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?
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2 Answers
62The 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 is the speed of the cyclist=20km/hr
Note that when they move in the same direction, Relative velocity=V-U
The time to cover the lag of VT distance (that separates the two buses) is t=18 mins= 18/60 hrs.
Distance= t.(V-U)
This will be equal to V.T above
Hence, t(V-U) = VT => VT=3/10.(V-U)
Now both
Also, Note that when they move in the opposite direction, Relative velocity=V+U
The time to cover the lag of VT distance (that separates the two buses) is t=6 mins= 6/60 hrs.
again you can equate them to get t(V+U) = VT
=> VT=1/10(V+U)
equating VT from the above 2,
3/10.(V-U) = 1/10(V+U)
3V-3U=V+U
2V=4U
V=2U
V=40Km/hr
Substituting,
40.T=1/10(40+20)
T=3/20 hrs, T=9 minutes!
See if i have made some calculation error.. But the method is the same :)
62The 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 is the speed of the cyclist=20km/hr
Note that when they move in the same direction, Relative velocity=V-U
The time to cover the lag of VT distance (that separates the two buses) is t=18 mins= 18/60 hrs.
Distance= t.(V-U)
This will be equal to V.T above
Hence, t(V-U) = VT => VT=3/10.(V-U)
Now both
Also, Note that when they move in the opposite direction, Relative velocity=V+U
The time to cover the lag of VT distance (that separates the two buses) is t=6 mins= 6/60 hrs.
again you can equate them to get t(V+U) = VT
=> VT=1/10(V+U)
equating VT from the above 2,
3/10.(V-U) = 1/10(V+U)
3V-3U=V+U
2V=4U
V=2U
V=40Km/hr
Substituting,
40.T=1/10(40+20)
T=3/20 hrs, T=9 minutes!
See if i have made some calculation error.. But the method is the same :)