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?
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
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.
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
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
On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h–1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
- Relative Velocity