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Two buses starting from two different places A and B simultaneously, travel towards each other and meet after a specified time. If the bus starting from A is delayed by 20 minutes, they meet 12 minutes later than the usual time. If the speed of the bus starting from A is 60km/hr, what is the speed of the second bus?

My attempt:

Let t be the usual time of meeting and s(A) and s(B) be the speeds of Bus A and Bus B respectively, then

Applying relative velocity concept (stopping bus B i.e. taking it's speed =0 and adding it to speed of A), i got:

s(A)= 60 (given)

(60 + s(B))t = d (distance between the buses)

Now if bus A is delayed by 20 mins, then B's travelling time is 20 mins more than A's to cover the same distance.

t(B) = t(A) + 20/60

hence, t(B)= t(A) + 1/5

distance covered by A = t(A) * 60

distance covered by B = t(B) * s(B) = (t(A) +1/5)*s(B)

The problem: How do i relate the delay of 12 mins in their meeting time?

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3 Answers 3

up vote 1 down vote accepted

ADDED. Solution using your equations and making the necessary corrections and additions.

(60 + s(B))t = d

distance covered by A = t(A) * 60

distance covered by B = t(B) * s(B) = (t(A) +1/5)*s(B)

(distance between buses = d = distance covered by A + distance covered by B)

t(B) = t(A) + 20/60

hence, t(B) = t(A) + 1/5

Correction: 20/60 = 1/3; hence, t(B) = t(A) + 1/3.

How do i relate the delay of 12 mins in their meeting time?

t(B) = t + 1/5.


Let $d$ be the distance from $A$ to $B$, $v_{A}=60$ km/hr be the speed of the bus departing from $A$ and $v_{B}$ the speed of the bus departing from $B$. If both buses start simultaneously, $v_{A}t$ is the distance traveled by bus $A$ and $v_{B}t$ is the the distance traveled by bus $B$, where $t$ is the usual time they spend to meet each other. We have

$$v_{A}t+v_{B}t=d.$$

(The units have to be consistent: e.g. speed in km/hr, time in hours and distance in km). If the bus starting from $A$ is delayed by 20 minutes =$\frac{1}{3}$ hour, then bus $A$ travels the distance $v_{A}(t-1/3+1/5)$ and bus $B$ travels $v_{B}(t+1/5)$, because they meet $12$ minutes = $\frac{1}{5}$ hour later than the usual time $t$.

$$v_{A}(t-1/3+1/5)+v_{B}(t+1/5)=d.$$

Equate both equations

$$v_{A}t+v_{B}t= v_{A}(t-1/3+1/5)+v_{B}(t+1/5),$$

and simplify

$$\frac{2}{15}v_{A}-\frac{1}{5}v_{B}=0.$$

Since $v_{A}$ is 60km/hr, we have

$$\frac{2}{15}\left( 60\right) -\frac{1}{5}v_{B}=0,$$

whose solution is $v_{B}=40$ km/hr.

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When the buses meet, Bus A has travelled $8$ fewer minutes than usual, and therefore $8$ fewer km.

These $8$ km were travelled by B in the $12$ extra minutes. So the speed of B is $\dfrac{8}{12}$ km per minute.

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So, the speed of $A$ is $1 KM/minute$

Let the speed of $B$ is $u KM/minute$

If they normally meet in $t$ minutes,

the distance covered by $A$ is $t$ KM and the distance covered by $B$ is $u\cdot t$ KM

So, total distance $u\cdot t+t$ KM

If $A$ delays 20 minute, $A$ travels for $t-20+12=t-8$ minutes,

the distance covered by $A$ is $t-8$ KM and the distance covered by $B$ is $u(t+12)$ KM

So, total distance becomes $t-8+u(t+12)$ KM

So, $t-8+u(t+12)=ut+t\implies 12u=8,$ or $u=\frac23$

So, the speed of $B$ bus is $\frac23 KM/minute=60\cdot \frac23 KM/Hour=40 KM/Hour$

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