Two people coming towards each other
Many times you will encounter questions wherein two people or objects will be coming towards each other. In this sub-module, we will discuss how to tackle such questions.
Such questions may be framed in various forms. Let us see all such possible cases/concepts and the corresponding formulae.
Case 1: People/Objects start at the same time
If two persons/objects A and B start at the same time from two points x and y towards each other and after crossing they take $T_1$ and $T_2$ hours in reaching y and x respectively, then:
$\frac{S_A}{S_B}$ = $√\frac{T_2}{T_1}$
Q. Car A starts on a journey from Bristol to London and at the same time another car B starts from London to go to Bristol. After they meet, they reach their destinations after 16 and 25 hours respectively. What is the ratio of the speeds of cars A and B?
Explanations :
If two persons/objects A and B start at the same time from two points x and y towards each other and after crossing they take $T_1$ and $T_2$ hours in reaching y and x respectively, then:
$\frac{S_A}{S_B}$ = $√\frac{T_2}{T_1}$ = √(25/16) = 5/4
Let the distance from Bristol to London be d km and they meet x km from Bristol.
So, $S_A/S_B$ = Distance travelled by car A/Distance travelled by car B = x/(d - x) (as the time for which the cars moved is the same) ….. (1)
$S_A$ = Distance travelled by car A after meeting/Time = (d - x)/16 ….. (2)
$S_B$ = Distance travelled by car B after meeting/Time = x/25 ….. (3)
Using (1) and (2), we get:
$S_A/S_B$ = 25 (d - x)/16 x
Or $S_A/S_B$ = (25/16) ($S_B/S_A$) (using equation 1)
Or ($S_A/S_B)^2$ = 25/16
Or $S_A/S_B$ = 5/4
Case 2: People/Objects end at the same time
Two persons A and B travel for $T_1$ and $T_2$ time towards each other from opposite ends on a linear track and then they meet. After this meeting both of them reach at their respective destinations at the same time, then:
Time taken by each of them to reach their respective destination after meeting each other, T = $√T_1 T_2$
Q. Car A starts on a journey from Bristol to London and another car B starts from London to go to Bristol. They meet after car A and car B have travelled for 16 and 25 hours respectively. What is the time taken by each of them to reach their respective destination after meeting each other, if they reach their destination at the same time?
Explanations :
Two persons A and B travel for $T_1$ and $T_2$ time towards each other from opposite ends on a linear track and then they meet. After this meeting both of them reach at their respective destinations at the same time, then:
Time taken by each of them to reach their respective destination after meeting each other, T = $√T_1 T_2$ = √(16 × 25) = 4 × 5 = 20 hours
Let the distance from Bristol to London be d km and they meet x km from Bristol.
$S_A$ = Distance travelled by car A before meeting/Time = x/16 ….. (1)
Distance travelled by car A after meeting = (d – x) = $S_A$ × T
Or d – x = (x/16) × T ….. from equation (1)
Or (d - x)/x = T/16 ….. (2)
$S_B$ = Distance travelled by car B before meeting/Time = (d - x)/25 ….. (3)
Distance travelled by car B after meeting = x = $S_B$ × T
Or x = [(d - x)/25] × T ….. from equation (3)
Or 25/T = (d - x)/x
Or 25/T = T/16 ….. from equation (2)
Or $T^2$ = 25 × 16
Or T = 5 × 4 = 20 hours
Case 3
A person/object leaves a point A at $t_1$ and reaches the point B at $t_2$. Time taken by A, $T_A$ = $t_2$ – $t_1$
Another person/object leaves the point B at $t_3$ and reaches the point A at $t_4$ Time taken by B, $T_B$ = $t_4$ – $t_3$
Then the time when they met = $t_1$ + $\frac{T_A (t_4 – t_1)}{T_A + T_B}$
Q. Car A starts on a journey from Bristol to London at 10 AM and reaches London at 2 PM. Similarly, car B starts on a journey from London to Bristol at 11 AM and reaches Bristol at 5 PM. What is the time when they must have met?
Explanations :
$t_1$ = 10 AM; $t_2$ = 2 PM; $t_3$ = 11 AM; $t_4$ = 5 PM
Time taken by A, $T_A$ = $t_2$ – $t_1$ = 2 PM – 10 AM = 4 hours
Time taken by B, $T_B$ = $t_4$ – $t_3$ = 5 PM – 11 AM = 6 hours
Then the time when they met = $t_1$ + $\frac{T_A (t_4 – t_1)}{T_A + T_B}$ = 10 AM + $\frac{(4 × 7)}{(4 + 6)}$
= 10 AM + (28/10) = 10 AM + 2.8 hours = 12 noon + 0.8 hours = 12 noon + (0.8 × 60) minutes = 12 noon + 48 minutes, i.e. 12:48 PM
Time taken by A in the whole journey, $T_A$ = 2 PM – 10 AM = 4 hours
Time taken by B in the whole journey, $T_B$ = 5 PM – 11 AM = 6 hours
As the distance travelled is the same, speed will be inversely proportional to the time taken.
So, $S_A/S_B$ = 6/4 = 3/2
Let $S_A$ = 3 km/hr and $S_B$ = 2 km/hr
So, the distance from Bristol to London = $S_A$ × 4 = 3 × 4 = 12 km
The distance between them at 11 AM = Total distance – Distance travelled by car A from 10 AM to 11 AM = 12 – 3 = 9 km
Time after which the cars will meet = 11 AM + [Relative distance/Relative speed] = 11 AM + [9/(3 + 2)]
= 11 AM + (9/5) = 11 AM + 1.8 hours = 12 noon + 0.8 hours = 12 noon + (0.8 × 60) minutes = 12 noon + 48 minutes, i.e. 12:48 PM