#### 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 the buses ply on the road?

#### Solution 1

Let *V* be the speed of the bus running between towns A and B.

Speed of the cyclist, *v* = 20 km/h

Relative speed of the bus moving in the direction of the cyclist

= *V* – *v* = (*V* – 20) km/h

The bus went past the cyclist every 18 min i.e `18/60 h` (when he moves in the direction of the bus)

Distance covered by the bus = `(V - 20)18/60 km` .....(i)

Since one bus leaves after every *T* minutes, the distance travelled by the bus will be equal to

`VxxT/60` ...(ii)

Both equation (i) and (ii) are equal

`(V-20) xx 18/60 = (VT)/60` ....(iii)

Relative speed of the bus moving in the opposite direction of the cyclist

= (*V* + 20) km/h

Time taken by the bus moving in the opposite direction of the cyclist.

=(V+20) km/h

Time taken by the bus to go past the cyclist = 6 min = `6/60 h`

`:.(V+20)6/60 = (VT)/60` ....(iv)

From equations (iii) and (iv), we get

`(V+20)xx6/60 = (V-20)xx 18/60`

V + 20 = 3V - 60

2V = 80

V = 40 "km/h"

Substituting the value of *V* in equation (iv), we get

`(40+20)xx6/60 = (40T)/60`

`T = 360/40`

T = 9 min

#### Solution 2

Let vb be the speed of each bus. Let vc be the speed of cyclist.

Relative velocity of the buses plying in the direction of motion of cyclist is vb – vc .

The buses playing in the direction of motion of the cyclist go past him after every

18 minute i.e.18/20 h.

:.Distance ccovered is `(v_b-v_c) xx 18/60`

Since a bus leaves after every T minute therefore distance is also to `v_b xxT/60`

`:. (v_b-v_c)cc18/60 = v_bxxT/60` ......(1)

Relative velocity of the buses plying opposite to the direction of motion of the cyclist is `v_b +v_c`.

In this case, the buses go past the cyclist affter every 6 minute.

`:.(v_b + v_c) cc 6/60 = v_b xx T/60` .....(2)

Dividing (1) by (2) we get `((v_b-v_c)18)/((v_b+v_c)6) = 1`

On simplification `v_b = 2v_c`

But v_c = 20 "km h"^(-1)`

`:. v_b = 40 "km h"^(-1)`

From equation (1)

`(40-20) xx18/60 = 40 xx T/60`

On simmplification T = 9 minutes