- 2 towns A and B are connected by a regular bus service with a bus leaving in either direction every T mins. a man cycling with speed 20kmph in A to B direction notice that a bus goes past him every 18 min in direction of his motion. and every 6 min in opposite direction. The period T of bus service is? Please explain the reason for each step professor!
Asked by imtiyazmulla68
4th March 2018,
9:17 PM
Answered by Expert
Answer:
Cyclist speed 20 km/hr. Let speed of Bus from A and B are V km/hr.
Relative velocity of Bus from A with respect to cyclist = (V-20) km/hr.
Relative velocity of Bus from B with respect to cyclist = (V+20) km/hr.
Cyclist is seeing every 18 minutes bus coming from A. Hence the distance S1 travelled by bus from A in 18 minutes is given by S1 = (V-20)×(18/60) km
Cyclist is seeing every 6 minutes bus coming from B. Hence the distance S2 travelled by bus from B in 6 minutes is given by S2 = (V+20)×(6/60) km
But S1 = S2; hence (V-20)×(18/60) = (V+20)×(6/60); solving for V, we get V = 40 km/hr
Hence the distance travelled by either bus = (40-20)×(18/60) = (40+20)×(6/60)= 6 km;
Time taken by cyclist to travel 6 km is same as interval of bus starting time at A and B.
Time taken by cyclist to travel 6 km = (6/20)×60 = 18 min
Hence buses are staring from A and B at an interval of 18 minutes.
Answered by Expert
6th March 2018,
10:35 PM
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