138_pdfsam_math 54 differential equation solutions odd

# 138_pdfsam_math 54 differential equation solutions odd -...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 3 Therefore, the velocity is given by v (t) = 6 − Ke−2t . Setting v = 1 when t = 0, we ﬁnd that 1=6−K ⇒ K = 5. Thus the equation for velocity v (t) is v (t) = 6 − 5e−2t . The limiting velocity of the sailboat under these conditions is found by letting time approach inﬁnity: t→∞ lim v (t) = lim 6 − 5e−2t = 6 (m/sec). t→∞ To determine the equation of motion we will use the equation of velocity obtained previously and substitute dx/dt for v (t) to obtain dx = 6 − 5e−2t , dt Integrating this equation we obtain x(t) = 6t + Setting x = 0 when t = 0, we ﬁnd 0=0+ 5 + C1 2 ⇒ 5 C1 = − . 2 5 −2t e + C1 . 2 x(0) = 0. Thus the equation of motion for the sailboat is given by x(t) = 6t + 5 −2t 5 e −. 2 2 23. In this problem, there are two forces acting on a boat: the wind force F1 and the water resistance force F2 . Since the proportionality constant in the water resistance force is diﬀerent for the velocities below and above of a certain limit (5 m/sec for the boat A and 6 m/sec for the boat B), for each boat we have two diﬀerential equations. (Compare with Problem 7.) Let x1 (t) denote the distance passed by the boat A for the time t, v1 (t) := dx1 (t)/dt. Then the equation describing the motion of the boat A before it reaches the velocity 5 m/sec is dv m1 dt 134 (A) (A) (A) (A) = F1 + F2 = 650 − (A) b1 v1 ⇒ dv1 dt (A) = 65 4 (A) − v1 . 6 3 (3.12) ...
View Full Document

## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

Ask a homework question - tutors are online