Time, Speed & Distance (TSD) is an important chapter of Quantitative Aptitude for various competitive exams like Banking, SSC, and other government exams. The problems on TSD not only include questions based on the basic concepts but also questions related to trains, boats & streams, circular motion, and races.
Having a good grasp on the basic concepts will help solve the questions easily. This article deals with all the concepts required from basics to advance level of TSD along with an organized method.
Let us start with the basic concepts of Time, Speed & Distance in one shot.
The Three Pillars: Time,Speed, Distance (TSD)
TSD is based on one basic equation:
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
From this, we get:
\(\therefore\)\(\text{Distance} = \text{Speed} \times \text{Time}\)
\(\therefore\)\(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)
e.g.: A car covers 150 km in 3 hours.
Speed = 150 / 3 = 50 km/h.
Unit conversion: (very imp)
\(1\ \text{km/h} = \frac{1000}{3600}\ \text{m/s} = \frac{5}{18}\ \text{m/s}\) \(\therefore\)\(\text{km/h} \rightarrow \text{m/s}: \times \frac{5}{18}\)
\(1\ \text{m/s} = \frac{3600}{1000}\ \text{km/h} = \frac{18}{5}\ \text{km/h}\) \(\therefore\ \text{m/s} \rightarrow \text{km/h}: \times \frac{18}{5}\)
So, 90 km/h = 90 × 5/18 = 25 m/s.
And 20 m/s = 20 × 18/5 = 72 km/h.
Ratio & Proportionality Rules
Ratio can save you from long calculations.
When Time is Constan :
Note : \(S_1 : S_2 = D_1 : D_2\) At Time Constant
Example: Two friends, A and B, run for the same amount of time at speeds of 4 km/h and 5 km/h, respectively.
When Distance is Constant :
Note : Speed and Time are inversely proportional.
When the distance is constant, \(D_1 = D_2\):
\(T_1 : T_2 = \frac{D_1}{S_1} : \frac{D_2}{S_2}\)
\(\therefore T_1 : T_2 = S_2 : S_1\)
Example: A covers 60 km at 30 km/h, while B covers the same distance at 20 km/h.
Speed ratio:
\(S_A:S_B=30:20=3:2\)
Since the distance is constant, time is inversely proportional to speed.
\(\therefore T_A:T_B=2:3\)
When Speed is Constant :
Note: Distance and Time are directly proportional.
When the speed is constant, \(S_1 = S_2\):
\(D_1 : D_2 = \frac{D_1}{S_1} : \frac{D_2}{S_2}\)
\(\therefore D_1 : D_2 = T_1 : T_2\)
Average Speed
Note : Average speed is not the average of speeds. It is:
ratio of total distance travelled to total time taken.
Special Case: Equal Distances :
If a person travels the same distance at speeds \(S_1\) and \(S_2\), then the average speed is:
for e.g: A person travels uphill at 30 km/h and downhill at 20 km/h, covering equal distances in both directions then avg. speed is
Different Distances : Just add distances and times.
Example:
Relative Speed
When two objects move:
Opposite Direction:
Same Direction:
Time to meet = \[ \frac{\text{Initial Distance}}{\text{Relative Speed}} \]
e.g. 1 (Opposite Direction):
Two trains are 300 km apart, travelling at speeds of 60 km/h and 40 km/h.
e.g. 2 (Same Direction):
A thief runs at 45 km/h, while the police travel at 65 km/h. The initial gap is 100 km.\[ \text{Relative Speed} = 65-45 = 20\text{ km/h} \]
Early, Late & Scheduled Time
If a person travels at two different speeds and reaches early or late, then,
(Note) :
If one case is early and the other is late, add the times.
If both are early or both are late, subtract the times.
Example:
At 60 km/h, A reaches 20 min early.
At 45 km/h, A reaches 10 min late.
Time difference = 20 + 10 = 30 min = 0.5 hr.
So the distance is 90 km.
Meeting Point Problems
When two Objects start from opposite ends and meet, time is the same for both.
So, Distance ratio = Speed ratio.
Example: A and B start from X and Y towards each other. Speeds: A = \(7\text{ km/h}\), B = \(3\text{ km/h}\). Total distance = \(200\text{ km}\).
Time taken by both is the same.
A covers:\[ 7\times20=140\text{ km} \]
B covers:\[ 3\times20=60\text{ km} \]
Same Direction & Head Start (This is specially for train problems)
Note :
Convert head start into distance gap.
Then use relative speed = difference concept
After-Meeting Formula
If two people meet and then continue to their destinations, and after meeting they take \(T_1\) and \(T_2\) hours respectively:
Example:
After crossing, A takes \(16\) hours to reach B’s start, and B takes \(9\) hours to reach A’s start.
So, speed ratio A : B = 3 : 4.
Segmented & Variable Speed Journeys
When a journey is split into parts with different speeds:
Example:
A bus covers \(120\) km at \(60\) km/h and the next \(120\) km at \(40\) km/h.
\text{Time} = \frac{120}{60}+\frac{120}{40} \Rightarrow 2+3=5\text{ hours}
\]
Note : If speed changes after an accident, the delay is caused only by the remaining distance. Set up equations using the reduced speed.
Circular Track Concepts
On a circular track of circumference \(D\):
Same Direction
-
First meeting time: \(\displaystyle \frac{D}{|S_1-S_2|}\)
-
Meeting at the starting point: \(\displaystyle \text{LCM}\left(\frac{D}{S_1},\frac{D}{S_2}\right)\)
Here, \(\frac{D}{S_1}\) and \(\frac{D}{S_2}\) are their individual lap times. Say (T1 and T2)
Example:
For \(D=600\), \(S_1=10\), \(S_2=15\):
\[
t_{\text{first meet}}=\frac{600}{|10-15|}=120\text{ s}
\]
Lap times:
\[
T_1=60\text{ s},\quad T_2=40\text{ s}
\]
\[
t_{\text{start}}=\operatorname{LCM}(60,40)=120\text{ s}
\]
So here both occur At 120s
\[
\text{first meeting} = \text{starting-point meeting} = \mathbf{120\text{ s}}
\]
Opposite Direction
First meeting time:
Example:
Track = 600 m. Speeds = 10 m/s and 15 m/s (opposite directions).
Note : Multiple Meetings
here,D is curcumference.
Opposite direction: nth meeting total distance =(2n-1)×D
Same direction (from same point): nth meeting total distance =2nD
Race Concepts
Beating by Distance :
If A beats B by \(x\) meters in a race of \(D\) meters, then:
When A finishes, A has covered \(D\) meters.
B has covered only \(D-x\) meters.
Since both run for the same time, distance is proportional to speed:
\[
S_A:S_B=D:(D-x)
\]
Equivalently:
\[
\frac{S_A}{S_B}=\frac{D}{D-x}
\]
For e.g. Race distance \(D=800\) m, A beats B by \(x=80\) m.
\[
800-80=720\text{ m}
\]
So:
\[
S_A:S_B=800:720=10:9
\]
Beating by Time :
If A beats B by \(x\) seconds, then B takes \(x\) seconds more to finish after A finishes.
Example:
In 1000 m race, A beats B by 100 m or 25 seconds then A's time
Let, \(D=1000\text{ m}\) \(S_A=\text{speed of A},\quad S_B=\text{speed of B}\) \(T_A=\text{time of A}, \quad T_B=\text{time of B}\)
When A finishes 1000 m, B has run:
\[
1000-100=900\text{ m}
\]
So B takes 25 s to cover the remaining 100 m.
\[
S_B=\frac{100}{25}=4\text{ m/s}
\]
B’s total time:
\[
T_B=\frac{1000}{4}=250\text{ s}
\]
Since A beats B by 25 s,
\[
T_A=T_B-25=250-25=225\text{ s}
\]
A’s time: \[
225\text{ s}=3\text{ min }45\text{ s}
\]
Start / Lead :
If A gives B a start of metres in a race of metres, then:
\(S_A:S_B = D:(D-x)\)
Combined Race Ratios :
If A beats B and B beats C, multiply ratios to find A : C.
For e.g.: In a 600 m race, A beats B by 60 m. In a 500 m race, B beats C by 50 m. Find by how many metres A beats C in a 400 m race.
Let, \[
A:B=600:540=10:9
\]
\[
\therefore
\]\[
B:C=500:450=10:9
\]
\[
\therefore
\]\[
A:C=\frac{10}{9}\times\frac{10}{9}=100:81
\]
In \(400\text{ m}\), C runs:
\[
\therefore
\]\[
400\times\frac{81}{100}=324\text{ m}
\]
So A beats C by:
\[
\therefore
\]\[
400-324=76\text{ m}
\]
\[
\mathbf{76\text{ m}}
\]
Final Thoughts
In the TestQuiz, you have already learned about the basics of Time, Speed & Distance, such as speed and time, relative speed, races, and circular tracks. Try to understand the underlying logic of each topic instead of rote learning formulas. Practice various kinds of questions regularly to be able to solve them quickly and accurately in IBPS, SBI, RRB, SSC, and CAT examinations.
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