Question

FOR THE BELOW QUESTION I AM UNABLE TO UNDERSTAND HOW TO FIND TIME FRND PLS SAY HOW TO FIND TIME ONLY I DON'T WANT OTHER STEPS....
Two towns A and B are connected by a regular bus service
either direction every T minutes. A man cycling with a speed of 20 km h-1 in the
direction A to B notices that a bus goes past him every 18 min in the direction of
his motion, and every 6 min in the opposite direction. What is the period T of the
bus service and with what speed (assumed constant) do the buses ply on the
road?
​

Answer 1

$\underline{\boxed{\bold\red{Question}}}$

Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km /h−1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do buses ply on the road?

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$\underline{\boxed{\bold\blue{Answer}}}$

As both the bus and cyclist are moving in same direction their relative velocity will be:

$\green{Relative \: velocity \: of \: bus= Vb-Vc= Vb +(-20) }$

$\green{Distance \: covered = Vt}$

$\pink{speed × time = distance }$

$\green{(Vb-20)×18 = Vbt....(1)}$

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case 2

Bus is moving from B to A

As the bus and cyclist moving in the opposite direction their relative velocity will be

$\green{Relative \: velocity \: of \: bus = Vb-Vc= Vb(-20)}$

$\green{Distance \: covered =Vt}$

$\pink{Speed × distance = time }$

$\green{(Vb+20)×6 = Vbt.....(2)}$

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$Solving (1)&(2),we get$

$\green{(Vb-20 )× 18 }$

$\green{Vb = \frac{40}{hr}}$

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$substitute \: the \: value \: of \: b \: in \: eqn \: \: \: (2)$

$\green{(40+20)×6 = 40T}$

$\green{60×6 = 40T}$

$\green{360=40T}$

$\green{T = \frac{360}{40}}$

T = 9

Speed of bus is 40km/hr and time interval after which each bus leaves is 9 minutes

Substuting is wrong

[Hope this helps you.../]