A train stopping at two stations 'd' distance apart takes time 't' on the journey from one station to the other.Assuming that its motion is first of uniform acceleration 'α' and then off immediately uniform retardation 'β'.The time interval 't' is
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r1 = ut + 0.5gt^2
r2 = u(t-3) + 0.5g(t-3)^2
Equating these two equations:
3u = 0.5g[(t-3)^2 - t^2] = -0.5g[(2t-3)(3)] => u = -0.5g(2t-3)
Putting values and assuming positive y-axis in vertically upward direction, we have,
49 j = 0.5*9.8*(2t-3) j
or, (2t-3) = 10 or t = 6.5
r2 = u(t-3) + 0.5g(t-3)^2
Equating these two equations:
3u = 0.5g[(t-3)^2 - t^2] = -0.5g[(2t-3)(3)] => u = -0.5g(2t-3)
Putting values and assuming positive y-axis in vertically upward direction, we have,
49 j = 0.5*9.8*(2t-3) j
or, (2t-3) = 10 or t = 6.5
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