Analytical Mechnics for Engineers

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1921 Excerpt: ... by solving the equations by the methods of calculus, or by a graphical method through the use of distance-time, velocity-time, and acceleration-time graphs. The use of distance-time and speed-time graphs in finding the speed and acceleration, respectively, have been discussed in Arts. 113 and 119. The more convenient of the two methods for any given problem depends upon the character of the problem. Calculus Method.--Equations (1) to (7) may be solved by the methods of calculus provided that certain relations between the variables are known. Thus, if s is expressed as a function of t, the velocity v may be obtained from equation (1). Likewise, if v is expressed as a function of s, the time required for a given displacement may be obtained from equation (5), and so on. ILLUSTRATIVE PROBLEMS 309. A point moves along a straight path according to the law, i=3+4, the units of distance and time being the foot and second, respectively. If 8=0 when t = 0, what is the value of s when t = 10 sec.? and, or, Hence, the gain in speed in the interval between the end of the fourth second and the end of the seventh second is 180 ft./sec. and, therefore, the speed at the end of the seventh second is, -7 = 10 + 180 = 190 ft./sec. Second Method.--Instead of using a definite integral as above, the problem may be solved by means of an indefinite integral as follows: (10+5)d/-5P+51+C. The constant of integration may be determined by means of the initial condition that v = 10 ft./sec. when t =4 sec. Hence, using this condition, the above equation becomes, 10 = 5X42+5X4+C. Therefore, C=--90 and the velocity at any time may be obtained from the equation, - = 5i, +5J-90. If t = 7 sec, the corresponding value of v is, !, = 5X7I+5 X7-90 = 190 ft./sec. 311. A point starts from.