# why was the engineer driving the train backwards answers

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.

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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