Question 3.9 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?
Chapter Motion In A Straight Line Page 56
Answers
Answered by
615
Let speed of bus = V
relative speed when cylist and bus both move same direction = (V - 20) km/h
We know ,
Distance = speed x time
A/C to question ,
Bus went past cyclist every 18min when in the motion of his direction .
e.g distance covered by bus = ( V-20) x 18/60 km
Every T time bus travels distance = VT
e.g. ( V - 20) x 18/60 = VT ----------(1)
Similarly ,
Relative speed of bus when bus and cyclist move in opposite direction = ( v+ 20) km/h
Bus went past every 6 min in opppsite direction of his motion .
then,
( V + 20) x 6/60 = VT ----------(2)
Solve both equation ,
( V -20)x 18/60 =( V +20) x 6/60
=> 3V - 60 = V + 20
=> 2V = 80
=> V = 40 Km/h put equation (1)
(40 - 20)x18/60 = 40T
T = 6/40 hour = 9 min
Hence,
V= 40 km/h and T = 9 min
Answered by
210
Hi friend,
The Distance between two buses just when one takes off is V.T
[where V is the velocity of the bus and T is frequency of the bus]
U = speed of the cyclist = 20km/hr
When they move in the same direction,
Relative velocity=V-U
The time to cover the lag of VT distance =
t=18 min = 18/60 hrs.
Distance= t.(V-U)
We know that ,this will be equal to V.T above
=> t(V-U) = VT
=> VT=3/10.(V-U).............1
When they move in the opposite direction,
Relative velocity=V+U
The time to cover the lag of VT distance
= t=6 min = 6/60 hrs.
Equating it to get t(V+U) = VT, we get;
=> VT=1/10(V+U) ..............2
Equating 1 and 2, we get;
3/10.(V-U) = 1/10(V+U)
=> 3V-3U=V+U
=> 2V=4U
=> V=2U
=> V= 2 x 20 = 40Km/hr
Substituting this value of V in equation 2,
40.T=1/10(40+20)
=> T=3/20 hrs
=> T=9 min
Hope it helps!
The Distance between two buses just when one takes off is V.T
[where V is the velocity of the bus and T is frequency of the bus]
U = speed of the cyclist = 20km/hr
When they move in the same direction,
Relative velocity=V-U
The time to cover the lag of VT distance =
t=18 min = 18/60 hrs.
Distance= t.(V-U)
We know that ,this will be equal to V.T above
=> t(V-U) = VT
=> VT=3/10.(V-U).............1
When they move in the opposite direction,
Relative velocity=V+U
The time to cover the lag of VT distance
= t=6 min = 6/60 hrs.
Equating it to get t(V+U) = VT, we get;
=> VT=1/10(V+U) ..............2
Equating 1 and 2, we get;
3/10.(V-U) = 1/10(V+U)
=> 3V-3U=V+U
=> 2V=4U
=> V=2U
=> V= 2 x 20 = 40Km/hr
Substituting this value of V in equation 2,
40.T=1/10(40+20)
=> T=3/20 hrs
=> T=9 min
Hope it helps!
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