Episode 8 — Aptitude and Reasoning / 8.11 — Speed Distance and Time
8.11.a Concepts and Formulas -- Speed, Distance, and Time
1. The Fundamental Relationship
The three quantities -- Speed, Distance, and Time -- are related by a single equation:
Distance = Speed x Time
Or equivalently:
Speed = Distance / Time (S = D / T)
Time = Distance / Speed (T = D / S)
The Triangle Memory Aid
--------
| D(ist) |
--------
/ \
/ \
-------- --------
| S(peed)| | T(ime) |
-------- --------
Cover what you need:
- Cover D --> S x T (Speed times Time)
- Cover S --> D / T (Distance over Time)
- Cover T --> D / S (Distance over Speed)
Units
| Quantity | Common Units |
|---|---|
| Distance | kilometres (km), metres (m), miles |
| Time | hours (h), minutes (min), seconds (s) |
| Speed | km/h (kmph), m/s, miles/h (mph) |
2. Unit Conversions
km/h to m/s
1 km/h = 1000 m / 3600 s = 5/18 m/s
Formula: Speed in m/s = Speed in km/h x (5/18)
Example:
72 km/h = 72 x (5/18) = 20 m/s
m/s to km/h
1 m/s = 3600/1000 km/h = 18/5 km/h
Formula: Speed in km/h = Speed in m/s x (18/5)
Example:
25 m/s = 25 x (18/5) = 90 km/h
Quick Conversion Table
km/h | m/s km/h | m/s
------|------ ------|------
18 | 5 72 | 20
36 | 10 90 | 25
54 | 15 108 | 30
45 | 12.5 126 | 35
Other Useful Conversions
1 mile = 1.609 km
1 km = 0.621 miles
1 hour = 60 minutes = 3600 seconds
3. Proportionality Rules
When Distance is Constant
If D is constant: S is inversely proportional to T
S1 / S2 = T2 / T1
If speed doubles, time halves.
If speed becomes 3/4, time becomes 4/3.
Example:
A person covers a distance at 40 km/h in 3 hours.
At what speed must he travel to cover the same distance in 2 hours?
S1 x T1 = S2 x T2
40 x 3 = S2 x 2
S2 = 120 / 2 = 60 km/h
When Speed is Constant
If S is constant: D is directly proportional to T
D1 / D2 = T1 / T2
When Time is Constant
If T is constant: D is directly proportional to S
D1 / D2 = S1 / S2
4. Relative Speed
Relative speed is the rate at which the distance between two objects changes.
Same Direction
-----> A (speed = Sa)
-----> B (speed = Sb)
Relative Speed = |Sa - Sb|
The faster object gains on the slower one at this rate.
Use case: Overtaking problems, races.
Opposite Direction
-----> A (speed = Sa)
<----- B (speed = Sb)
Relative Speed = Sa + Sb
The two objects approach (or separate) at this combined rate.
Use case: Meeting problems, objects approaching each other.
Key Formulas
Time to meet (opposite direction) = Total Distance / (Sa + Sb)
Time to overtake (same direction) = Distance gap / |Sa - Sb|
5. Average Speed
IMPORTANT: Average Speed is NOT the average of speeds
Average Speed = Total Distance / Total Time
Case 1: Equal Distances at Different Speeds
When the same distance D is covered at speed S1 and then at speed S2:
Time for 1st half = D / S1
Time for 2nd half = D / S2
Total Distance = 2D
Total Time = D/S1 + D/S2 = D(S1 + S2) / (S1 x S2)
Average Speed = 2D / [D(S1 + S2)/(S1 x S2)]
+-----------------------------------------+
| |
| Average Speed = 2 x S1 x S2 / (S1+S2) |
| |
| (Harmonic Mean of the two speeds) |
| |
+-----------------------------------------+
Example:
A car goes from A to B at 40 km/h and returns at 60 km/h.
Average speed for the entire journey:
Avg Speed = 2 x 40 x 60 / (40 + 60)
= 4800 / 100
= 48 km/h
NOTE: The simple average would give 50 km/h, which is WRONG.
Case 2: Equal Times at Different Speeds
When equal time T is spent at each speed:
Distance at S1 = S1 x T
Distance at S2 = S2 x T
Total Distance = (S1 + S2) x T
Total Time = 2T
Average Speed = (S1 + S2) / 2 (Simple arithmetic mean)
Case 3: Three Different Speeds for Equal Distances
Average Speed = 3 x S1 x S2 x S3 / (S1.S2 + S2.S3 + S1.S3)
6. Meeting Problems
Two Objects Starting Simultaneously from Opposite Ends
A ---------> <--------- B
|<-------- D km ------->|
Sa = speed of A, Sb = speed of B
Time to meet = D / (Sa + Sb)
Distance covered by A when they meet = Sa x D / (Sa + Sb)
Distance covered by B when they meet = Sb x D / (Sa + Sb)
Ratio of distances covered:
Da : Db = Sa : Sb
Multiple Meetings (Objects bouncing back and forth)
1st meeting: They cover D together.
2nd meeting: They cover 3D together (total from start).
3rd meeting: They cover 5D together (total from start).
nth meeting: They cover (2n - 1) x D together.
Delayed Start
If A starts 't' hours before B:
Distance covered by A in 't' hours = Sa x t
Remaining distance = D - Sa x t
Time after B starts to meet = (D - Sa x t) / (Sa + Sb)
7. Overtaking Problems
One Object Catching Another
A -------> Gap = G km B ------->
(Sa) (Sb) where Sa > Sb
Time to overtake = G / (Sa - Sb)
Head Start (Time)
If B starts 't' hours before A:
Gap when A starts = Sb x t
Time for A to catch B = (Sb x t) / (Sa - Sb)
Head Start (Distance)
If B has a head start of 'd' km:
Time for A to catch B = d / (Sa - Sb)
8. Problems Involving Stoppages
A bus has a speed of S km/h without stoppages.
With stoppages, its effective speed is S' km/h.
Due to stoppages, it covers LESS distance per hour.
Distance lost per hour = S - S'
Stoppage time per hour = (S - S') / S hours
= (S - S') / S x 60 minutes
Example:
Without stoppages: 60 km/h
With stoppages: 40 km/h
Stoppage per hour = (60 - 40)/60 = 20/60 = 1/3 hour = 20 minutes
9. Circular Track Problems
Two People Running on a Circular Track
.----.
/ \
/ L km \ L = circumference (track length)
| (track) |
\ /
\ /
'----'
Start
Same Direction
Time for 1st meeting = L / |S1 - S2|
They meet again at the starting point when:
Time = LCM(L/S1, L/S2)
Opposite Direction
Time for 1st meeting = L / (S1 + S2)
10. Speed, Distance, and Time with Ratios
If speeds are in ratio a : b
For same distance: Times are in ratio b : a (inverse)
For same time: Distances are in ratio a : b (direct)
Example:
Speeds of A and B are in ratio 3 : 4.
They travel the same distance.
Ratio of times taken = 4 : 3.
If A takes 20 minutes MORE than B:
Difference in ratio = 4 - 3 = 1 part = 20 minutes
A's time = 4 x 20 = 80 minutes
B's time = 3 x 20 = 60 minutes
11. Problems with Partial Journeys
Covering fractions of distance at different speeds
Total distance = D
First (1/n) at speed S1
Remaining ((n-1)/n) at speed S2
Total Time = (D/n)/S1 + ((n-1)D/n)/S2
= D/n x [1/S1 + (n-1)/S2]
Example:
A person covers 1/3 of a journey at 20 km/h
and the remaining 2/3 at 30 km/h.
Let distance = 60 km (LCM of 20, 30 for easy calculation)
Time for 1st part = 20/20 = 1 hour
Time for 2nd part = 40/30 = 4/3 hours
Total time = 1 + 4/3 = 7/3 hours
Average speed = 60 / (7/3) = 60 x 3/7 = 180/7 = 25.71 km/h
12. Summary of Key Formulas
+-----------------------------------------------------------+
| S = D / T |
| D = S x T |
| T = D / S |
| |
| km/h to m/s: multiply by 5/18 |
| m/s to km/h: multiply by 18/5 |
| |
| Relative Speed (same dir): |S1 - S2| |
| Relative Speed (opposite dir): S1 + S2 |
| |
| Avg Speed (equal dist): 2.S1.S2 / (S1 + S2) |
| Avg Speed (equal time): (S1 + S2) / 2 |
| |
| Meeting time: D / (S1 + S2) |
| Overtaking time: Gap / (S1 - S2) |
| |
| Circular (same dir): L / |S1 - S2| |
| Circular (opp dir): L / (S1 + S2) |
| |
| Stoppage/hr = (S - S') / S hours |
+-----------------------------------------------------------+