why was the engineer driving the train backwards answers

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The point of the outer rim of the train wheel that is directly  below the inner wheel rail contact point on the rail track, is  going backward at an instant in time. Based on the theory of the  instantaneous center (IC) of rotation the contact point of the  train wheel at the rail track is the instant center. Train wheels  have an outer rim that is a greater diameter than the inner part of  the wheel that contacts the rail to keep the wheels on the track.  At an instant in time all points on the wheel measured by a  straight line from the IC indicate the position of that point on  the wheel and its path of motion.

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At each such point construct the  direction vector that is at 90 degrees to that straight line from  the IC. The vector should point in the direction of the wheel  rotation. It can then be seen that such a vector that is at the  outer wheel rim when that point is directly below the inner rim  contact point of the wheel on the track is pointing backwards.    Also if the train is moving at speed V, the only part of the  rotating train wheel going speed V is the center of the axel of the  wheel. Other parts of the wheel have a velocity relative to the  distance from the IC. For example the bottom of the wheel in  contact with the track has zero speed relative to the track at that  instant in time unless there is slip atb that point.

The point on  the inner track contact rim that is directly above the IC is moving  at 2 times the train speed being 2 times the radius from the IC. To  calculate speed at points on the wheel the 2D angular velocity  equation is applied as V = (straight distance from IC) x 2 pi x  revolutions/time. One revolution = 2 pi radians.    Additional reading on IC:    http://en. wikipedia. org/wiki/Instant_centre_of_rotation
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