Chapter 2:Motion along a straight line

Translational Motion and Rotational Motion

Today Later

Current information

1. Please read carefully all the instructions on the coursewebsite

http://sibor.physics.tamu.edu/teaching/phys201/

2. The problems in the syllabus are the same as in

Mastering Physics on the web.

3. Have a notebook, solve the problems first and then submit

them on Mastering Physics site.

4. There are due dates for the problems on Mastering Physics,

so don’t procrastinate.

5. Enroll in WebAssign, since all tasks for the labs are given

and submitted through this system. Prelab part is due 15 min before

the start of the recitation.

Modified Mastering Physics Course: schuessler00785

Settings in Mastering Physics:

4 attempts are given (no subtractions within this attempts),

after 4 attempts – no credit. Sometimes need use reset button.

Graded tutorials are due soon after the lecture.

Other problems are due on the date of the respective exam.

Practical hint: perform calculations to 4 sign. figures

and round as needed.

Describing Motion …

Coordinates

Position (displacement)

Velocity

Acceleration

a) Motion with zero acceleration

b) Motion with non-zero acceleration

Kinematics in One Dimension: Displacement

0xxΔx

Average speed and velocity

)_(

)_(_

timetotalt

ntdisplacemetotalxvelocityaveragev

dt

dx

timet

ntdisplacemexvelocityv

t

)(

)(lim0

• Instantaneous velocity, velocity at a given instant

SpeedSpeed is just the magnitude of is just the magnitude of velocityvelocity!! The “how fast” without accounting for the direction.The “how fast” without accounting for the direction.

• Average velocity = total displacement covered Average velocity = total displacement covered per total elapsed time,per total elapsed time,

• Average speed = total distance covered per total elapsed time,Average speed = total distance covered per total elapsed time,

Average velocity

Δt

Δxv

Instantaneous velocity

0 ΔtwhenΔt

Δxv lim

Graphic representation: Velocity

Slope is average velocity during t=t2-t1

Slope is instantaneous velocity at t1

Ex2-3: Engine traveling on rail

Average Velocity (example)x (meters)

t (seconds)

2

6

-2

4

What is the average velocity over the first 4 seconds ?

A) -2 m/s D) not enough information to decide.

C) 1 m/sB) 4 m/s

1 2 430

x (meters)

t (seconds)

2

6

-2

4

What is the instantaneous velocity at the fourth second ?

A) 4 m/s D) not enough information to decide.

C) 1 m/sB) 0 m/s

1 2 43

Instantaneous Velocity

Acceleration

• We say that things which have changing velocity are “accelerating”

• Acceleration is the “Rate of change of velocity”

• You hit the accelerator in your car to speed up– (Ok…It’s true you also hit it to stay at constant velocity, but

that’s because friction is slowing you down…we’ll get to that later…)

Average acceleration

Δt

Δva

Unit of acceleration:

(m/s)/s=m/s2

Meters per second squared

Average Velocity

• For constant acceleration atvv 0 v

t

t

vvav

v0

vv2

1v 0av

The area under the graph v(t)is the total distance travelled

0( ) / 2avx v t t v v

Kinematics in one dimension

Motion with constant acceleration.From the formula for average acceleration

t

vva 0

atvv 0

atvatvvvvv2

1))((

2

1)(

2

10000

0x xAverage velocity v

t

2000 2

1)

2

1( attvtatvtvxx

We find

On the other hand

Then we can find

20 0

1

2x x v t at

Recap• So for constant acceleration we find:

atvv 0

200 at

2

1tvxx

a const

x

a

v t

t

t

Examples

• Can a car have uniform speed and non-constant velocity?

• Can an object have a positive average velocity over the last hour, and a negative instantaneous velocity?

Conceptual Example

• If the velocity of an object is zero, does it mean that the acceleration is zero?– Example?

• If the acceleration is zero, does that mean that the velocity is zero?– Example?

Example of graphic solution: Catching a speeder (Example 2.9)

m/s

Kinematics equations for constant acceleration

atvv 0

200 2

1attvxx

)(2 020

2 xxavv

tvv

xx )2

( 00

All objects fall with the same constant acceleration!!

Testing Kinetics for a=9.80m/s2

Experiments on the motion of objects falling from leaning tower of Pisa under the action of the force of gravity

Acceleration of gravity on different planets, =9.81m/s2

Eartha

Important to have the correct interpretation of the results! The optimal selection of the reference frame helps.

Problem: Vertical motion

Stone is thrown vertically upward

height of the cliff h=50m

9

t1 = (15 + (15^2 + 4*4.9*50)^0.5)/(2*4.95) = 5.0727 = 5.07 s

t2 = (15 - (15^2 + 4*4.9*50)^0.5)/(2*4.95) = -1.9912 = -1.99 s

1-D motions in the gravitational field

When you brake on dry pavement, your maximum acceleration is about three times greater than when you brake on wet pavement. For a given initial speed, how does your stopping distance xdry on dry pavement compare with your stopping distance xwet on wet pavement?a)xdry = 1/3xwet

b)xdry = 3xwet

c)xdry = 1/9xwet

d)xdry = 9xwet

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