Chap36 revision notes on rectilinear motions

8/11/2019 Chap36 revision notes on rectilinear motions

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Chapter 3Kinematics I: Rectilinear Motion


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Displacement

Displacement is the net change in position:r = r 2 r 1 = ( x2 x1)i + ( y2 y1) j + ( z2 z1)kr

2is the position at t

2and r

1is the position at t

1with t

2occurring after t 1. Displacement can have a positive ornegative sign.

Note that displacement is not the same as total distancetraveled (|d| ).

In one-dimension (rectilinear motion):r = ( x2 x1)i


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iivt

xt t x x

=

=

12

12av

Average Velocity

Velocity is a vector and has a magnitude and direction. Inone dimension, the only direction is the positive ornegative direction. But in more than one dimension, thedirection is more complicated. This direction doesntreally affect the problem in one dimension, but will bevery important when we work in two or more dimensions.

t t t =

= rrrv

12

12av

In one dimension:

Writing only the x component: t x

t t

x x

v

=

=

12

12

av


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For an object traveling in one direction along a straight

line, the average speed is the magnitude of the averagevelocity.

t d

=Speed Average

Average speed


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A) Your average speed and average velocity are the same,and neither is zero.

B) Your average speed and average velocity are the same,and both are zero.

C) Your average velocity is zero, and your average speedis 4 m/s.

D) Your average speed is zero, and your average velocityis 4 m/s.

You jog around a 400 m track in 100 seconds, returningto the place where you started. Which of the following

statements is true?

Interactive Question


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The absolute value of the magnitude of the instantaneousvelocity is the instantaneous speed.For example, the speedometer in your car gives your

instantaneous speed, but not instantaneous velocity.

( ) ( )dt dr

t t t t t

t t =

=

+=

rrrv

00limlim

Instantaneous Velocity and Speed

When discussing velocity and speed, we will alwaysmean instantaneous velocity or speed, unless explicitlystated otherwise.

Instantaneous velocity is defined as:

iivdt dx

t x

t =

= 0

limIn one-dimension:


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Which physical quantity is not correctly paired with its SIunit and dimension?

Quantity Unit Dimension

A) velocity m/s [L]/[T]

B) path length m [L]C) speed m/s [L]/[T]D) displacement m/s2 [L]/[T] 2

E) speed time m [L]



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Relative Velocity

Velocity is always measured relative to some fixed axis.Often, in everyday life, that axis is the earth. Relativevelocity usually refers to the velocity of two objectsrelative to each other when they are both moving relativeto a third object, say, the earth.


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1. Think about the problemA. Read the problem twice carefully.B. Draw a detailed picture of the situation.C. Write down what the problem is asking for.D. Think about the physics principles and determine the

approach to use.2. Draw a physics diagram and define variables.

A. Write down what is given in the problem.B. Determine which equations can be used.

3. Do the calculation.

A. It is a good idea to start with the target variableB. Do algebra before using numbersC. Check the units.

4. Think about the answer.A. Is it reasonable? (Order of magnitude)

Problem Solving Steps


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Problem: You are going to meet some friends at the

Arbuckle mountains, about 110 km south of Norman.Your friends merge onto I-35 exactly three minutes beforeyou. They are traveling 105 km/hr. If you are traveling

115 km/hr, will you catch up to them before they reach theArbuckles? If so, where will you catch them? If not, howmuch later than them will you arrive?


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Focus the Problem:

My car Friends car

Speed = 105 km/hr 110 km

Speed = 115 km/hr distance

What is the problem asking for:Will I reach my friends car before 110 km. If not, when will Iarrive compared with them?Outline the Approach: Determine what point the two cars willmeet, and compare it to 110 km. If it is after 110 km, thencompare the location of cars when my friend arrives andcalculate the time it takes for me to go the last few miles.

3 minutes aheadof me


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Describe the Physics:Draw physics diagrams and define all quantities uniquely

My car Friends Car

t 1 = 0 s xF1 = ?

+ xt 1 = 0 s xM1 = 0 m

t 2 = ? s xM2 = xF2 = x2 = ?

We have 3 unknown quantitiesWhich defined quantity is your target variable? x2Write equations you will use to solve this problem.v = ( x2 x1)/(t 2 t 1)

vF = 105 km/hr vM = 115 km/hr

t 0 = 3 min = .050 hr xF0 = 0 m


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PLAN the SOLUTIONConstruct Specific Equations (Same Number as Unknowns)

x2 xM1 = vM(t 2 t 1) x2 = vMt 2 (1)

Unknowns x2, t 2

x2 xF1 = vF(t 2 t 1) x2 xF1 = vFt 2 (2)

We have three equations and three unknowns. The problem isnow just algebra:

xF1

xF1 xF0 = vF(t 1 t 0) xF1 = vF t 0 (3)


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Check Units:[L] ={[L]/[T]}[T]) 1 = [L]

Units are good!

From (1): t 2 = x2/vM (4)

Plug (4) and (3) into (2) x2 = vFt 2 + xF1 x2 = vF x2/vM vF t 0 x2 vF x2/vM = vF t 0 x2 (1 vF/vM) = vF t 0 x2 = vF t 0 (1 vF/vM)

x2 = vMt 2 (1) x

2 x

F1= v

Ft 2

(2) xF1 = vF t 0 (3)


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EXECUTE the PLAN x

2= v

Ft 0

(1 vF/v

M)

= (105 km/hr)(0.050 hr) {1(105 km/hr)/(115 km/hr)}= 60.4 km

EVALUATE the ANSWER Is Answer Properly Stated?

No. The answer is that I will catch my friend 60.4 km from

NormanIs Answer Unreasonable?We can check that this works.

t 2 = x2/vM = 60.4 km 115 km/hr = 0.525 hr t 2 t 0 = ( x2 x0)/vFt 2= x2/vF + t 0 = (60.4 km105 km/hr) 0.050 hr = 0.525 hr

Is Answer Complete?Yes, we would meet 60.4 km from Norman


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Lets solve this problem using the concept of relativevelocity. Focusing the Problem and Describing thePhysics are the same. There are extra equations to use.


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Unknowns

We have three equations and three unknowns.


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You are going to meet some friends at the Arbucklemountains, about 110 km south of Norman. Your friendsmerge onto I-35 exactly two minutes before you. They

are traveling 105 km/hr. You are traveling 115 km/hr.Which graph below describes this situation?

x

t

(A) x

t (E)

x

t (D)

x

t

(C)

x

t

(B)


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Since acceleration is a vector with a magnitude anddirection, then an object is accelerating if it changes speedand/or direction.Acceleration is the rate at which velocity changes, whilevelocity is the rate at which position changes.

Average Acceleration

iiat

vt t vv x x x ==12

12av

t t t =

= vvva

12

12av

In one dimension:

Writing only the x

component: t v

t t

xv

a x x x

x

=

=

12

12

av


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Problem: You are driving 35 mi/hr when a dog runsacross your path. You slam on your brakes and in 2.0seconds slow to 10 mi/hr. What was your averageacceleration in m/s2?


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dt d

t t vv

a ==

0lim

Instantaneous acceleration

When discussing acceleration, we will always meaninstantaneous acceleration, unless explicitly stated

otherwise.


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A) The velocity of the car is constant.B) The acceleration of the car must be non-zero.C) The first 45 miles must have been covered in 30

minutes.D) The speed of the car must be 90 miles per hour

throughout the entire trip.E) The average velocity of the car is 90 miles per hour in

the direction of motion.

A car travels in a straight line covering a total distance of90.0 miles in 60.0 minutes. Which statement concerningthis situation is true?



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Suppose that an object is moving with constantacceleration. Which of the following is an accuratestatement concerning its motion?

A) In equal times its speed increases by equal amountsB) In equal times its velocity changes by equal amounts.

C) In equal times it moves equal distances.D) All of the above are true.E) None of the above are true.



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r 1 = x1i + y1 j + z1k .r 2 = x2i + y2 j + z2k ,

Multi-Dimensional Variables and Differences

r = r 2 r 1 = ( x2 x1)i + ( y2 y1) j + ( z2 z1)k .

+ y

+ x

( x1, y1)

( x2, y2)r 1

r 2

r


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If the velocity changes from v1 to v2 in a time interval t ,then the average acceleration is given by .

t t t =

= vvva

12

12av

k jik jia z y x

z y x aaadt

dv

dt

dv

dt

dv ++=+= +

The instantaneous acceleration is given by,


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Motion Diagrams

A snapshot of an object at different times. A motiondiagram will usually include quantitative information

about the objects position, velocity, and acceleration.


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The picture below shows snapshots of an object taken atequal time intervals. Which statement is true?

A) The object is definitely moving to the right

B) The object is definitely moving to the leftC) The object is definitely speeding upD) The object is moving at a constant speed

E) None of the above is necessarily true



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The picture below shows snapshots of four cars taken atequal time intervals. If the cars are moving forward,which car has the greatest magnitude of acceleration?

A) Car 1 B) Car 2B) Car 3 D) Car 4E) Car 1 and 3 tie

1)2)

3)

4)



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The picture below shows snapshots of four cars taken atequal time intervals. If the cars are moving forward,which car has the smallest magnitude of acceleration?

A) Car 1 B) Car 2B) Car 3 D) Car 4E) Car 1 and 3 tie

1)

2)

3)

4)



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The picture below shows snapshots of four cars taken at equaltime intervals. If the cars are moving forward, which car has anegative acceleration and which car is slowing down?

Negative Acceleration Slowing DownA) Car 3 only Car 3 onlyB) Car 1 only Car 2 onlyC) Car 2 only Car 1 onlyD) Car 1 and 3 Car 2 and 3E) Car 2 and 3 Car 1 and 3

1)

2)

3)


+ x


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Consider the two cars shown with four pictures taken at thesame equal time intervals for each car . At which point(s) dothe two cars have equal speeds?

A) Point 1 onlyB) Point 1 and 4C) Point 2D) Point 3E) Somewhere between point 2 and 3


(1) (2) (3) (4)

(1) (2) (3) (4)


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Consider two balls, one rolling down a ramp and one rollingalong the floor. At which point(s) do the two balls have equalspeeds?

A) Point 2 onlyB) Points 2 and 5C) Somewhere between points 2 and 3D) Somewhere between points 3 and 4E) Either C or D


(1) (2) (3) (4) (5)

( 1) ( 2) ( 3) ( 4) ( 5)


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Look carefully at what the axes on the graph represent. Look carefully at what is constant, and what is changing

linearly (at a constant rate). Determine what the slope represents. Example: If the vertical axis is displacement x and

the horizontal axis is the time t , then the slope is x/t = v, or dx/dt = v (the slope is the velocity). If the vertical axis is velocity v, and the horizontal

axis is time, t , then the slope is given by v/t = a, ordv/dt = a ( (the slope is the acceleration).

Analyzing Motion on a Graph

i Q i


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A) AB C) CD E) AB and CDB) BC D) DE

An object is movingalong a straight line.The graph at the rightshows its positionfrom the starting

point as a function oftime.

t (seconds)0 1 2 3 4 5 6

40

x ( m

)30

2010A

B C

D

E


In what section of the graph does the object have thefastest speed?

I i Q i


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An object is movingalong a straight line.

The graph at theright shows its position from the

starting point as afunction of time.

A) +6.0 m/s C) +10.0 m/s E) +40 m/sB) +8.0 m/s D) + 13.3 m/s

t (seconds)0 1 2 3 4 5 6

40

x ( m )

30

2010

A

B C

D

E


What was the instantaneous velocity of the objectat t = 4 seconds?

I t ti Q ti


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Consider the plot of x vs t at theright, at which point(s) is themotion slowest?

A) AB) BC) DD) EE) More than one of the above answers

x

A

B

CD

E t


I t ti Q ti


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Consider the plot of x vs t at theright, at which point(s) is theobject moving at a constant non-zero velocity?

A) A and CB) A, C and D

C) C onlyD) D onlyE) B and D

A

BC

D

E t


x



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Consider the plot of x vs t at theright, at which point(s) is the

object moving in the negative xdirection?

A) AB) BC) CD) DE) E

A

B

CD

E t


x



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Consider the plot of x vs t at theright, at which point(s) is the

object turning around?

A) A

B) BC) CD) DE) E

A

BC

D

E t


x



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An object is movingalong a straight line.The graph at the rightshows its velocity asa function of time.

t (seconds)0 1 2 3 4 5 6

20

v ( m / s )15

105

0

During which interval(s) of the graph does the object

travel equal distances in equal times ?


A) 0 to 2 s D) 0 to 2 s and 3 s to 5 sB) 2 s to 3 s E) 0 to 2 s, 3 s to 5 s, and 5 s to 6 sC) 3s to 5 s



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The graph shows position as a function of time for twotrains running on parallel tracks. Which is true:

A) At time t B both trains have thesame velocity.

B) Both trains speed up all thetime.

C) Both trains have the samevelocity at some time before t B.

D) Somewhere on the graph, bothtrains have the sameacceleration.

p o s

i t i o n

t B time

A

B


E) More than one of the above is true.



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Consider the graph to the right.Which graph below representsthe same motion?


t

x

t

v

(A)

t

v

(E)t

v

(D)

t

v

(C)t

v

(B)



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t

a

t

v

(A)

t

v

(E)t

v

(D)

t

v

(C)t

v

(B)



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t

a

t

x

(A)

t

x

(E)t

x

(D)

t

x

(C)t

x

(B)

Equations for Constant Acceleration


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a = (v2 v1)/(t 2 t 1)v2 = v 1 + a (t 2 t 1) = v 1 + at (1)

Equations for Constant Acceleration

When the acceleration is a constant, the average velocityis simply midway between the initial and final velocities.(On a v vs. t graph, the acceleration is just the slope of aline. For constant acceleration, that slope never changes).

vav = (v2 + v1)/2 (A)

vav = (x2 x1)/(t2 t1) = (v2 + v1)/2


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vav ( x2 x1)/(t 2 t 1) (v2 v1)/2 x2 x1 = (1/2)(v2 + v1)(t 2 t 1) = (1/2)(v2 + v1)t (2)From equation (1), v2 = v 1 + at

x2 x1 = (1/2) (v 1 + at + v 1)t x2 x1 = v1t + (1/2)a t 2 (3)

Now start with (2), and substitute t from (1) x

2

x1

= (1/2)(v2

+ v1)t

x2 x1 = (1/2)(v2 + v1)(v2 v1 )/a x2 x1 = (1/2)(v22 v12)/av

2

2 = v1

2 + 2a ( x2

x1

) (4)

Substituting v1 = v 2 a (t 2 t 1) from (1) into (2), x

2

x1

= v2

t (1/2)a t 2 (5)

x2 x1 v1 v2 a t 2t 1


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2 1 1 2 2 1

v2 = v1 + at ! ! ! !v2

2 = v12 + 2a x ! ! ! ! x = (1/2)(v2 + v1)t ! ! ! ! x = v1t + (1/2)at 2 ! ! ! ! x = v2t (1/2)at 2 ! ! ! !

Also vav = (v1 + v2)/2

If a=0, then there is only one equation, x = (1/2)(v2 + v1)t, with v2 = v1 x = v t distance equals velocity times time

These equations only work when acceleration is constant!

Often these are written by setting the subscript on the


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x v0 v a t

v = v0 + at ! ! ! !v2 = v02 + 2ax ! ! ! !

x = (1/2)(v + v0)t ! ! ! ! x = v0t + (1/2)at 2 ! ! ! ! x = vt (1/2)at 2 ! ! ! !

y g pinitial point to 0, having no subscript on the final point,and setting t 0=0 and x0=0.

This is simpler to write, but can create confusion if thereare more than two important points in a problem.

One advantage of writing the equations this way is you


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g g q y ycan remember one equation and derive others readily

Start with this equation: x = v0t + (1/2)at 2 (1)Take the derivativedx/dt = v0 dt /dt + (1/2)a d (t 2)/dt v = v0 + at (2)Square the resultv2 = v02 + 2v0at + a 2t 2 = v02 + 2a (v0t + (1/2)at 2)v2 = v02 + 2ax (3)

Problem: A sports car can accelerate at 4.5 m/s2. It starts


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pat rest and accelerates in the negative x direction. After8.0 seconds what is its speed and how far has it gone?

a = 4.5 m/s2

Time = 8.0 s

+ x

Initial velocity is 0

Problem: A spacecraft is traveling with a speed of 3250


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gm/s, and it slows down by firing its retro rockets, so that itdecelerates at a rate of 10 m/s2. What is the velocity ofthe spacecraft after it has traveled 215 km?

Speed = 3250 m/sDeceleration = 10 m/s2Distance = 325 km

Problem: You are attending the OU-Texas football game.


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During the game, the Texas quarterback is tackled behindthe line of scrimmage and fumbles the football. An OUdefensive lineman picks up the fumbled football on the 20yard line and runs toward the end zone at a speed of 7.3m/s. A Texas running back is standing on the 23 yard lineand wants to catch up to the lineman before he scores atouchdown. If the running back can accelerate at aconstant rate, what must be his minimum acceleration tocatch the lineman before OU scores a touchdown. Whatwill be the result of the play?

Focus the Problem:


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TexasRunningBack

OULineman

20 yd = 18.29 m

Speed = 7.3 m/s

23 yd = 21.03 m

What is the problem asking for:


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(1) What acceleration is necessary for the running back to just catch the lineman when the lineman reaches thegoal line?

(2) What will be the result of this play?

Outline the Approach: I will use the kinematic equations

for constant acceleration. The lineman will have aconstant velocity, so no acceleration. I will determine thetime it takes for the lineman to reach the goal line and

from that determine the running backs acceleration.Then I will evaluate my answer and try to determine theresult of the play.

Describe the Physics:


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Draw physics diagrams and define all quantities uniquelyTexasRunningBack

OULineman

+ x x = 0 x = 21.03 m

Which of your defined quantities is your Target


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variable(s)?

Quantitative Relationships: Write equations you will useto solve this problem.

v2 = v1 + a (t 2 t 1)v2

2 = v12 + 2a ( x2 x1)

x2 x1 = (1/2)(v2 + v1) (t 2 t 1) x2 x1 = v1(t 2 t 1) + (1/2)a (t 2 t 1)2 x2 x1 = v2(t 2 t 1) (1/2)a (t 2 t 1)2

PLAN the SOLUTIONf ( b


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Construct Specific Equations (Same Number asUnknowns) Unknowns

We have two equations and two unknowns. The problem

is now just algebra:


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Check Units:

EXECUTE the PLAN

EVALUATE the ANSWER I A P l S d?


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Is Answer Properly Stated?

Is Answer Unreasonable?

Is Answer Complete?

Objects in a Uniform Gravitational Field


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One of the most important cases of uniformaccelerated motion is the case of objects that are near toearth which undergo free falling motion. In the late 1500sGalileo showed that even objects of different weights allfell at the same rate if air resistance is neglected.

All objects near the surface of the earth fall with aconstant acceleration of about 9.80 m/s2 which we call g,the acceleration due to gravity. Therefore, all of theequations we have derived for constant accelerationapply to an object in free fall, neglecting air resistance.

The reason that some things actually fall slower in the airis beca se air resistance p shes against the mo ing object


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is because air resistance pushes against the moving object.If there were no air resistance then even a piece of paperwould drop at the same rate as a heavy ball. Astronautson the moon, where there is no air, dropped a feather anda hammer and they actually fell at the same rate. When anobject is dropped, its velocity increases by 9.80 m/s everysecond. It continues to go faster. If there were no airresistance, this would continue every second. Becausethere is air resistance, the object eventually reaches aterminal velocity and doesnt go any faster. But for manyapplications, we can neglect air resistance. Then,everything near the surface of the earth accelerates towardthe center of the earth with a constant acceleration.

Problem: A stone is thrown upward with a speed of 10.0m/s from the top of a building 40 0 m high How long


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m/s from the top of a building 40.0 m high. How longwill it take for the stone to reach the ground?



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Ball A is dropped from a window. At the same instant, ball B is thrown downward and ball C is thrown upwardfrom the same window. Which statement concerning the

balls is necessarily true if air resistance is neglected?A) At one instant, the acceleration of ball C is zero.B) All three balls strike the ground at the same time.C) All three balls have the same velocity at any instant.D) All three balls have the same acceleration at any

instant.E) All three balls reach the ground with the same

velocity.



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If you drop a brick from a building in the absence of airresistance, it accelerates downward at 9.8 m/s2. If insteadyou throw it downward, its downward acceleration after

release isA) less than 9.8 m/s2

B) 9.8 m/s2

C) more than 9.8 m/s2D) impossible to determine with the information given


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Two balls are thrown straight up. The first is thrown withtwice the initial speed of the second. Ignore airresistance. How much higher will the first ball rise?

A) 2 times as high.B) Twice as high.C) Three times as high.D) Four times as high.E) Eight times as high

T b ll th t i ht Th fi t i th ithInteractive Question


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Two balls are thrown straight up. The first is thrown withtwice the initial speed of the second. Ignore airresistance. How much longer will it take for the first ballto return to earth?

A) 2 times as high.B) Twice as high.C) Three times as high.D) Four times as high.E) Eight times as high



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Two balls are thrown straight up. The first one takestwice as long to return to earth as the second one. Ignoreair resistance. How much faster was the first ball thrown?

A)2 times as fast.B) Twice as fast.C) Three times as fast.D) Four times as fast.E) Impossible to tell without knowing the exact times.



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Two rocks are dropped into two different deep wells. Thefirst one takes three times as long to hit bottom as thesecond one. Ignore air resistance. How much deeper is

the first well than the second?

A) 3 times as deep.

B) Three times as deep.C) Four and a half times as deep.D) Six times as deep.

E) Nine times as deep.

Problem: You are part of a citizens group evaluating thesafety of a high school athletic program. To help judge


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safety of a high school athletic program. To help judgethe diving program, you would like to know how fast adiver hits the water in the most complicated dive. Thecoach has his best diver perform for your group. Thediver jumps from the high dive and performs her routine.During her dive, she passes near a lower diving board,which is 3.0 m above the water. With a stopwatch, youdetermine that it took 0.20 seconds to enter the water fromthe time the diver passed the lower board. From this youdetermine how fast she was going when she hit the water.After you have done this, another judge wonders out loudhow high the high dive is. You quickly calculate theheight and, to the judges surprise, announce it.

Focus the Problem:


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Distance = 3 mTime = 0.20 s

What is the problem asking for:

Outline the Approach:

Describe the Physics:Draw physics diagrams and define all quantities uniquely


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p y g q q y

Target variable(s)?Quantitative Relationships:



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Unknowns

Check Units:

EXECUTE the PLAN: Calculate Target Quantity(ies)


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EVALUATE the ANSWER Is Answer Properly Stated?

Is Answer Unreasonable?

Answer Complete?


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Ball A is dropped from the top of a building. One secondlater, ball B is dropped from the same building. Neglecting air resistance, as time progresses the distance

between themA) increases.

B) remains constant.C) decreases.D) depends on the size of the balls.


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dxTaking the integral again:


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( )

( )

022

f

2

1

2

1 xvt at cvt at x

dt vat dx

dt vat dx

vat dt v

++=++=

+=

+=+==

!!

Definite Integrals


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A definite integral specifically indicates starting andending values:

( )( )

( ) ( )20010210

1712157525

72

743

2323

51

23

5

12

==

++++=

++=

++= ! x x x

dx x x y

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Chap36 revision notes on rectilinear motions