The document outlines a physics lecture series focusing on mechanics, specifically covering topics such as velocity, acceleration, and motion equations. It details the schedule of lectures, reading materials, and homework assignments from early January to mid-February 2012. Key concepts include graphical interpretation of velocity and acceleration, average and instantaneous values, and examples demonstrating these principles in real-world scenarios.

1. Physics - Mechanics
Lecture 2
Velocity and Acceleration
September , 2013

2. Physics 114A - Lecture 3 2/19
Physics 114A - Introduction to Mechanics - Winter-2012
Textbook: Physics, Vol. 1 (UW Edition), James S. Walker
Week Date L# Lecture Topic Pages Slides Reading HW Due Lab
1
2-Jan-12 H1 New Year Holiday
No Lab 1st week
3-Jan-12 1 Introduction to Physics 12 21 Chapter 1
5-Jan-12 2 Position & Velocity 8 22 2-1 to 2-3 No HW
6-Jan-12 3 Velocity & Acceleration 10 25 2-4 to 2-5
2
9-Jan-12 4 Equations of Motion 9 20 2-6 to 2-7
1-D Kinematics
10-Jan-12 5 Vectors 8 24 3-1 to 3-3
12-Jan-12 6 r, v & a Vectors 5 24 3-4 to 3-5 HW1
13-Jan-12 7 Relative Motion 3 18 3-6
3
16-Jan-12 H2 MLK Birthhday Holiday
Free Fall & Projectiles
17-Jan-12 8 2D Motion Basics 5 19 4-1 to 4-2
19-Jan-12 9 2D Examples 13 22 4-3 to 4-5 HW2
20-Jan-12 E1 EXAM 1 - Chapters 1-4
Lecture Schedule (Part 1)
We are here.

3. Velocity &
Acceleration

4. 4/19
Graphical Interpretation of
Average & Instantaneous Velocity

5. 5/19
Velocity & Slope
The position vs. time graph of a particle moving
at constant velocity has a constant slope.
The position vs. time graph
of a particle moving with a
changing velocity has a
changing slope.
3.0 s
4.5 m
slope = velocity = 4.5 m/3.0 s = 1.5 m/s

6. 6/19
Constant Acceleration
0 0 avx x x xv v v v a t     
av if is constantx x xa a a
Acceleration characterizes
the change in velocity with time:
v/t.
If the acceleration is
constant, then the velocity is
changing at a constant rate.
Graphically, if we plot the
velocity vs. time, it will fall on a
straight line with a slope
determined by the acceleration.

7. 7/19
Acceleration
, ,(so t)
fx ixx
av x x av x
f i
v vv
a v a
t t t

    
 
0
( ) lim x
x
t
v
a t
t 



Average acceleration:
Instantaneous acceleration:
Acceleration units: (m/s)/s = m/s2

8. 8/19
Position, Velocity, & Acceleration
Velocity negative;
acceleration negative.
Velocity positive;
acceleration zero.
Velocity positive;
acceleration
negative.
Velocity positive;
acceleration positive.
Velocity zero;
acceleration zero.

9. 9/19
Acceleration
Average acceleration:
Eqn. (2-5)

10. 10/19
Graphical Interpretation of Average and
Instantaneous Acceleration:
Acceleration

11. 11/19
Example: An Accelerating Train
A train moving in a straight line
with an initial velocity of 0.50 m/s
accelerates at 2.0 m/s2 for 2.0 s,
coasts with zero acceleration for
3.0 s, and then accelerates at -1.5
m/s2 for 1.0 s.
(a) What is the final velocity vf of
the train?
(b) What is the average acceleration
aav of the train?
2(3.0 m/s) (0.5 m/s)
0.42 m/s
(6.0 s) (0 s)
f i
av
f i
v vv
a
t t t
 
   
  
1 1 2 2 3 3
2 2 2
(0.50 m/s) (2.0 m/s )(2.0 s) (0 m/s )(3.0 s) ( 1.5 m/s )(1.0 s)
3.0 m/s
f i iv v v v a t a t a t         
    


12. 12/19
Acceleration (increasing speed) and deceleration
(decreasing speed) should not be confused with
the directions of velocity and acceleration:
Acceleration vs. Deceleration
Accelerating
Accelerating
Decelerating
Decelerating

13. 13/19
Motion with Constant Acceleration
If the acceleration is constant, the velocity
changes linearly:
(2-7)
Constant
Acceleration
Changing
Acceleration1
02
( )avv v v 
Slope Constant Slope Changing

14. 14/19
Motion with Constant Acceleration
Velocity vs. time: (2-7)
Average velocity: (2-9)
Position as a function of time:
(2-10)
(2-11)
Velocity as a function of position:
(2-12)

15. 15/19
Motion with Constant Acceleration
The relationship between position and time
follows a characteristic curve.
Parabola

16. 16/19
Motion with Constant Acceleration

17. 17/19
A park ranger driving at 11.4 m/s in
back country suddenly sees a deer
“frozen” in the headlights. He applies
the brakes and slows with an
acceleration of 3.80 m/s2.
(a) If the deer is 20.0 m from the
ranger’s car when the brakes
are applied, how close does
the ranger come to hitting
the deer?
(b) What is the stopping
time?
2 2 2 2
0
2
(0) (11.4 m/s)
17.1 m
2 2( 3.80 m/s )
v v
x
a
 
   

20.0 m 17.1 m 2.9 md   
0
0 2
(11.4 m/s)
0 3.00 s
( 3.80 m/s )
v
v v at t
a
        

Example: Hit the Brakes!