat travels 30 km downstream in 2 hours and 18 km upstream in 3 hours. Find the speed of the boat in still water and the speed of the stream. A:

  1. Calculate Downstream Speed (Ds): Ds = Distance / Time = 30 km / 2 hours = 15 km/h
  2. Calculate Upstream Speed (Us): Us = Distance / Time = 18 km / 3 hours = 6 km/h
  3. Calculate Speed of Boat in Still Water (B): B = (Ds + Us) / 2 = (15 + 6) / 2 = 21 / 2 = 10.5 km/h
  4. Calculate Speed of Stream (S): S = (Ds - Us) / 2 = (15 - 6) / 2 = 9 / 2 = 4.5 km/h

Shortcuts & Tricks

  • Direct Calculation: Always remember B = (Ds+Us)/2 and S = (Ds-Us)/2. These are derived from the basic equations and save time by avoiding solving simultaneous equations.
  • Same Distance Round Trip: If a boat travels a certain distance upstream and returns the same distance downstream, and the total time is given, you can use the concept of average speed for the round trip: Average Speed = (2 * Ds * Us) / (Ds + Us). Then, Total Distance = Average Speed * Total Time.
  • Ratio Method: If the ratio of upstream time to downstream time for the same distance is given, say T_us : T_ds = x : y, then Us : Ds = y : x. From this, you can find B and S ratios.

Common Mistakes

  1. Confusing Upstream and Downstream: Students often mix up B+S and B-S. Always remember 'downstream' means 'with the flow' (adds speed), 'upstream' means 'against the flow' (subtracts speed).
  2. Incorrectly Applying B-S vs S-B: The boat's speed (B) must always be greater than the stream's speed (S) for upstream travel to be possible. If S > B, the boat cannot move upstream.
  3. Unit Conversion Errors: Ensure all speeds, distances, and times are in consistent units (e.g., km/h, meters/second). Convert if necessary.
  4. Algebraic Errors: When solving equations involving time and distance, especially in round trip problems, be careful with algebraic manipulations.

Derivation (brief)

Let's quickly see where the core formulas come from:

  1. Downstream Speed (Ds): When a boat moves with the current, its effective speed is the sum of its own speed and the stream's speed. Imagine walking on a moving walkway – your speed adds to the walkway's speed. Hence, Ds = B + S.
  2. Upstream Speed (Us): When a boat moves against the current, the stream's speed reduces the boat's effective speed. Imagine walking against a moving walkway – your speed is reduced. Hence, Us = B - S.
  3. Speed of Boat (B) and Stream (S) from Ds & Us: We have two equations: (i) Ds = B + S (ii) Us = B - S Adding (i) and (ii): Ds + Us = (B + S) + (B - S) = 2B. Therefore, B = (Ds + Us) / 2. Subtracting (ii) from (i): Ds - Us = (B + S) - (B - S) = 2S. Therefore, S = (Ds - Us) / 2.

Advanced Examples

Q: A boat takes 6 hours to travel a certain distance upstream and 4 hours to travel the same distance downstream. If the speed of the stream is 3 km/h, find the speed of the boat in still water. A: Let the distance be D km. Let the speed of the boat in still water be B km/h.

  • Upstream Speed (Us) = B - S = B - 3 km/h
  • Downstream Speed (Ds) = B + S = B + 3 km/h

Given, Time_upstream = 6 hours and Time_downstream = 4 hours. Since the distance is the same: D = Us * Time_upstream => D = (B - 3) * 6 D = Ds * Time_downstream => D = (B + 3) * 4

Equating the distances: (B - 3) * 6 = (B + 3) * 4 6B - 18 = 4B + 12 6B - 4B = 12 + 18 2B = 30 B = 15 km/h So, the speed of the boat in still water is 15 km/h.

Variation Types

  1. Round Trip Problems: A boat travels to a point and returns to the starting point. Total time or total distance is given. These often involve setting up equations for T_total = D/Ds + D/Us.
  2. Fixed Time, Different Distances: A boat travels different distances upstream and downstream in the same amount of time. This implies D_us / Us = D_ds / Ds.
  3. Fixed Distance, Different Times: (As seen in the advanced example above) A boat covers the same distance upstream and downstream but takes different times. This implies Us * T_us = Ds * T_ds.
  4. Ratio Problems: The ratio of speeds (B:S, Ds:Us) or times is given, and you need to find specific speeds or distances.

Time-Saving Methods

  • Option Elimination: For complex problems, if options are given, try substituting them into the equations. This can be faster than solving a quadratic equation.
  • Mental Math for Basic Conversions: Quickly calculate Ds and Us from B and S, or vice-versa, without writing down every step.
  • Focus on Ratios: Many problems can be simplified by working with ratios of speeds or times rather than absolute values, especially when distances are equal.
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Start Lesson

Master train problems by understanding relative speed and total distance (train length + object length). Convert units km/hr to m/s using 5/18 for quick solutions.

Core Formula

Train problems are a direct application of the fundamental formula: Distance = Speed × Time. The key is correctly identifying the 'Distance' and 'Speed' in various scenarios.

  1. Unit Conversion: Speeds are often given in km/hr and lengths in meters. It's critical to convert units to be consistent, usually to meters/second (m/s).

    • km/hr to m/s: Multiply by 5/18
    • m/s to km/hr: Multiply by 18/5

  2. Total Distance: This is the total length a train must cover.

    • Crossing a point object (pole, standing man, tree): The distance covered is simply the length of the train (L_train).
    • Crossing a linear object (platform, bridge, tunnel, another train): The distance covered is the length of the train (L_train) + length of the object (L_object).

  3. Relative Speed: When two moving objects are involved (e.g., two trains, train and a moving man).

    • Moving in the same direction: Relative Speed = |Speed1 - Speed2|
    • Moving in opposite directions: Relative Speed = Speed1 + Speed2

The core formula for time taken to cross is: Time = Total Distance / Relative Speed

Worked Example 1

A train 180 meters long is running at a speed of 72 km/hr. How much time will it take to cross an electric pole?

Step-by-step Solution:

  1. Identify Distance: The train is crossing a pole (point object), so the distance is the length of the train = 180 meters.
  2. Convert Speed: Speed = 72 km/hr. Convert to m/s: 72 * (5/18) = 4 * 5 = 20 m/s.
  3. Calculate Time: Time = Distance / Speed = 180 meters / 20 m/s = 9 seconds.

Worked Example 2

A train 200 meters long crosses a platform 300 meters long in 25 seconds. What is the speed of the train in km/hr?

Step-by-step Solution:

  1. Identify Total Distance: The train is crossing a platform, so total distance = Length of train + Length of platform = 200m + 300m = 500 meters.
  2. Calculate Speed (m/s): Speed = Distance / Time = 500 meters / 25 seconds = 20 m/s.
  3. Convert Speed (km/hr): Convert 20 m/s to km/hr: 20 * (18/5) = 4 * 18 = 72 km/hr.

Shortcuts & Tricks

  • Memorize Conversions: Quickly convert common speeds: 18 km/hr = 5 m/s, 36 km/hr = 10 m/s, 54 km/hr = 15 m/s, 72 km/hr = 20 m/s, 90 km/hr = 25 m/s. This saves calculation time for 5/18 and 18/5.
  • Ratio Method: If distance is constant, Speed is inversely proportional to Time. If S1/S2 = a/b, then T1/T2 = b/a. Useful for comparing scenarios.
  • Mental Math for Relative Speed: For same direction, quickly subtract; for opposite, quickly add. Then apply 5/18 once.

Common Mistakes

  1. Forgetting Unit Conversion: The most frequent error. Always ensure distance is in meters and speed in m/s (or vice-versa) before calculating time.
  2. Incorrect Total Distance: Not adding the length of the platform/bridge/another train when required, or adding it when only a point object is crossed.
  3. Confusing Relative Speed Directions: Using S1+S2 for same direction or |S1-S2| for opposite direction. Double-check the scenario.

Derivation (brief)

The concept of 'distance covered' for a train is crucial. When a train crosses a point object (like a pole), the train effectively covers its own length from the moment its front end reaches the pole until its rear end leaves the pole. When crossing a linear object (like a platform), the train's front end enters the platform, travels the platform's length, and then travels its own length until its rear end clears the platform. Hence, L_train + L_object.

Relative speed arises when two objects are in motion. If they move towards each other (opposite directions), they are closing the distance between them faster, so their speeds add up. If they move in the same direction, the faster object 'gains' on the slower one by the difference in their speeds.

Advanced Examples

Scenario: Two trains, Train A (120m long, 60 km/hr) and Train B (180m long, 48 km/hr).

1. Time to cross each other if moving in opposite directions:

  • Total Distance = Length A + Length B = 120m + 180m = 300m.
  • Relative Speed (opposite) = Speed A + Speed B = 60 km/hr + 48 km/hr = 108 km/hr.
  • Convert Relative Speed: 108 * (5/18) = 6 * 5 = 30 m/s.
  • Time = Total Distance / Relative Speed = 300m / 30 m/s = 10 seconds.

2. Time for Train A to completely overtake Train B if moving in the same direction:

  • Total Distance = Length A + Length B = 120m + 180m = 300m.
  • Relative Speed (same) = |Speed A - Speed B| = |60 km/hr - 48 km/hr| = 12 km/hr.
  • Convert Relative Speed: 12 * (5/18) = 2 * (5/3) = 10/3 m/s.
  • Time = Total Distance / Relative Speed = 300m / (10/3 m/s) = 300 * (3/10) = 90 seconds.

Variation Types

  • Train passing a man sitting in a moving train: Here, the distance is the length of the first train. The relative speed is calculated between the first train and the second train (or the speed of the man if he's moving relative to the second train, though usually it's the speed of the second train itself). Be careful with which speeds to use for relative speed.
  • Meeting Point Problems: Two trains start from different points and move towards each other. The time until they meet can be found using their relative speed (sum of speeds) and the initial distance between them.
  • Finding Unknown Length/Speed: Often, time and other parameters are given, and you need to find a train's length or speed by rearranging the formula.

Time-Saving Methods

  • Option Elimination: Sometimes, units or magnitudes in options can help eliminate incorrect choices before detailed calculation.
  • Approximation: For complex numbers, sometimes rounding can give a close enough answer to pick from options, especially if options are far apart.
  • Practice Common Values: Knowing 5/18 and 18/5 conversions for multiples of 18 km/hr (e.g., 18, 36, 54, 72, 90, 108 km/hr) and 5 m/s (e.g., 5, 10, 15, 20, 25, 30 m/s) by heart is a massive time saver.
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Start Lesson

Master Speed, Distance, Time (SDT) fundamentals: S=D/T, unit conversions (km/hr to m/s), and average speed. These are critical for quick problem-solving in SSC CGL.

Core Formula

Speed, Distance, and Time are fundamentally related by the formula:

Speed = Distance / Time

This can be rearranged to find any variable:

  • Distance = Speed × Time
  • Time = Distance / Speed

Units must be consistent. Common units are km/hr, m/s, km, m, hr, min, s.

Worked Example 1

A car travels a distance of 300 km in 5 hours. What is its average speed?

Solution:

  1. Identify given values: Distance (D) = 300 km, Time (T) = 5 hours.
  2. Apply the formula: Speed (S) = D / T
  3. S = 300 km / 5 hours
  4. S = 60 km/hr

Worked Example 2

A train is moving at a speed of 90 km/hr. How much distance will it cover in 10 seconds?

Solution:

  1. Identify given values: Speed (S) = 90 km/hr, Time (T) = 10 seconds.
  2. Crucial Step: Unit Conversion. The speed is in km/hr and time is in seconds. Convert speed to m/s for consistency.
    • To convert km/hr to m/s, multiply by 5/18.
    • S = 90 × (5/18) m/s = 5 × 5 m/s = 25 m/s.
  3. Apply the formula: Distance (D) = Speed × Time
  4. D = 25 m/s × 10 s
  5. D = 250 meters

Shortcuts & Tricks

  • Unit Conversion:
    • To convert km/hr to m/s: Multiply by 5/18.
    • To convert m/s to km/hr: Multiply by 18/5.
  • Constant Distance: If distance is constant, Speed is inversely proportional to Time (S₁/S₂ = T₂/T₁). If you increase speed, time taken decreases proportionally.
  • Average Speed (for two equal distances): If a person travels a certain distance at speed 'a' and returns the same distance at speed 'b', the average speed for the entire journey is (2ab) / (a+b).

Common Mistakes

  1. Unit Inconsistency: Failing to convert units (e.g., using km/hr with seconds) leads to incorrect answers. Always ensure all units are consistent (e.g., km, hr, km/hr OR m, s, m/s).
  2. Incorrect Average Speed: For a journey with two different speeds, students often calculate average speed as (S₁ + S₂) / 2. This is incorrect. Use the formula (2ab)/(a+b) for equal distances or Total Distance / Total Time for general cases.
  3. Confusing Direct/Inverse Proportionality: Misapplying the relationship between S, D, T when one is constant. For example, if time is constant, Distance is directly proportional to Speed (D ∝ S), but if distance is constant, Speed is inversely proportional to Time (S ∝ 1/T).

Derivation (brief)

The fundamental concept of speed arises from observing how quickly an object covers a certain distance. If an object covers a distance 'D' in time 'T', its speed 'S' is simply the rate at which it covers that distance. Hence, Speed = Distance / Time. This is a definitional relationship, not derived from more complex principles, but rather the basis for understanding motion.

Advanced Examples

1. Relative Speed (Opposite Direction): Two trains, 120 m and 180 m long, are moving in opposite directions at speeds of 54 km/hr and 36 km/hr respectively. How long will they take to cross each other?

Solution:

  1. Total Distance (D): The distance to be covered for them to cross each other is the sum of their lengths. D = 120 m + 180 m = 300 m.
  2. Relative Speed (S_rel): When moving in opposite directions, their speeds add up. S_rel = 54 km/hr + 36 km/hr = 90 km/hr.
  3. Unit Conversion: Convert S_rel to m/s: 90 × (5/18) = 25 m/s.
  4. Time (T): T = D / S_rel = 300 m / 25 m/s = 12 seconds.

2. Relative Speed (Same Direction): A thief is spotted by a policeman from a distance of 200 m. The thief starts running at 10 km/hr and the policeman chases him at 12 km/hr. How far will the thief have run before he is caught?

Solution:

  1. Initial Distance: 200 m (this is the distance the policeman needs to cover to catch the thief).
  2. Relative Speed (S_rel): When moving in the same direction, their speeds subtract. S_rel = 12 km/hr - 10 km/hr = 2 km/hr.
  3. Unit Conversion: Convert S_rel to m/s: 2 × (5/18) = 10/18 = 5/9 m/s.
  4. Time to catch (T): T = Initial Distance / S_rel = 200 m / (5/9 m/s) = 200 × (9/5) = 40 × 9 = 360 seconds.
  5. Distance run by thief: In 360 seconds, the thief runs at 10 km/hr.
    • Convert thief's speed to m/s: 10 × (5/18) = 50/18 = 25/9 m/s.
    • Distance = Speed × Time = (25/9 m/s) × 360 s = 25 × 40 = 1000 meters or 1 km.

Variation Types

  • Problems with varying speed/time for fixed distance: These often involve ratios. If Speed increases by X%, Time decreases by (X/(100+X))%. E.g., if speed increases by 25% (1/4), time decreases by (1/5).
  • Problems with varying distance/time for fixed speed: These are direct proportionality problems. If speed is constant, Distance is directly proportional to Time (D₁/T₁ = D₂/T₂).
  • Meeting Point Problems: These use relative speed to determine when and where two objects moving towards or away from each other will meet.

Time-Saving Methods

  • Ratio Method: For problems where distance is constant, if the ratio of speeds is S₁:S₂, then the ratio of times taken will be T₁:T₂ = S₂:S₁. This avoids complex algebraic equations.
  • Using Options: For certain problems, especially those involving finding a specific speed or time, plugging in the options can be faster than solving the full equation. Always check units before plugging in.
  • Memorize Common Conversions: Knowing that 18 km/hr = 5 m/s, 36 km/hr = 10 m/s, 54 km/hr = 15 m/s, 72 km/hr = 20 m/s, 90 km/hr = 25 m/s can save precious seconds.
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