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www.flickr.com/photos/internetarchivebookimages/tags/book... . 5 = 1734 : 1109 :: 13 : 8*31 = distance of point of rever-sion from main track. Now since x= ^2E.G, we have bysubstituting %/-J x 1109 x 8*31 =
c e given by the same tormula 14. The distance travelled by a body moving along a
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11 Tot (3 50s 4=' 5. A car accelerates from rest at constant rate for t
