At any instant the horizontal position of the weather balloon in Fig. 12-18a is defined by x=(8) ft, where is in seconds. If the equation of the path is y = x/10, determine the magnitude and direction of the velocity and the acceleration when t = 2 s. SOLUTION Velocity. The velocity component in the x direction is (81) = 8 ft/s To find the relationship between the velocity components we will use the chain rule of calculus. When t = 2s, x = 8(2) = 16 ft, Fig. 12-18a, and so vyy 1 (x/10) = 2xi/10 = 2(16)(8)/10 = 25.6 ft/s When 2s, the magnitude of velocity is therefore v = (8 ft/s) + (25.6 ft/s) = 26.8 ft/s The direction is tangent to the path, Fig. 12-18b, where Ans. (a) tan - Vy Vx 25.6 =tan 72.6 -16 ft- B 4v 26.8 ft/s Ans. 8 = 72.6 B Acceleration. The relationship between the acceleration components is determined using the chain rule. (See Appendix C.) We have Thus, = 0 ay = vy = (2x/10)=2(x)/10 + 2x(x)/10 = dt = 2(8)2/10 + 2(16)(0)/10 = 12.8 ft/s a 12.8 ft a = (0) + (12.8) = 12.8 ft/s Ans. 8-90 The direction of a, as shown in Fig. 12-18c, is (c) 12.8 =tan = 90 Ans. Fig. 12-18 0 NOTE: It is also possible to obtain v, and a, by first expressing y = f(t) = (8)/10 = 6.4 and then taking successive time derivatives.