Speed Time Distance FOR Placement

Meta Description: Master Speed Time Distance questions for placement with our 2026 guide. Learn formulas, tricks, and solve practice problems with solutions for TCS, Infosys, and more.

Introduction

Aptitude quant sections are the gateway to cracking campus drives, and mastering Speed Time Distance questions for placement is absolutely non-negotiable. Almost every major recruiter—from TCS, Infosys, and Wipro to Accenture and Cognizant—heavily tests this chapter because it evaluates logical reasoning, unit fluency, and real-time numerical agility under time pressure. The difficulty spectrum ranges from straightforward formula application to intricate multi-concept problems involving moving objects, fluid dynamics, and relative trajectories.

For 2026 placement cycles, companies are increasingly blending traditional concepts with twist-heavy scenarios. You’ll frequently encounter trains crossing platforms of varying lengths, boats navigating against river currents, runners on circular tracks, and vehicles with intermittent stoppages. Understanding core principles, internalizing relative motion logic, and practicing with a structured approach will help you secure high accuracy. This comprehensive guide covers every essential formula, provides graded practice problems, and shares exam-tested shortcuts to help you solve even the trickiest scenarios in under 40 seconds.

Key Formulas & Concepts

Speed, time, and distance form an interdependent trio. Memorize these foundational relationships and unit conversions to eliminate calculation errors:

• Core Relationship:
  • Speed (S) = Distance (D) / Time (T)
  • Time (T) = Distance (D) / Speed (S)
  • Distance (D) = Speed (S) × Time (T)
• Unit Conversion:
  • km/hr to m/s → Multiply by 5/18
  • m/s to km/hr → Multiply by 18/5
• Relative Speed (Two Moving Objects):
  • Same direction: Relative Speed = |S₁ − S₂|
  • Opposite direction: Relative Speed = |S₁ + S₂|
• Trains & Moving Objects:
  • Crossing a pole/standing person: Distance = Length of train (L)
  • Crossing platform/bridge/tunnel: Distance = Length of train + Length of platform
  • Crossing another moving train: Distance = L₁ + L₂ (opposite) or |L₁ − L₂| (same direction, if overtaking)
• Boats & Streams:
  • Downstream speed U = b + s
  • Upstream speed D = b − s
  • b = (U + D)/2 (Speed in still water)
  • s = (U − D)/2 (Speed of stream)
• Circular Tracks:
  • Time to meet for first time (same direction) = Circumference / |S₁ − S₂|
  • Time to meet for first time (opposite direction) = Circumference / (S₁ + S₂)
  • Number of meeting points = LCM of reduced speeds / Relative speed factor (practically, use |v₁ - v₂| and v₁ + v₂ ratios)
• Average Speed (Round Journey):
  • Avg Speed = 2xy / (x + y) when distances are equal.
  • General: Total Distance / Total Time

Solved Examples (Basic Level)

Q1. A cyclist covers 45 km in 3 hours. What is his speed in m/s? Solution: Speed = Distance/Time = 45 km / 3 hr = 15 km/hr. Convert to m/s: 15 × (5/18) = 75/18 = 4.16 m/s (≈ 4.17 m/s).

Q2. A car travels 120 m in 10 seconds. How far will it travel in 5 minutes at the same speed? Solution: Speed = 120/10 = 12 m/s. Time = 5 min = 300 seconds. Distance = 12 × 300 = 3600 m or 3.6 km.

Q3. Two runners move in the same direction at 8 m/s and 5 m/s. After how many seconds will they be 150 m apart? Solution: Relative speed (same direction) = 8 − 5 = 3 m/s. Time = Distance / Relative Speed = 150 / 3 = 50 seconds.

Q4. A boat goes 16 km downstream and returns in 4 hours. If the boat's speed in still water is 10 km/hr, find stream speed. Solution: Let stream speed = x km/hr. Downstream = 10 + x, Upstream = 10 − x. Total time: 16/(10+x) + 16/(10−x) = 4 Simplify: 16(10−x + 10+x) / (100 − x²) = 4 → 320 / (100 − x²) = 4 100 − x² = 80 → x² = 20 → x = √20 ≈ 4.47 km/hr. (Wait, this is medium difficulty. I'll adjust to a basic one.) Adjusted Q4: A train moving at 36 km/hr crosses a pole in 15 seconds. Find its length. Solution: 36 km/hr = 36 × (5/18) = 10 m/s. Length = Speed × Time = 10 × 15 = 150 m.

Q5. A person walks at 4 km/hr for 30 minutes. How much distance is covered? Solution: Time = 30/60 = 0.5 hours. Distance = 4 × 0.5 = 2 km.

Practice Questions (Medium Level)

Q1. A 150 m long train crosses a platform in 30 seconds while crossing a pole in 12 seconds. Find platform length. Solution: Speed from pole: 150 m / 12 s = 12.5 m/s. Distance covered while crossing platform = Speed × Time = 12.5 × 30 = 375 m. Platform length = 375 − 150 = 225 m.

Q2. A man rows 12 km upstream in 4 hours and the same distance downstream in 2 hours. Find speed in still water. Solution: Upstream speed = 12/4 = 3 km/hr. Downstream speed = 12/2 = 6 km/hr. Boat speed = (6 + 3)/2 = 4.5 km/hr.

Q3. Two cars start from opposite ends and travel toward each other at 60 km/hr and 80 km/hr. They meet after 1.5 hours. What was initial distance? Solution: Relative speed (opposite) = 60 + 80 = 140 km/hr. Distance = Relative Speed × Time = 140 × 1.5 = 210 km.

Q4. Runners A and B (speeds 5 m/s and 3 m/s) run on a 400 m circular track starting from same point. After how many seconds will they meet first if running in same direction? Solution: Relative speed = 5 − 3 = 2 m/s. Time to meet = Track length / Relative speed = 400 / 2 = 200 seconds.

Q5. A bus travels first 60 km at 30 km/hr and next 60 km at 60 km/hr. Find average speed. Solution: Using harmonic mean for equal distances: Avg = 2×30×60 / (30+60) = 3600/90 = 40 km/hr. Alternatively, Time1 = 2 hr, Time2 = 1 hr. Total = 120/3 = 40 km/hr.

Q6. A train 200 m long overtakes a motorcycle moving at 60 km/hr in same direction in 15 seconds. Find train’s speed. Solution: Convert motorcycle speed: 60 × (5/18) = 16.67 m/s. Let train speed = v m/s. Relative speed = v − 16.67. Distance to overtake = 200 m. So, (v − 16.67) = 200/15 = 13.33. v = 30 m/s = 30 × (18/5) = 108 km/hr.

Q7. A swimmer covers 6 km downstream and returns upstream in 4 hours. Still water speed = 4 km/hr. Find current speed. Solution: Let current = c km/hr. Time downstream = 6/(4+c), upstream = 6/(4−c). Sum = 4. 6/(4+c) + 6/(4−c) = 4 → 24 = 4(16 − c²) → 6 = 16 − c² → c² = 10 → c ≈ 3.16 km/hr. (Note: Simplified for exam context, often designed to yield integers. If c=2: 6/6 + 6/2 = 1+3=4. So current = 2 km/hr works perfectly. I'll use c=2 km/hr as intended design). Correct solution: Testing integer values or solving quadratic properly: 12(4) = 4(16−c²) → 48 = 64 − 4c² → 4c² = 16 → c² = 4 → c = 2 km/hr.

Q8. Three friends cycle around a 1200 m track at 10, 15, and 20 m/s. When will they all meet at starting point again? Solution: Time to complete one round: 1200/10=120s, 1200/15=80s, 1200/20=60s. Meet at start = LCM(120, 80, 60) = LCM(80,120) = 240 seconds (4 minutes).

Tricky Questions (Advanced Level)

Q1. A train passes two bridges of lengths 300 m and 200 m in 28 sec and 20 sec respectively. Find train speed and length. Solution: Let length = L, speed = v. (L + 300)/v = 28 → L + 300 = 28v ...(i) (L + 200)/v = 20 → L + 200 = 20v ...(ii) Subtract (ii) from (i): 100 = 8v → v = 12.5 m/s. Substitute v in (ii): L + 200 = 250 → L = 50 m. Speed = 12.5 m/s = 45 km/hr.

Q2. Points A and B are 800 km apart. Train X leaves A at 70 km/hr at 9 AM. Train Y leaves B at 90 km/hr at 11 AM toward A. When and where do they meet? Solution: By 11 AM, Train X covers 2 × 70 = 140 km. Remaining distance = 660 km. Now moving towards each other: Relative speed = 70 + 90 = 160 km/hr. Meeting time after 11 AM = 660/160 = 4.125 hours = 4 hr 7.5 min. Meeting time = 11:00 AM + 4h07m = 3:07 PM approx. Distance from A = 140 + (70 × 4.125) = 140 + 288.75 = 428.75 km.

Q3. Two runners on 400 m track start from same point opposite directions at 6 m/s and 2 m/s. How many times do they meet in first 2400 seconds? Solution: Relative speed = 6 + 2 = 8 m/s. First meeting = 400/8 = 50 sec. They meet every 50 seconds thereafter. Number of meetings = Total time / Interval = 2400 / 50 = 48 times.

Q4. A car travels a distance with speed v. Returning, it increases speed by 20% and takes 15 min less. Find original time if distance is constant. Solution: Original speed = v, New = 1.2v = 6v/5. Speed ratio = 5:6. Time ratio (inversely) = 6:5. Difference in time = 1 part = 15 min. Original time = 6 parts = 6 × 15 = 90 minutes = 1.5 hours.

Q5. A boat takes 8 hrs downstream and 12 hrs upstream for a round trip. Stream = 3 km/hr. Find one-way distance. Solution: Let still water speed = b. Distance = D. D/(b+3) = 8, D/(b−3) = 12. Equating D: 8b + 24 = 12b − 36 → 4b = 60 → b = 15 km/hr. Distance D = 8 × (15+3) = 8 × 18 = 144 km.

Common Mistakes to Avoid

• Ignoring Unit Consistency: Mixing km/hr with meters/seconds without converting leads to completely wrong answers. Always check units before calculating.
• Forgetting Object Lengths: In train problems, students often only use platform/train length. Remember: Train crossing platform = L_train + L_platform.
• Relative Speed Direction Errors: Adding speeds when moving in the same direction or subtracting when opposite is a classic trap. Memorize: Same direction → Subtract, Opposite → Add.
• Misapplying Average Speed Formula: Using (S₁ + S₂)/2 when distances are equal is wrong. Must use 2S₁S₂/(S₁+S₂) or stick to Total Distance/Total Time.
• Overlooking Stoppage Time: Many questions hide actual moving time. Subtract stoppage minutes from total elapsed time before applying D = S × T.
• Circular Track Meeting Point Confusion: Assuming meeting time = Track/V1 or using wrong LCM. Always use relative speed logic for first meeting, and LCM of individual round times for meeting at start.

Shortcut Tricks

1. Average Speed for Equal Distances: If an object travels same distance at speed x and y, average speed is 2xy/(x+y). Example: 40 km/hr & 60 km/hr → Avg = 4800/100 = 48 km/hr (instantly).
2. Stoppages Formula: If stoppage rate is t min/hr, and moving speed S, then t = 60 × (1 − V_avg / S). Reverse: V_avg = S × (60 − t)/60.
3. Train Platform Shortcut: If train crosses pole in t₁ and platform in t₂, then Platform Length = Train Length × (t₂ − t₁)/t₁. Example: Train 200m, t₁=10s, t₂=25s → Lp = 200 × 15/10 = 300 m.
4. Upstream/Downstream Quick Split: If downstream time T₁ and upstream time T₂ for same distance, speed of boat = Stream × (T₁ + T₂)/(T₂ − T₁). Works without calculating distance.
5. Circular Track Meeting at Start: Time to meet at starting point together = LCM of (C/S₁, C/S₂, C/S₃). Simplify fractions first to avoid huge numbers.

Previous Year Questions from Top Companies

Q1. (TCS NQT Style) A car travels 1/4th of distance at 30 km/hr, 1/3rd at 40 km/hr, and remaining at 45 km/hr. If total time is 5 hours, find total distance. Solution: Let distance = 12D (LCM of 4,3,1). Time1 = 3D/30 = D/10, Time2 = 4D/40 = D/10, Time3 = 5D/45 = D/9. Total = D(1/10 + 1/10 + 1/9) = D(9+9+10)/90 = 28D/90 = 5 → D = 450/28. Total distance = 12D = 12 × 450/28 = 192.8