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Two cities A and B are connected by a regular bus service with buses leaving in either direction every T seconds . The speed of each bus is uniform and equal to V b A cyclist cycles from A to B with a uniform speed V c (V b >V c ) . A bus goes fast the cyclist in T 1 second in the direction A to B and every T 2 second in the direction B to A .then (a) T 1 = V b T /V b +V c (b) T 2 =V b T /V b -V c (c) T 1 = V b T/V b --V c (d) T 2 = V b T/V b --V c Please explain this question without using relative velocity concept

Two cities A and B are connected by a regular bus service with buses leaving in either direction every T seconds . The speed of each bus is uniform and equal to Vb  A cyclist cycles from A to B with a uniform speed Vc  (Vb >Vc ) . A bus goes fast the cyclist in T1 second in the direction A to B and every T2 second in the direction B to A .then 
(a) T1 = VbT /Vb +Vc  (b) T2 =VbT /Vb-Vc  (c) T1= VbT/Vb --Vc  (d) T2 = VbT/Vb --Vc 
Please explain this question without using relative velocity concept

Grade:11

1 Answers

Vikas TU
14149 Points
4 years ago
Hiiii 
Every bus is running in either direction from Ato B in every T second , 
So distnce from A to B would be = VbT 
an when bus is going from A to B , The relative velocity = Vb – Vc 
So, T1 = VbT / (Vb- Vc )
so, C is th correct ans.

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