A truck leaves station X for Y at 8 AM traveling at an average speed of
100 kilometer per hour.30 minutes later a bus leaves station X for Y at
an average speed of 120 km/h (kilometers/hour). At 9 AM a car leaves
station X for Y traveling at an average speed of 150 km/h. The car reaches
station Y one hour earlier than the truck and the bus reaches station Y
30minutes earlier than the truck.

(1) What is the distance between X and Y.
(2) What are the arrival timings of these three vehicles?
(3) At what distance from Y would the car overtake the bus?
(4) At what distance from Y would the bus would overtake
the truck?

t_T_X=time truck leaves station X = 8:00AM
t_T_Y=time truck arrives at station Y
t_B_X=time bus leaves station X = 8:30AM
t_B_Y=time bus arrives at station Y = t_T_Y - 30 min
t_C_X=time car leaves station X = 9:00AM
t_C_Y=time car arrives at station Y = t_T_Y - 1
v_T=speed of truck = 100
v_B=speed of bus = 120
v_C=speed of car = 150
d=distance between X and Y

(1) d = 100(t_T_Y - 8:00AM) = 120(t_T_Y - 30min - 8:30AM) = 150(t_T_Y - 1 - 9:00AM)
d = 100(t_T_Y - 8:00AM) = 120(t_T_Y - 9:00AM) = 150(t_T_Y - 10:00AM)
d = 100(m+2) = 120(m+1) = 150 m
200=20m+120=50m m=4
d = 100(6) = 120(5) = 150(4) = 600 km = distance between X and Y
(2)
t_T_Y = 2:00PM Time truck arrives at Y
t_B_Y = 1:30PM Time bus arrives at Y
t_C_Y = 1:00PM Time car arrives at Y
(3)
t_B_X=8:30AM=time bus leaves station
t_C_X=9:00AM=time car leaves station X
120T = 150(T - 1/2) T=30/75=6/15=2/3
600-2(80)=600-160=440
The car overtakes the bus 440 km from station Y.
(4)
t_T_X= 8:00AM=time truck leaves station X
t_B_X=8:30AM=time bus leaves station
100T = 120(T - 1/2) T=60/20=3
600-3(100)=300
The bus overtakes the truck 300 km from station Y
(extra)
100T = 150(T - 1) T=3
600-3(100)=300
The car overtakes the truck 300 km from station Y