ch04-p065 - 65. We first note that a1 (the acceleration at...

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

View Full Document Right Arrow Icon
65. We first note that a 1 (the acceleration at t 1 = 2.00 s ) is perpendicular to a 2 (the acceleration at t 2 =5.00 s ), by taking their scalar (dot) product.: 22 2 2 12 ˆˆ ˆ ˆ [(6.00 m/s )i+(4.00 m/s )j] [(4.00 m/s )i+( 6.00 m/s )j]=0. aa ⋅= GG Since the acceleration vectors are in the (negative) radial directions, then the two positions (at t 1 and t 2 ) are a quarter-circle apart (or three-quarters of a circle, depending on whether one measures clockwise or counterclockwise). A quick sketch leads to the conclusion that if the particle is moving counterclockwise (as the problem states) then it
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 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online