Time, Speed and Distance: Basics, Late-Early, Relative Speed and Stoppages
Time, Speed and Distance: Basics, Late-Early, Relative Speed and Stoppages
Every question in this chapter is one formula, Distance = Speed × Time, used with the right units and the right relative speed. NTPC asks unit conversion, late-and-early problems, fractions of the usual speed, two people meeting or chasing, stoppage time and journeys in parts.
1. Formula Box
Distance = Speed × Time 1 km/h = 5/18 m/s 1 m/s = 18/5 km/h
Same distance: speed and time are inversely proportional (speed ratio a : b means time ratio b : a)
Relative speed: opposite directions = u + v; same direction = u − v
| km/h | 18 | 36 | 54 | 72 | 90 | 108 |
|---|---|---|---|---|---|---|
| m/s | 5 | 10 | 15 | 20 | 25 | 30 |
2. Shortcut Rules
| Situation | Rule |
|---|---|
| Late by t₁ at speed u, early by t₂ at speed v | Distance = uv × (t₁ + t₂)/(v − u) (times in hours) |
| Moving at a/b of usual speed makes one t minutes late | Usual time = t × a/(b − a) |
| Moving at a/b of usual speed (a > b) makes one early | Usual time = t × a/(a − b) |
| Stoppages: speed without s₁, with s₂ | Stoppage per hour = (s₁ − s₂)/s₁ × 60 minutes |
| Two start towards each other D apart | Meet after D/(u + v) |
| Chase with a gap D | Caught after D/(u − v) |
| Equal distances at u and v | Average speed 2uv/(u + v) |
✗ 54 km/h for 12 s, so distance = 54 × 12 = 648 m | ✓ Convert first: 54 km/h = 15 m/s, then distance = 15 × 12 = 180 m
✗ At 3/4 of usual speed one is late by 20 min, so usual time = 20 × 3/4 | ✓ Time becomes 4/3 of usual, so the extra 1/3 = 20 min and usual time = 60 min
हिंदी नोट: चाल, समय और दूरी में इकाइयाँ पहले मिलाइए: km/h को m/s में बदलने के लिए 5/18 से गुणा कीजिए। आमने-सामने चलने पर चालें जुड़ती हैं, एक ही दिशा में घटती हैं।
Exam Pointer: Verified NTPC patterns: speed from distance and time in mixed units (June 2022 CBT-2, September 2025 UG CBT-1), late-early at two speeds (April 2016) and speed change versus time change (June 2022 CBT-2), a fraction of the usual speed (February 2021), two people meeting or a police-thief chase (April 2016), stoppage time (January 2021, June 2022 CBT-2) and journeys in parts or "covered is a fraction of remaining" (January 2021, two June 2025 Graduate CBT-1 shifts).
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Unit conversion and the basic formula
[PYQ: NTPC CBT-2 12-Jun-2022 Shift-1 | NTPC UG CBT-1 8-Sep-2025 Shift-1]
EXAM LEVEL
Q. Convert 72 km/h into m/s and find the distance covered in 45 minutes at this speed.
72 × 5/18 = 20 m/s. In 45 minutes (3/4 hour) the distance = 72 × 3/4 = 54 km.
Answer: 20 m/s; 54 km
EXAMATLAS LEVEL
Q. A car covers 43.2 km in 36 minutes. Find its speed in m/s and the time it takes to cover 1.8 km at that speed.
36 minutes = 0.6 hour, so speed = 43.2/0.6 = 72 km/h = 20 m/s. Time for 1,800 m = 1,800/20 = 90 seconds.
Answer: 20 m/s; 90 seconds
Pattern 2: Late at one speed, early at another
[PYQ: NTPC CBT-1 22-Apr-2016 Shift-3 | NTPC CBT-2 15-Jun-2022 Shift-3]
EXAM LEVEL
Q. At 40 km/h a man reaches his office 15 minutes late; at 50 km/h he reaches 9 minutes early. Find the distance.
The two times differ by 24 minutes = 0.4 hour. d/40 − d/50 = 0.4 gives d/200 = 0.4 and d = 80 km. Formula: 40 × 50 × 0.4/10 = 80.
Answer: 80 km
EXAMATLAS LEVEL
Q. If a train's speed were 10 km/h more, it would take 1 hour less for a journey; if it were 10 km/h less, it would take 1.5 hours more. Find the distance and the usual speed.
d × 10/(s(s + 10)) = 1 and d × 10/(s(s − 10)) = 1.5. Dividing, (s + 10)/(s − 10) = 1.5, so s + 10 = 1.5s − 15 and s = 50 km/h. Then d = 50 × 60/10 = 300 km. Check: 300/50 = 6 h, 300/60 = 5 h, 300/40 = 7.5 h.
Answer: 300 km at 50 km/h
Pattern 3: A fraction of the usual speed
[PYQ: NTPC CBT-1 10-Feb-2021 Shift-1]
EXAM LEVEL
Q. Walking at 3/4 of his usual speed, a man reaches 20 minutes late. Find his usual time.
At 3/4 speed the time becomes 4/3 of usual, an extra 1/3 of the usual time = 20 minutes. Usual time = 60 minutes.
Answer: 60 minutes
EXAMATLAS LEVEL
Q. A train running at 7/8 of its usual speed reaches 15 minutes late. How early would it reach at 5/4 of its usual speed?
At 7/8 speed, time = 8/7 of usual, so 1/7 of usual = 15 minutes and usual time = 105 minutes. At 5/4 speed, time = 4/5 × 105 = 84 minutes, which is 21 minutes early.
Answer: 21 minutes early
Pattern 4: Meeting and chasing (relative speed)
[PYQ: NTPC CBT-1 3-Apr-2016 Shift-1 | NTPC CBT-1 29-Apr-2016 Shift-2]
EXAM LEVEL
Q. Two towns are 360 km apart. Two cars start at the same time from the towns towards each other at 50 km/h and 70 km/h. When and where do they meet?
Relative speed = 120 km/h, so they meet after 3 hours, 50 × 3 = 150 km from the first town.
Answer: After 3 hours, 150 km from the first town
EXAMATLAS LEVEL
Q. A thief is 500 m ahead of a policeman. The thief runs at 9 km/h and the policeman chases at 11 km/h. How far will the thief have run when he is caught?
The gap closes at 2 km/h, so 0.5 km closes in 0.25 hour. The thief runs 9 × 0.25 = 2.25 km in that time (the policeman runs 2.75 km).
Answer: 2.25 km
Pattern 5: Stoppage time per hour
[PYQ: NTPC CBT-1 18-Jan-2021 Shift-2 | NTPC CBT-2 16-Jun-2022 Shift-3]
EXAM LEVEL
Q. Excluding stoppages a train runs at 60 km/h; including stoppages it averages 48 km/h. For how many minutes per hour does it stop?
It loses 12 km of running each hour, which at 60 km/h takes 12 minutes. Stoppage = 12/60 × 60 = 12 minutes per hour.
Answer: 12 minutes
EXAMATLAS LEVEL
Q. A bus stops for 10 minutes every hour, and its average speed including stoppages is 45 km/h. Find its speed excluding stoppages.
It runs only 50 minutes in each hour, so 45 = s × 50/60 and s = 54 km/h.
Answer: 54 km/h
Pattern 6: Journeys in parts and "covered versus remaining"
[PYQ: NTPC Graduate CBT-1 23-Jun-2025 Shift-2 | NTPC Graduate CBT-1 23-Jun-2025 Shift-3 | NTPC CBT-1 19-Jan-2021 Shift-1]
EXAM LEVEL
Q. A man covers half a journey at 30 km/h and the other half at 45 km/h, taking 5 hours in all. Find the total distance.
Average speed over equal halves = 2 × 30 × 45/75 = 36 km/h. Distance = 36 × 5 = 180 km. Check: 90/30 + 90/45 = 3 + 2 = 5 hours.
Answer: 180 km
EXAMATLAS LEVEL
Q. On a 54 km walk, a man notices after 4 hours that the distance covered is 4/5 of the distance remaining. Find his speed, and the speed he needs to finish the rest in 3 hours.
Covered : remaining = 4 : 5, so covered = 54 × 4/9 = 24 km and his speed = 6 km/h. Remaining 30 km in 3 hours needs 10 km/h.
Answer: 6 km/h; 10 km/h
60-Second Revision
- D = S × T; km/h × 5/18 = m/s.
- Late-early: distance = uv × (total time gap)/(v − u).
- a/b of usual speed: usual time = late minutes × a/(b − a).
- Opposite directions add speeds; same direction subtract.
- Stoppage minutes per hour = (s₁ − s₂)/s₁ × 60.