physics ii: lecture 1, pg 1 ap physics “mechanics for physicists and engineers” agenda for today...

Physics II: Lecture 1, Pg 1

AP Physics AP Physics “Mechanics for Physicists and Engineers”“Mechanics for Physicists and Engineers”

Agenda for TodayAgenda for Today

1-D Kinematics (review).1-D Kinematics (review).Average & instantaneous velocity and accelerationMotion with constant acceleration

Introduction to calculus applicationsderivatives and slopesIntegrals and area


Kinematics ProblemsKinematics Problems 1-D Kinematics 1-D Kinematics

Average & instantaneous velocity (Chapter2 1,4,5,11-13,15-17) and acceleration (18,21)

Motion with constant acceleration(23,24,27,31,35,37,39,40-(23,24,27,31,35,37,39,40-1,43)1,43)

Free Fall (44,47,49,51,53,56,61,63)Free Fall (44,47,49,51,53,56,61,63) Motion Graphs (66,67,69,70) Review Phun!! Review Phun!!


KinematicsKinematics

Location and motion of objects is described using Kinematic Variables:

Some examples of kinematic variables.position rr vectorvelocity vv vector

Kinematic VariablesKinematic Variables: : Measured with respect to a reference frame. (x-y axis)Measured using coordinates (having units).Many kinematic variables are VectorsVectors, which means

they have a directiondirection as well as a magnitudemagnitude.Vectors denoted by boldface VV or arrow

v


Motion in 1 dimensionMotion in 1 dimension In general, position at time t1 is usually denoted rr(t1).

In 1-D, we usually write position as x(t1 ). Since it’s in 1-D, all we need to indicate direction is + or .

Displacement in a time t = t2 - t1 is x = x(t2 ) - x(t1 ) = x2 - x1

t

x

t1 t2

x

t

x1

x2some particle’s trajectory

in 1-D

See text : 2-1


1-D kinematics1-D kinematics

vx t x t

t t

x

tav

( ) ( )2 1

2 1

t

x

t1 t2

x

x1

x2trajectory

Velocity v is the “rate of change of position” Average velocity vav in the time t = t2 - t1 is:

t

Vav = slope of line connecting x1 and x2.

See text : 2-1


Instantaneous velocity v is defined as:

1-D kinematics...1-D kinematics...

v tdx t

dt( )

( )

t

x

t1 t2

x

x1

x2

t

so V(t2 ) = slope of line tangent to path at t2.

See text : 2-2


1-D kinematics...1-D kinematics...

av t v t

t t

v

tav

( ) ( )2 1

2 1

Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is:

And instantaneous acceleration a is defined as:

a tdv t

dt

d x t

dt( )

( ) ( ) 2

2

See text : 2-3


RecapRecap

If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time!

adv

dt

d x

dt

2

2

vdx

dt

x x t ( )

x

a

vt

t

t


More 1-D kinematicsMore 1-D kinematics

We saw that v = x / t so therefore x = v t ( i.e. 60 mi/hr x 2 hr = 120 mi )

In “calculus” language we would write dx = v dt, which we can integrate to obtain:

x t x t v t dtt

t( ) ( ) ( )2 1

1

2

Graphically, this is adding up lots of small rectangles:

v(t)

t

+ +...+

= displacement

v

t1 2

60


High-school calculus:

Also recall that

Since a is constant, we can integrate this using the above rule to find:

Similarly, since we can integrate again to get:

1-D Motion with constant acceleration1-D Motion with constant acceleration

t dtn

t constn n

1

11

adv

dt

v adt a dt at v 0

vdx

dt

x vdt at v dt at v t x ( )02

0 01

2

See text : 2-4


RecapRecap So for constant acceleration we find:

v v at 0

x x v t at 0 021

2

a const

x

a

v t

t

tv v a x x

v v vav

22

12

2 1

1 2

2

1

2

( )

( )

From which we can derive:

See text : Table 2-1 (p. 33)


Problem 1Problem 1

A car traveling with an initial velocity vo. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab

x = 0, t = 0ab

vo


Problem 1...Problem 1...

A car traveling with an initial velocity vo. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel ??

x = xf , t = tf

v = 0

x = 0, t = 0ab

vo



Above, we derived: (a)

(b)

Realize that a = -ab

Using (b), realizing that v = 0 at t = tf :

find 0 = v0 - ab tf or tf = vo /af

Plugging this result into (a) we find the stopping distance:

x x v t at 0 021

2v v at 0

x vv

aa

v

a

v

af oo

bb

o

b

o

b

1

2

1

2

2 2



So we found that

Suppose that vo = 65 mi/hr x .45 m/s / mi/hr = 29 m/s Suppose also that ab = |g| = 9.8 m/s2.

Find that tf = 3 s and xf = 43 m

tv

ax

v

afo

bf

o

b ,

1

2

2


Tips:Tips:

Read !Before you start work on a problem, read the problem

statement thoroughly. Make sure you understand what information in given, what is asked for, and the meaning of all the terms used in stating the problem.

Watch your units !Always check the units of your answer, and carry the units

along with your numbers during the calculation.

Understand the limits !Many equations we use are special cases of more general

laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).


Recap of kinematics lecturesRecap of kinematics lectures 1-D Kinematics 1-D Kinematics

Average & instantaneous velocity (Chapter3- 1,3,7,9,11) and and acceleration

Motion Graphs (14,15,17,19) Motion with constant acceleration(Ch3 (Ch3 21,23,27,29,31,35,37

41)) Free Fall (Ch3-Free Fall (Ch3-41,43,47,49,51,52)) Review Phun!! (Review Phun!! (67,69,70 ))

physics ii: lecture 1, pg 1 ap physics “mechanics for physicists and engineers” agenda for today...

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