Chapter 2 Motion in One Dimension Physics Review Chapter 2 Motion in One Dimension This is one of the most important chapters in the book!!! If you fall off the boat here…..you’re in trouble.

∆y=yf-yi ∆x=xf-xi Section 1 Displacement Displacement: change in position; strait line distance between start and end point ∆y=yf-yi ∆x=xf-xi

Section 1 Displacement vs. Distance *Displacement can be both positive and negative *Displacement is not always equal to the distance traveled.

Vavg= ∆x/∆t = xf-xi / tf-ti = displacement/time=m/s (units) Section 1 Velocity Vavg= ∆x/∆t = xf-xi / tf-ti = displacement/time=m/s (units) Object 1 - positive slope Object 2 - stationary Object 3 - negative slope Slope of line = rise/run *Velocity is NOT the same as speed →average speed=distance/time

Section 1 Instantaneous Velocity Instantaneous Velocity (red line) velocity of an object at some instant along the object’s path Tangent Line (dotted line) a line that can calculate the I.V. by only touching it in one spot at an instant in time

Section 1 Sample Problem #4 pg 47 Two Students walk in the same direction along a strait path, at a constant speed- one at 0.90 m/s and the other at 1.90 m/s. How much sooner does the faster student arrive at the destination 780 m away? Known: V1= 0.90 m/s Vavg= ∆x/∆t V2=1.90 m/s Rearrange for unknown Xi= 0 m/s ∆t1= ∆x/ Vavg Xf= 780 m/s ∆t1= Xf-Xi/ Vf-Vi Unknown: ∆t1=? ∆t2 =? ∆t1= 867 s ∆t1= 780m-0m/ 0.90m/s-0m/s ∆t2 =411 s ∆t2 = 780m-0m/ 1.90m/s- 0m/s ∆t =867s-411s= 456s or 7.6 min= ∆t

Section 1: Displacement and Velocity ~Vocabulary Frame of Reference: what we are comparing motion to Displacement: the change in position of an object Average Velocity: the total displacement divided by the time interval during which the displacement occurred Instantaneous Velocity: the velocity of an object at some instant or at a specific point in the object’s path

Section 2 Acceleration a= ∆v/∆t Acceleration: rate of change of velocity a= ∆v/∆t a= Vf- Vi/∆t = m/s/s= m/s² *Acceleration has units of . meters per second squared

Section 2 Acceleration Changes in Velocity Velocity Time Graph *When velocity is constant there is no acceleration. *When velocity in the positive direction is increasing the acceleration is positive. *When the velocity in the decreasing the acceleration is negative

Section 2 Acceleration Motion with Constant Acceleration ∆x =½ (Vi+Vf)∆t no ‘a’ Vf =Vi+a∆t no ‘∆x’ ∆x =Vi∆t+½ a(∆t)² no ‘Vf’ Vf ² =Vi² +2a∆x no ‘∆t’

Section 2 Sample problem pg 53 #1 ∆x =½ (Vi+Vf)∆t An airplane accelerates uniformly from rest to a speed of 6.6m/s in 6.5s. Find the distance the airplane travels during this time. Known: Vi=0m/s Vf=6.6m/s ∆t= 6.5s Unknown: ∆x=? ∆x =½ (Vi+Vf)∆t ∆x = ½(0m/s+6.6m/s) 6.5s ∆x =21.45 or 21m

Sample problem pg 55 #3 part one Vf =Vi+a∆t A car starts from rest and travels for 5.0s with a constant acceleration of -1.5m/s². What is the final velocity of the car? Known: Vi=0m/s ∆t=5.0s a= -1.5m/s² Unknown: Vf=? Vf =Vi+a∆t Vf= 0m/s+ (-1.5m/s²)(5.0s) Vf=(-1.5m/s²)(5.0s)= -7.5m/s Vf= -7.5m/s

Sample problem pg 55 #3 part 2 ∆x =Vi∆t+½ a(∆t)² A car starts from rest and travels for 5.0s with a constant acceleration of -1.5m/s². How far does the car travel in this time interval? Known: Vi=0m/s ∆t=5.0s a= -1.5m/s² Vf=-7.5m/s Unknown: ∆x=? ∆x =Vi∆t+½ a(∆t)² ∆x =½ a(∆t)² ∆x =½ -1.5m/s² (5) ² ∆x= -18.75m or 19m

Sample problem pg 58 #4 Vf ² =Vi² +2a∆x A motorboat accelerates uniformly from a velocity of 6.5m/s to west to a velocity of 1.5m/s to the west. If its acceleration was 2.7m/s² to the east, how far did it travel during the acceleration? Known: Vi=6.5m/s Vf=1.5m/s a=-2.7m/s² Unknown ∆x= Vf² =Vi² +2a∆x Rearrange for unknown ∆x= Vf² -Vi²/2a (1.5m/s)² - (6.5m/s) ² / 2 (-2.7m/s²) ∆x= 7.4m

Section 3 Falling Objects Free Fall- motion of a body when only the force due to gravity acts upon it. NO AIR RESISTANCE! Acceleration due to gravity (Ag) Ag= -9.81 m/s² * Once an object is in the air it ALWAYS has an acceleration of -9.81 m/s²

Section 3 Objects in air *At the very top of its path, the ball’s velocity is zero, but the ball’s acceleration is -9.81 m/s² at every point-both when it is moving up (a) and when it is moving down (b)

Section 3 Falling Objects Sample Problem pg 64 #2 A flowerpot falls from a windowsill 25.0 m above the sidewalk. How fast is it moving when it strikes the ground? Known: ∆y=25.0m a= -9.81 m/s² Vi=0 m/s Unknown: Vf=? Vf ² =Vi² +2a∆y Vf ² =2a∆y Vf =√2a∆y Vf= √2(-9.81m/s²)(25.0m) Vf= 22.2m/s

Chapter 2 Practice Problem #44 pg 881 A ball is hit upward with a speed of 7.5m/s. How long does the ball take to reach maximum height? Known: Vi=7.5m/s Vf=0m/s a= -9.81m/s² Unknown: ∆t=? Vf =Vi+a∆t Rearrange for unknown ∆t= Vf-Vi/a ∆t= 0m/s-7.5 m/s/-9.81m/s² ∆t=.76s

Equations in Chapter 2!! Look for equation that has the variables you want and have. You will use these equations ALL year. Know, understand, and utilize them!!!!

Physics Advice Physics gets easier later on as long as you put the time in now!!! (you’ll get it around chapter 7 or 8) WORK HARD You can do it!