1. 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.

2. ∆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

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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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’

11. 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

12. 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

13. 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= m or 19m

14. 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

15. 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= m/s²
* Once an object is in the air it ALWAYS has an acceleration of m/s²

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

17. 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= 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

18. 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

19. 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!!!!

20. 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!