CHAPTER 3 - 8

# CHAPTER 3 - 8 - a =(3.44m/s(2.25m/s =4.11m/s av 22 2 2 1/2...

This preview shows pages 1–2. Sign up to view the full content.

d v 0 v = a d t 0 t ; v = (4.0 m/ s 2 ) i + (3.0 m/ s 2 ) j d t 0 t = (4.0 m/ s 2 ) t i + (3.0 m/ s 2 ) t j . a av  = [(3.44 m/s 2 ) 2  + (2.25 m/s 2 ) 2 ] 1/2  =          4.11 m/s 2 . We find the direction from tan  θ  = (2.25 m/s 2 )/(3.44 m/s 2 ) = 0.654, which gives           = 33.2° north of east . ( c ) Because we do not know the distance traveled, the average speed is         unknown . 22. ( a ) For the vertical component we have a V  = (3.80 m/s 2 ) sin 30.0° =         1.90 m/s 2  down . ( b ) Because the elevation change is the vertical displacement, we find the time from the  vertical motion, taking down as the positive direction: y  =  v 0 y t  +  ! a V t 2 ; 250 m = 0 +  ! (1.90 m/s 2 ) t 2 , which gives           t  = 16.2 s . 23. The acceleration is  a  = (4.0 m/s 2 ) i  + (3.0 m/s 2 ) j . ( a ) We find the velocity by integrating: ( b ) The speed of the particle is

This preview has intentionally blurred sections. Sign up to view the full version.

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

## This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

### Page1 / 2

CHAPTER 3 - 8 - a =(3.44m/s(2.25m/s =4.11m/s av 22 2 2 1/2...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online