Chapter 2: One-Dimensional Kinematics

• Outline• 2-1 Position, Distance, and Displacement• 2-2 Average Speed and Velocity• 2-3 Instantaneous Velocity• 2-4 Acceleration• 2-5 Motion with Constant Acceleration• 2-6 Applications of the Equations of Motion• 2-7 Freely Falling Objects

What we will do today:

• The most important things to understand in Chapter 2.

• How to use this information to solve problems.

TA’s Information

• Name: Tatiana Brusentsova

• Email:

• Office hours

Major Concepts

• Position, distance and displacement• Speed and velocity

– Average– Instantaneous– Constant

• Acceleration– Average– Instantaneous– Constant

• Graphs of position versus time, velocity versus time, and acceleration versus time

• Equations of motion with constant acceleration• Free fall

Warning Areas:Speed vs. VelocityDistance vs. DisplacementConstant Velocity vs. Constant Acceleration

How the physics is described, step by step.

t

xvavg

Constant speed: vtxx 0

t

vaavg

Constant acceleration: atvv 0

tvx

ttvxx

avg

avg

0

00

tavv avg 0

Constant speed, acceleration.

Constant Speed

20vv

vavg

vtxx 0

t, sec

x, M

t

xSlope = v

Constant Acceleration atvv 0

t, sec

v, m

/s

t

vSlope = a Special case of constant a.

Figure 2-26Problem 2-21

Find the average velocity for each segment of the “walk”, and for the total “walk”.

Now, compare DISPLACEMENT and DISTANCE.Compare, average VELOCITY and SPEED.

Figure 2-28Problem 2-32

What is the average acceleration for each segment, what is the average acceleration for the whole motorcycle ride?

Constant Acceleration: The “master” 1D formula:

200

000

00

0

2

12

12

)(

attvx

tatvvx

tvv

x

tvxx avg

20vv

vavg

Uses:

atvv 0Uses:

Table 2-4Constant-Acceleration Equations of Motion

Variables Related Equation Number

Velocity, time, acceleration

v = v0 + at 2-7

Initial, final, and average velocity

vav = ½(v0 + v) 2-9

Position, time, velocity

x = x0 + ½(v0 + v)t 2-10

Position, time, acceleration

x = x0 + v0t + ½ at2 2-11

Velocity, position, acceleration

v2 = v02 + 2a(x – x0) = v0

2 + 2ax 2-12

These equations are relatedx = x0 + v0t + ½ at2

v = v0 + atUse:

or a=(v-v0)/t

Then:

x = x0 + v0t + ½ at2

= x0 + v0t + ½(v-v0)t2/t

= x0 +½(v+v0)t

= x0+vavgt

Everything is derived from definitions of average and constant acceleration and velocity.

(the “Master equation”)

Figure 2-29Problem 2-33

CAREFUL!

What is the displacement for each segment of the graph shown?

Problem solving:1. Draw a picture (given)2. What are they asking for?3. What is given?4. What are the mathematical relationships?

Example 2-6Put the Pedal to the Metal

The formula that most correctly describes the graph of Drag Racing motion is:

1. v2 = v02 + 2a(x – x0)

2. x = x0 + ½ at2

3. x = x0 + ½(v0 + v)t

4. v = v0 + at

Figure 2-16Velocity as a Function of Position for the Ranger in Example 2-8

Notice that ½ the speed is lost in last ¼ of stopping distance.

CONSTANT DECELERATION

v2 = v02 + 2a(x – x0)

Hit the Brakes!

Vo=11.4 m/s a=-3.8m/s2

Name that graph!t t x

Name that graph! (in right order)

1. Velocity, position, acceleration2. Position, acceleration, velocity3. Velocity, acceleration, velocity

Vo=11.4 m/s a=-3.8m/s2

t xt

Constant acceleration problem:A vehicle is traveling at an initial speed v. It brakes with constant acceleration a (which can be negative).

What is the total time to come to a stop?What is the total distance traveled?What is the shape of position vs. time?What is the shape of velocity vs. time?

-5

0

5

10

15

20

0 1 2 3 4

x

v

a

V0=11.4 m/sa=-3.8 m/s^s

t, sec

m (x)m/s (v)

Objects in “free-fall”

• Acceleration is gravity, a = -g = -9.8 m/s^2

Q: A ball is dropped from the top of the library. At the same time, a ball is thrown from the ground so that it just reaches the top of the library. Which of the following statements is true?

Draw picture.What is asked?What is given?What are relationships?Solve for unknown.

Help, I’m in free fall!

1. The dropped ball reaches the ground sooner, because it picks up speed as it drops.

2. The thrown ball reaches the top sooner, because it started with higher speed.

3. They reach the endpoints at the same time.

4. Can’t answer—depends on initial speed of ball.

This problem can be quickly solved by looking at the time symmetry of the situation.

Let’s do this the “long way”, just to see how it’s done.

P 2-27: The pop fly.A baseball is hit straight upwards. The round-trip time in the air is 4.5 seconds. How high did the ball go? What was the initial speed?

Draw a picture.Assign “knowns” to variables.Assign variable names to “unknowns”.Work backwards from “knowns” to “unknowns”.

P. 2-91: Seagull flying upwards with seashell, which drops.

Seagull flying upward 5.2 m/s, drops shell at 12.5 m. What is acceleration at time of drop? What is maximum height? When does it return to 12.5 m? What is its speed at 12.5m?

P. 2-94. Ball thrown vertically with initial speed v.

Ball one thrown upwards with speed v. Reaches maximum height h, when ball two is thrown with same speed. Draw x vs. t for each ball. At what height do balls cross paths?