134_pdfsam_math 54 differential equation solutions odd

134_pdfsam_math 54 differential equation solutions odd -...

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

View Full Document Right Arrow Icon
Chapter 3 The negative sign is present because air resistance acts in opposition to the motion of the object. Therefore the net force acting on the shell is F = F 1 + F 2 = mg (0 . 1) v 2 . We now apply Newton’s second law to obtain m dv dt = ± mg +(0 . 1) v 2 ² . Because the initial velocity of the shell is 500 m/sec, a model for the velocity of the rising shell is expressed as the initial-value problem m dv dt = ± mg . 1) v 2 ² ,v ( t = 0) = 500 , (3.11) where g =9 . 81. Separating variables, we get dv 10 mg + v 2 = dt 10 m and so Z dv 10 mg + v 2 = Z dt 10 m 1 10 mg tan 1 ³ v 10 mg ´ = t 10 m + C. Setting m =3, g . 81 and v = 500 when t = 0, we Fnd C = 1
Background image of page 1
This is the end of the preview. Sign up to access the rest of the 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