A passenger is late for a train. He stood at a railway station while the accelerating train passed him car by car. The passenger noticed that the last train car passed him for 8 seconds and the previous train car for 10 seconds. How long was the passenger late for the train? You should consider that the train moved with uniform acceleration and the length of the train cars is the same.
The position-time equation of motion of the train is given by the formula
xf = xi + vi∙t + (a∙t^2)/2
The length of the train cars could be denoted by d. Then the equation for the last train car is
xf – xi = d = v1∙t1 + (a∙t1^2)/2 (1)
where v1 – the initial velocity of the last train car from the point of view of the passenger and t1 – the time it takes the last train car to pass the passenger.
The equation for the previous train car is
d = v2∙t2 + (a∙t2^2)/2 (2)
where v2 – the initial velocity of the previous train car and t2 – the time it takes the previous train car to pass the passenger.
The equation for velocity as a function of time is given by the formula
vf = vi + a∙t
Obviously, that the train was not moving at the beginning, vi = 0. The final velocity vf can become the initial velocity v1 of the last train car from the point of view of the passenger and the equation is
vf = v1 = a∙(t + t2) (3)
where t – a period of time by which the passenger was late and t2 – the time it takes the previous train car to pass the passenger.
The equation for initial velocity v2 of the previous train car is
v2 = a∙t (4)
Now, we substitute equation (3) into (1) and (4) into (2).
d = a∙(t + t2)∙t1 + (a∙t1^2)/2 (5)
d = a∙t∙t2 + (a∙t2^2)/2 (6)
We equate the right sides of equations (5) and (6) and by dividing by a obtain
(t + t2)∙t1 + t1^2/2 = t∙t2 + t2^2/2
We can write the last equation as
t = (t1∙t2 + 0.5∙(t1^2 – t2^2))/(t2 – t1)
Finally, we can substitute values and calculate
t = (8∙10+0.5∙(8^2 -10^2))/(10-8) = 31 s
If you want to know how to solve such problems. Please try Baby Steps In Physics.
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