118 



PROCEEDINGS OF THE AMERICAN ACADEMY. 



Now let us take the case a-o>0, /o = — »o/2k. The constant /3 is zero, 

 and 



x = Xo — v (t - t) — j- (to - r) 2 + 8 (r — r) 5 + , 



«o 



4k 

 5 



8 = «o + ;T.(to-t) + ^ S(r -r)' + ... 



The value of 5, which does not vanish, may readily be found, if desired. 



This solution brings out the fact very clearly that the motion runs 

 into a point d'arret. The curve, to be sure, has a cusp, but the expan- 

 sion is in powers of V To — r, which means that the time cannot exceed 

 r . If we could imagine that time satisfies the same possibilities of 

 reversal of direction as space, we could interpret this cuspidal motion, 

 but as time can flow only in one direction, the motion must be inter- 

 preted as having a point d'arret. This does not mean that the two 

 parts of the geometric locus cannot be interpreted, but that they must 

 be representative of two different motions terminating in the same 

 point d'arret. 



We may indeed trace the motion back from the point d'arret along 

 the two branches. We have 



v 1 , 



J = — ^r =*= n~ "v^ 2 — 4jcVX. 

 2k Ik 



The acceleration is negative for both signs before the radical; x has 

 increased up to .To at t = to and v has decreased to v . In following 

 the motion backward through time we find that x decreases and v 



Figure 1. 



increases. For the motion with the positive sign before the radical 

 the retardation is less than for the motion with the negative sign. 

 Hence in going backward, the former motion gives larger values of x 



