In 29 and 30, the linear acceleration a = dv/dt of a moving particle is given by a

Chapter 6, Problem 6.3.30

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In 29 and 30, the linear acceleration a = dv/dt of a moving particle is given by a formula dv/dt = f(t, v), where the velocity v = dy/dt is the derivative of the function y = y(t) giving the position of the particle at time t. Suppose that the velocity v(t) is approximated using the Runge-Kutta method to solve numerically the initial value problem dv dt = f(t, v), v(O) = Vo (19) That is, starting with to = 0 and vo, the formulas in Eqs. (5) and (6) are applied-with t and v in place of x and y-to calculate the successive approximate velocity values VI> Vz, V 3 , ... , Vm at the successive times t" t 2 , t3 , ... ,tm (with tn+' = tn + h). Now suppose that we also want to approximate the distance y(t) traveled by the particle. We can do this by beginning with the initial position y(O) = Yo and calculating (20) (n = 1, 2, 3, ... ), where an = f(tn, vn) v'(tn) is the particle's approximate acceleration at time tn. Theformula in (20) would give the correct increment (from Yn to Yn+') if the acceleration an remained constant during the time interval [tn, tn+,]. Thus, once a table of approximate velocities has been calculated, Eq. (20) provides a simple way to calculate a table of corresponding successive positions. This process is illustrated in the project for this section, by beginning with the velocity data in Fig. 6.3.8 (Example 3) and proceeding tofollow the skydiver's position during her descent to the ground. Now consider again the crossbow bolt of Example 3 in Section 1 .8. It still is shot straight upward from the ground with an initial velocity of 49 mis, but because of air resistance proportional to the square of its velocity, its velocity function v(t) satisfies the initial value problem dt = -(O.OOl l)vlvl - 9.8, v(O) = 49. Beginning with this initial value problem, repeat parts (a) through (c) of (except that you may need n = 200 subintervals to get four-place accuracy in part (a) and n = 400 subintervals for two-place accuracy in part (b. According to the results of 17 and 18 in Section 1 .8, the bolt's velocity and position functions during ascent and descent are given by the following formulas.Ascent: v(t) = (94.388) tan(0.478837 - [0. 103827]t), y(t) = 108.465 + (909.091) In (cos(0.478837 - [0. 103827]t ; Descent: v(t) = -(94.388) tanh(0. 103827[t - 4.61 19]), y(t) = 108.465 - (909.091) In (cosh(O. l 03827[t - 4.61 19]) .