Relative speed = (84 + 60) km/h
= 144 km/h = 144 x (5/18) m/s
= 8 x 5 = 40 m/s
Distance covered in passing each other
= 512 + 528 = 1040 m
? Required time = 1040/40 = 26 s
Length of the train = 280 m
Length of the platform = 280 x 3 = 840 m
Taken time = 6 min 40 s
= (360 + 40) = 400 s
? Required speed = Distance / Time
= (840 + 280)/400
= 1120/400
= 28 m/s
Speed of the train relative to man
= (240 + 24) km/h = 264 km/h
= 264 x (5/18) m/s = 220/3 m/s
Distance covered in passing the man = 440 m
? Time taken = (440/220) x 3 = 6 s
Distance covered by the train in crossing the platform
= (45 x 30) / 3600
= 3/8 km
= 375 metres
? Length of train = (375 - 100) = 275 meters
? Time taken to cross the pole
= 275 ÷ (375/30)
= (275 x 30) /375
= 30 sec
Relative speed of both trains
= (50 - 30) = 20 km/hr
= 20 x (5/18) m/sec.
= (50/9) m/sec.
Let the length of the faster train be L
Then, (L x 9)/50 = 18
? L = (18 x 50)/9
= 100 meters
? Speed of the train = 54 x (5/18) m/sec = 15 m/sec
? Time taken by the train to cross the tunnel
= Time taken by it to cover (120 + 130) m
= (250/15) sec
= 162/3
Speed of first train = (100/10) = 10 m/sec.
Let the speed of second train be x m/sec
? 200/ 10 + x = 8
? 200 = 80 + 8x
? x = 15
? Speed of second train = 15 m/sec
= (15 x 18/5)
= 54 km/hr.
A's speed / B's speed = ?(Time taken by B to reach X) / (Time taken by A to reach Y)
? 45/B's speed = ?(10/3) x (5/24) = 5/6
? B's speed = 45 x (6/5) km/hr = 54 km/hr.
Basically they will exchange their speeds just after half of the time required for the whole journey. It means after covering 210 km distance they will exchange their speeds. check it out graphically for more clarification.
Let the distance between Mokama and Hazipur be D km
Time taken by train P to cover D km = 1 h
Time taken by Q to cover D km = 3/2 h
? Speed of train P = D km/h
Speed of train Q = 2D/3 km/h .
Let they cross each other T h after 5 : 00 pm.
Then, DT + 2DT/ 3 = D
? T + 2T/3 = 1
? T(1 + 2/3) = 1
? T = 3/5h = (3/5) x 60 = 3 x 12 = 36 min.
Hence, the two trains meet 36 min after 5 : 00 pm.
? Meeting time = 5:36 pm
Let speed in the return journey = V
? Speed in onward journey = (125/100) x V = 5V/4 km/h
Average speed = (2x 5V/4 x V) / (5V/4 + V) = 10V/9 km/h
? 1600 x (9/10V) = 30 [? halts for 2 h ]
? V = 1600 x 9/30 x 10 = 48 km/h
? Speed in onward journey = 5V/4 = 5/4 x 48 = 60 km/h
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