Time Distance

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Time and Distance problems test your ability to calculate speed, time, or distance when two or more of these quantities are related. These problems frequently involve trains, cars, pedestrians, and circular tracks. The fundamental relationship is:

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$ $$\text{Distance} = \text{Speed} \times \text{Time}$$ $$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

Key Units Conversion:

Always convert to consistent units:

• km/hr to m/s: multiply by $\frac{5}{18}$
• m/s to km/hr: multiply by $\frac{18}{5}$

Average Speed:

When a journey is divided into parts with different speeds: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$

For two equal distances at speeds $v_1$ and $v_2$: $$\text{Average Speed} = \frac{2v_1v_2}{v_1 + v_2}$$

For two equal times at speeds $v_1$ and $v_2$: $$\text{Average Speed} = \frac{v_1 + v_2}{2}$$

⚡ Exam Tip: The average speed formula $\frac{2v_1v_2}{v_1+v_2}$ applies ONLY when equal distances are covered at different speeds. If equal time is spent at each speed, average = $\frac{v_1+v_2}{2}$. Many students incorrectly apply the harmonic mean in both situations.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding and problem-solving practice.

Relative Speed:

When two objects move in opposite directions: $$\text{Relative Speed} = v_1 + v_2$$

When two objects move in same direction: $$\text{Relative Speed} = |v_1 - v_2|$$

Train Problems:

Length of train passing:

• A stationary person: Time = $\frac{\text{Train Length}}{\text{Train Speed}}$
• A platform/pole of length L: Time = $\frac{\text{Train Length} + L}{\text{Train Speed}}$

Two trains moving in opposite directions: Time to cross each other = $\frac{\text{Length}_1 + \text{Length}_2}{v_1 + v_2}$

Two trains moving in same direction: Time to cross = $\frac{\text{Length}_1 + \text{Length}_2}{|v_1 - v_2|}$

Meeting Point Problems:

When two persons start from A and B (towards each other) and meet at point M:

• Time taken by A to reach M = Time taken by B to reach M
• Distance covered by A : Distance covered by B = Speed of A : Speed of B

Example: A and B start simultaneously from two points 200 km apart, towards each other. Speeds are 40 km/hr and 60 km/hr. When and where do they meet?

• Relative speed = 100 km/hr
• Time to meet = 200/100 = 2 hours
• Distance from A = 40 × 2 = 80 km from A (120 km from B)

Upstream and Downstream:

If boat’s speed in still water = b km/hr and stream’s speed = s km/hr: $$\text{Downstream speed} = b + s$$ $$\text{Upstream speed} = b - s$$ $$\text{Speed in still water} = \frac{\text{Downstream} + \text{Upstream}}{2}$$ $$\text{Stream speed} = \frac{\text{Downstream} - \text{Upstream}}{2}$$

Circular Track Problems:

When two runners start from same point on circular track:

• Same direction: They meet every $\frac{L}{|v_1 - v_2|}$ seconds where L = track length
• Opposite directions: They meet every $\frac{L}{v_1 + v_2}$ seconds

When runners start from opposite ends of diameter:

• They meet at one point (diametrically opposite) after covering half the circumference

⚡ SSC CGL-Specific Tip: In circular track problems, track length is crucial. If two runners are running at v₁ and v₂ m/s on a track of L metres, running in the same direction, the faster runner laps the slower one every $\frac{L}{v_1 - v_2}$ seconds.

Common Student Mistakes:

• Forgetting to add platform length in train problems
• Mixing up average speed formulas for equal distance vs equal time cases
• Not converting units (km/hr to m/s or vice versa)

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Round Trip Problems:

When a person travels from A to B at speed $v_1$ and returns from B to A at speed $v_2$:

• Total distance = 2d (where d is one-way distance)
• Total time = $t_1 + t_2 = \frac{d}{v_1} + \frac{d}{v_2} = d\left(\frac{v_1 + v_2}{v_1 v_2}\right)$
• Average speed for round trip = $\frac{2d}{d\left(\frac{v_1 + v_2}{v_1 v_2}\right)} = \frac{2v_1 v_2}{v_1 + v_2}$ (harmonic mean)

Pursuit Problems:

When a faster person starts later and pursues a slower person: Example: A starts at 9 AM with speed 40 km/hr. B starts at 11 AM with speed 60 km/hr chasing A. When does B catch A?

• Head start distance = 40 × 2 = 80 km (distance A covers before B starts)
• Relative speed = 60 - 40 = 20 km/hr
• Time to catch = 80/20 = 4 hours after B starts = 3 PM

Boat and Stream — Extension:

A boat goes downstream from A to B and returns to A. If still water speed = b and stream speed = s: $$\text{Time downstream} = \frac{d}{b+s}$$ $$\text{Time upstream} = \frac{d}{b-s}$$ $$\text{Total time} = d\left(\frac{1}{b+s} + \frac{1}{b-s}\right) = \frac{2bd}{b^2 - s^2}$$

If the boat’s speed in still water is given as v and stream flows at s km/hr, upstream effective speed is v - s (positive only if v > s; otherwise upstream journey impossible).

Speed Ratio Problems:

When two persons A and B cover the same distance but A takes $t_1$ hours and B takes $t_2$ hours:

• Speed ratio $v_A : v_B = t_2 : t_1$ (inverse ratio of time)

When they start from opposite ends and meet somewhere on the path:

• Time to meeting is same for both
• Distance covered proportional to speeds
• If distance between A and B is D, and they start simultaneously: $$d_A = \frac{v_A}{v_A + v_B} \times D, \quad d_B = \frac{v_B}{v_A + v_B} \times D$$

Problems with Rest Periods:

When a person covers distance at speed $v$ with rest periods of $r$ hours after every $t$ hours of travel: Total distance D covered in time T means: actual travel time = T - (number of rest periods) × r But number of rest periods depends on how many complete travel intervals were made.

Acceleration Problems:

If a train accelerates from rest at rate $a$ m/s² and reaches speed $v$ m/s:

• Distance covered while accelerating: $s = \frac{v^2}{2a}$
• Time to reach speed $v$: $t = \frac{v}{a}$

SSC CGL Tier II Advanced:

Questions in Tier II may involve:

• Problems with changing speeds (speed changes at different points of journey)
• Escalator problems (person walking on moving escalator)
• Wind speed affecting aircraft speed (air speed + wind speed = ground speed)

Sample SSC CGL Questions:

Q1. A train 150m long passes a man running at 9 km/hr in the same direction in 10 seconds. Find train speed. Solution: Relative speed = train speed - 9 km/hr (same direction) Time = 10 sec, Distance = 150m = 0.15 km $10 \text{ sec} = \frac{10}{3600} \text{ hr} = \frac{1}{360} \text{ hr}$ Relative speed = $\frac{0.15}{1/360} = 54$ km/hr Train speed = 54 + 9 = 63 km/hr

Q2. Two trains 200m and 150m long run at 45 km/hr and 30 km/hr respectively. Time to completely pass each other (opposite directions). Solution: Relative speed = 45 + 30 = 75 km/hr = $\frac{75 \times 1000}{3600}$ = 20.83 m/s Total length = 200 + 150 = 350m Time = 350/20.83 = 16.8 seconds

⚡ Advanced Tip: For escalator problems, the effective speed of a person walking on a moving escalator is $v_p + v_e$ (same direction) or $v_p - v_e$ (opposite direction). The number of steps on an escalator is fixed. If a person takes N steps to walk up and the escalator takes M steps during this time, and total steps on escalator is S: $S = N + M$ when person walks in same direction as escalator.

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