Lesson 2-2b Uniform Acceleration (1)

8/3/2019 Lesson 2-2b Uniform Acceleration (1)

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Lesson 2-2b Uniform Acceleration

Homework

Read pages 51-59 in Holt textbook.

Complete lesson 1-4 in Homework

Helpers: Physics.

Check Moodle foradditional assignmentsfor this lesson.


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Section Objectives

Beable to isolate forany of the possible

unknown quantities in our kinematics

equations.

Apply kinematicsequations to calculatedistance, time, or velocity under

conditions of constant acceleration.


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General Steps for Problem Solving

Problem Solving

Sketch the Problem.

List your given quantities.

Identify the unknown quantity.

Select a formula that contains the givens and theunknowns.

Isolate the unknown in the formula.

Solve the equation.

Round your answer and check your units.


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Do you solve problems logically?

What is the first thing that you do when you encounter

a new word-problem to solve?

How do you identify the quantities that have been

given to you?

How do you identify the unknown quantity?

How do you select the correct formula to use?


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Equations for Motion with Constant Acceleration

0:When0

{R 0:When 0 !R

at!0

RR

2

00

2

1attxx ! R

)(20

2

0

2xxa ! RR

at!R

2

0

2

1atxx !

)(20

2xxa !R


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Another formula that it is useful,although

not as commonly used, isshown below.

t)1/2(x 0 (!( RR


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Example 1. A man on a rocket sled that is initially

at rest experiencesa uniform acceleration of 5.0m/s2. How longwill it take the man to travel 200.0

m?


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Sketch the problem,list thegiven quantities,

identify the unknown andselect a formula thathasall of ourgiven quantitiesand our unknown

quantity.


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m?

Given:

Find:

Original Formula:

V0 = 0 m/s a = 5.0 m/s2 (x = 200.0 m

(t

2

0 2

1atxx !


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Isolation:

Original Formula:

Subtract X0 from both sides:

2

0

2

1atxx !

2

000

2

1atxxxx !


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Isolation:

Formula:

Substitute (X for X X0 :2

2

1atx !(

2

000

2

1atxxxx !


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Isolation:

Formula:

Multiply both sides by 2 :

2

2

1atx !(

2

2

1

22 atxx

!(

22 atx !(


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Isolation:

Formula:

Divide both sides by a :

22 atx !(

a

ta

a

x

!

(2

2

22t

a

x!

(


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Isolation:

Formula:

Take the square rootof both sides:

22t

a

x!

(

22 ta

x!(

a

xt

(!

2


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m?

Given:

Find:

Working Formula:

V0 = 0 m/s a = 5.0 m/s2 (x = 200.0 m

(t

a

xt

(!

2


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m?

Given:

Find:

Solution:

V0 = 0 m/s a = 5.0 m/s2 (x = 200.0 m

(t

ss

sm

m

a

xt 9.880

/0.5

)0.200(22 22

!!

!

(!


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Click on the imageabove to check youranswer.

Repeat the calculation with an acceleration of 15.0

m/s2. (requires Interactive Physics)


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Example 2. A car is travelingat 30.0 m/swhen

thedriver,seeinga van stopped in his path 40.0m away, hits the brakes in an attempt to try to

stop in time to avoida collision. If the car

deceleratesat a constant rate of 5.00 m/s2will

thedriver beable to stop in time?Hint. There is more than one

way to solve this problem,so

the method that I show here

may differ from your own work.


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quantity.


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Example 2. A car is travelingat 30.0 m/swhen

thedriver,seeinga van stopped in his path 40.0m away, hits the brakes in an attempt to try to

stop in time to avoida collision. If the car

deceleratesat a constant rate of 5.00 m/s2will

thedriver beable to stop in time?Given:

Find:

V0 = 30.0 m/s a = -5.00 m/s2 (x = 40.0 m

V

Original Formula: )(2 02

0

2

xxa ! RR


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Original Formula:

Substitute (x for x x0:

)(2 02

0

2

xxa ! RR

xa(! 22

0

2RR

Isolation:

)2(2

0

2

xa(! RRTake the square

root of each side:

)2(2

0xa(! RR

We get the

working formula:


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Given: V0 = 30.0 m/s a = -5.00 m/s2 (x = 40.0 m

Find: VFormula:

Solution:

So, theanswer to our original question is, no, the carwill

not stop on time! It will hit the van with a velocity of 22.4

m/s.

)2(2

0xa(! RR

smmsmsm /4.22))0.40)(/00.5)(2()/0.30(( 22 !!R


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Click on the imageabove to check youranswer. Repeat

The calculation with an initial velocity of 20.0 m/s.

(requires Interactive Physics)


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Example 3. Going back to our last example, ifa

car hasan initial velocity of 30.0 m/sand it candecelerateat a constant rate of 5.0 m/s2, how much

stoppingdistancewould it require?


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quantity.


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Given:

Find:

Original Formula:

V0 = 30.0 m/s a = -5.0 m/s2 V = 0 m/s

(x

)(2

0

2

0

2xxa ! RR


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Original Formula:

Substitute (x for x x0:

)(2 02

0

2

xxa ! RR

xa(! 22

0

2RR

Isolation for(x:

Subtract V02 from

both sides:

We get the formula:

xa(! 2

2

0

2

0

2

0

2

RRRR

xa(! 22

0

2RR


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We get our working

formula:

Now, divide both

sides by 2a:

We get the formula: xa(! 22

0

2RR

a

xa

a

(!

2

2

2

2

0

2RR

a

x

2

2

0

2RR

!(


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Given:

Find:

Working Formula:

V0 = 30.0 m/s a = -5.0 m/s2 V = 0 m/s

(x

a

x

2

2

0

2RR

!(


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Given:

Find:

Solution:

V0 = 30.0 m/s a = -5.0 m/s2 V = 0 m/s

(x

m10x9.0orm90.

)m/s0.5(2

m/s)(30.0-m/s)(0.0

a2

-

1

2

222

0

2

!

!!(RR

x


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Solution:

m10x9.0orm90.)m/s0.5(2

m/s)(30.0-m/s)(0.0

a2

-

1

2

222

0

2

!

!!(

RR

x

Check: We can put 90. m back into the

equation we used forexample 2.

m/s0.0m)))(90.0m/s(2)(-5.0m/s)((30.0)2a( 222

0!!(! xRR


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SummaryProblem Solving

Sketch the Problem.

List your given quantities.

Identify the unknown quantity.

Select a formula that contains the givens and the

unknowns. Isolate the unknown in the formula.

Solve the equation.

Round your answer and check your units.


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Summary of Formulas:

0:When 0 {R 0:When 0 !R

at!0

RR

2

00

2

1attxx ! R

)(20

2

0

2xxa ! RR

at!R

2

0

2

1 atxx !

)(20

2xxa !R

t)1/2(x0

(!( RR

Documents

Lesson 2-2b Uniform Acceleration (1)