Himanshu
Last Activity: 8 Years ago
Let velocity of train = VT = Vcosx i^ + Vsinx j^
Velocity of man1 = VM = 4 i^
Velocity of train with respect to man = VTM j^ = VT – VM = Vcosx i^ + Vsinx j^ – 4 i
= (Vcosx – 4)i^ + Vsinx j^
Now, Vcosx – 4 should be zero as VTM has only vertical direction.
So, Vcosx = 4 ….i)
Velocity of man2 = VM = – j^ m/s
Velocity of train with respect to man = VTM = VT – VM = Vcosx i^ + Vsinx j^ – (– j^)
= Vcosx i^ +(Vsinx + 1)j^
According to the statement as angle is 45` so Vsinx + 1 = Vcosx ...ii)
Solving i) and ii).
We get velocity of train = 5 m/s 37` above the +ve x-direction.
