Chapter 2

Chapter 2

Motion in One DimensionMotion in One Dimension

2.1 position, Velocity, and Speed2.1 position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed2.2 Instantaneous Velocity and Speed 2.3 Acceleration2.3 Acceleration 2.4 Freely Falling Objects2.4 Freely Falling Objects 2.5 Kinematic Equations Derived from 2.5 Kinematic Equations Derived from

Calculus.Calculus.

KinematicsKinematics

Kinematics describes motion while ignoring Kinematics describes motion while ignoring the agents that caused the motionthe agents that caused the motion

For now, we will consider motion in one For now, we will consider motion in one dimensiondimension Along a straight lineAlong a straight line

We will use the particle modelWe will use the particle model A particle is a point-like object, that has mass but A particle is a point-like object, that has mass but

infinitesimal sizeinfinitesimal size

PositionPosition

Position is defined in Position is defined in terms of a terms of a frame of frame of referencereference For one dimension the For one dimension the

motion is generally along motion is generally along the the xx- or - or yy-axis-axis

The object’s position is The object’s position is its location with respect its location with respect to the frame of referenceto the frame of reference

Position-Time GraphPosition-Time Graph

The position-time The position-time graph shows the graph shows the motion of the motion of the particle (car)particle (car)

The smooth curve is The smooth curve is a guess as to what a guess as to what happened between happened between the data pointsthe data points

DisplacementDisplacement

DisplacementDisplacement is defined as the change in is defined as the change in position during some time intervalposition during some time interval Represented asRepresented as xx x = xx = xff - x - xii

SI units are metersSI units are meters ((mm),), xx can be positive or can be positive or negativenegative

DisplacementDisplacement is different than is different than distancedistance. . Distance is the length of a path followed by a Distance is the length of a path followed by a particle.particle.

Vectors and ScalarsVectors and Scalars

Vector quantities that need both -Vector quantities that need both -magnitude (size or numerical value) and magnitude (size or numerical value) and direction to completely describe themdirection to completely describe them We will useWe will use + + and and –– signs to indicate vector signs to indicate vector

directionsdirections Scalar quantities are completely described Scalar quantities are completely described

by magnitude onlyby magnitude only

Average VelocityAverage Velocity

The The average velocityaverage velocity is the rate at which the is the rate at which the displacement occursdisplacement occurs

The dimensions areThe dimensions are length / timelength / time [L/T][L/T] TheThe SISI units are units are m/sm/s Is also the slope of the line in the position – Is also the slope of the line in the position –

time graphtime graph

f iaverage

x xxv

t t

Average SpeedAverage Speed

SpeedSpeed is a scalar quantity is a scalar quantity same units as velocitysame units as velocity total distance / total timetotal distance / total time

The average speed is not (necessarily) The average speed is not (necessarily) the magnitude of the average velocitythe magnitude of the average velocity

Instantaneous VelocityInstantaneous Velocity

Instantaneous velocity is the limit of the Instantaneous velocity is the limit of the average velocity as the time interval average velocity as the time interval becomes infinitesimally short, or as the becomes infinitesimally short, or as the time interval approaches zerotime interval approaches zero

The instantaneous velocity indicates The instantaneous velocity indicates what is happening at every point of timewhat is happening at every point of time


The general equation for instantaneous The general equation for instantaneous velocity isvelocity is

The instantaneous velocity can be The instantaneous velocity can be positive, negative, or zeropositive, negative, or zero

0limxt

x dxv

t dt


The instantaneous The instantaneous velocity is the slope velocity is the slope of the line tangent to of the line tangent to thethe x x vsvs t t curvecurve

This would be theThis would be the greengreen lineline

The blue lines show The blue lines show that asthat as tt gets gets smaller, they smaller, they approach the green approach the green lineline

Instantaneous Instantaneous SpeedSpeed

The instantaneous speed is the The instantaneous speed is the magnitude of the instantaneous velocitymagnitude of the instantaneous velocity

Remember that the average speed is Remember that the average speed is notnot the magnitude of the average the magnitude of the average velocityvelocity

Average AccelerationAverage Acceleration

Acceleration is the rate of change of the Acceleration is the rate of change of the velocityvelocity

Dimensions areDimensions are L/TL/T22

SI units areSI units are m/s²m/s²

xf xixx

v vva

t t

Instantaneous AccelerationInstantaneous Acceleration

The instantaneous acceleration is the limit of The instantaneous acceleration is the limit of the average acceleration asthe average acceleration as tt approachesapproaches 0 0

2

20lim x x

xt

v dv d xa

t dt dt

Instantaneous AccelerationInstantaneous Acceleration

The The slopeslope of the of the velocity vs. time velocity vs. time graph is the graph is the accelerationacceleration

The The greengreen line line represents the represents the instantaneous instantaneous accelerationacceleration

The The blueblue line is the line is the average accelerationaverage acceleration

Acceleration and VelocityAcceleration and Velocity

When an object’s velocity and acceleration When an object’s velocity and acceleration are in the same direction, the object is are in the same direction, the object is speeding upspeeding up

When an object’s velocity and acceleration When an object’s velocity and acceleration are in the opposite direction, the object is are in the opposite direction, the object is slowing downslowing down


The car is moving with constant positive The car is moving with constant positive velocity (shown by red arrows maintaining the velocity (shown by red arrows maintaining the same size)same size)

Acceleration equals zeroAcceleration equals zero


Velocity and acceleration are in the same directionVelocity and acceleration are in the same direction Acceleration is uniform (blue arrows maintain the same Acceleration is uniform (blue arrows maintain the same

length)length) Velocity is increasing (red arrows are getting longer)Velocity is increasing (red arrows are getting longer) This shows positive acceleration and positive velocityThis shows positive acceleration and positive velocity


Acceleration and velocity are in opposite directionsAcceleration and velocity are in opposite directions Acceleration is uniform (blue arrows maintain the Acceleration is uniform (blue arrows maintain the

same length)same length) Velocity is decreasing (red arrows are getting Velocity is decreasing (red arrows are getting

shorter)shorter) Positive velocity and negative accelerationPositive velocity and negative acceleration

if

if

tt

vvaa

atvv if

1D motion with constant 1D motion with constant accelerationacceleration

ttff – t – tii = t = t

1D motion with constant acceleration1D motion with constant acceleration

In a similar manner we can rewrite equation In a similar manner we can rewrite equation for average velocity:for average velocity:

and than solve it for xand than solve it for x ff

Rearranging, and assumingRearranging, and assuming

if

ifavg tt

xxvv

tvxx avgif

)(2

10 favg vvv


Using Using

and than substituting into equation for final and than substituting into equation for final position yieldsposition yields

20 2

1attvxx if

atvv if

(1)

(2)

Equations (1) and (2) are the basic kinematics Equations (1) and (2) are the basic kinematics equationsequations

(1)

(2)


These two equations can be combined to These two equations can be combined to yield additional equations. yield additional equations. We can eliminate We can eliminate tt to obtain to obtain

Second, we can eliminate the acceleration a Second, we can eliminate the acceleration a to produce an equation in which acceleration to produce an equation in which acceleration does not appear:does not appear:

)(222ifif xxavv

tvvxx ff )(2

100

Kinematics with constant acceleration Kinematics with constant acceleration - Summary- Summary

xa

2vv

attvΔx

2

vvv

atvv

2i

2f

22

1i

fi

if

Kinematic Equations - summaryKinematic Equations - summary

Kinematic EquationsKinematic Equations

The kinematic equations may be used to The kinematic equations may be used to solve any problem involving one-dimensional solve any problem involving one-dimensional motion with a constant accelerationmotion with a constant acceleration

You may need to use two of the equations to You may need to use two of the equations to solve one problemsolve one problem

Many times there is more than one way to Many times there is more than one way to solve a problemsolve a problem

Kinematics - Example 1Kinematics - Example 1

How long does it take for a train to come to rest if How long does it take for a train to come to rest if it decelerates at it decelerates at 2.0m/s2.0m/s22 from an initial velocity of from an initial velocity of 60 km/h60 km/h??

Kinematics - Example 1

How long does it take for a train to come to rest if it decelerates at 2.0m/s2.0m/s22 from an initial velocity of 60 km/h60 km/h? Using we rearrange to solve

for t:

Vf = 0.0 km/h, vi=60 km/h and a= -2.0 m/s2.

tavv if

a

vv if t

s3.80.2

)3600/1000x60(0

t

A car is approaching a hill at A car is approaching a hill at 30.0 m/s30.0 m/s when its engine when its engine suddenly fails just at the bottom of the hill. The car moves suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of with a constant acceleration of –2.00 m/s–2.00 m/s22 while coasting up while coasting up the hill. (a) Write equations for the position along the slope the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking and for the velocity as functions of time, taking x = 0x = 0 at the at the bottom of the hill, where bottom of the hill, where vvii = 30.0 m/s = 30.0 m/s. (b) Determine the . (b) Determine the

maximum distance the car rolls up the hill.maximum distance the car rolls up the hill.

(a) Take at the bottom of the hill where xxii=0=0, vvii=30m/s=30m/s, a=-2m/sa=-2m/s22. Use these values in the general equation

0it

212fi ix x v t at

2 210 30.0 m s 2.00 m s

2fx t t

230.0 mfx t t

(a) Take at the bottom of the hill where xi=0, vi=30m/s, a=-2m/s2. Use these values in the general equation

0it

230.0 m s 2.00 m sfiv v at t

30.0 2.00 m sfv t

The distance of travel, xThe distance of travel, xff, , becomes a maximumbecomes a maximum,, maxx

whenwhen 0fv (turning point in the motion). (turning point in the motion). Use the expressions found in part (a) for Use the expressions found in part (a) for

0fv whenwhen t=15sec.t=15sec.

22max 30.0 m 30.0 15.0 15.0 225 mx t t

smtv f /00.200.30

Graphical Look at Motion: Graphical Look at Motion: displacement-timedisplacement-time curve curve

The slope of the The slope of the curve is the velocitycurve is the velocity

The curved line The curved line indicates the indicates the velocity is changingvelocity is changing Therefore, there is Therefore, there is

an accelerationan acceleration

Graphical Look at Motion: Graphical Look at Motion: velocity-timevelocity-time curve curve

The slope gives the The slope gives the accelerationacceleration

The straight line The straight line indicates a constant indicates a constant accelerationacceleration

The zero slope The zero slope indicates a constant indicates a constant accelerationacceleration

Graphical Look at Motion: Graphical Look at Motion: acceleration-timeacceleration-time curve curve

Freely Falling ObjectsFreely Falling Objects

A A freely falling objectfreely falling object is any object is any object moving freely under the influence of moving freely under the influence of gravity alone.gravity alone.

It does not depend upon the initial It does not depend upon the initial motion of the objectmotion of the object Dropped – released from restDropped – released from rest Thrown downwardThrown downward Thrown upwardThrown upward

Acceleration of Freely Falling ObjectAcceleration of Freely Falling Object

The acceleration of an object in free fall is The acceleration of an object in free fall is directed downward, regardless of the initial directed downward, regardless of the initial motionmotion

The magnitude of free fall acceleration isThe magnitude of free fall acceleration is g = 9.80 m/sg = 9.80 m/s22

g g decreases with increasing altitudedecreases with increasing altitude gg varies with latitude varies with latitude 9.80 m/s9.80 m/s22 is the average at the Earth’s surface is the average at the Earth’s surface

Acceleration of Free FallAcceleration of Free Fall

We will neglect air resistanceWe will neglect air resistance Free fall motion is constantly accelerated Free fall motion is constantly accelerated

motion in one dimensionmotion in one dimension Let upward be positiveLet upward be positive Use the kinematic equations withUse the kinematic equations with

aayy = g = -9.80 m/s = g = -9.80 m/s22

Free Fall ExampleFree Fall Example

Initial velocity at Initial velocity at AA is upward ( is upward (++) ) and acceleration is and acceleration is gg ( (-9.8 m/s-9.8 m/s22))

At At BB, the velocity is , the velocity is 00 and the and the acceleration is acceleration is gg ( (-9.8 m/s-9.8 m/s22))

At At CC, the velocity has the same , the velocity has the same magnitude as at magnitude as at AA, but is in the , but is in the opposite directionopposite direction

A student throws a set of keys vertically upward to her A student throws a set of keys vertically upward to her sorority sister, who is in a window sorority sister, who is in a window 4.00 m4.00 m above. The keys above. The keys are caught are caught 1.50 s1.50 s later by the sister's outstretched hand. later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were What was the velocity of the keys just before they were caught?caught?


212fi iy y v t at 24.00 1.50 4.90 1.50iv

10.0 m s upwardiv

(a)(a)


(b)(b) 10.0 9.80 1.50 4.68 m sfiv v at

4.68 m s downwardfv

A ball is dropped from rest from a height A ball is dropped from rest from a height hh above the above the ground. Another ball is thrown vertically upwards from the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a speed of the second ball if the two balls are to meet at a height height hh/2/2 above the ground. above the ground.

A ball is dropped from rest from a height A ball is dropped from rest from a height hh above the above the ground. Another ball is thrown vertically upwards from the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a speed of the second ball if the two balls are to meet at a height height hh/2/2 above the ground. above the ground.

21

12

y h gt

22

12iy v t gt 2-nd

ball:

1-st ball:1 2

hy 21

2 2h

h gt ht

g

12 2ih h h

v gg g

iv gh

A freely falling object requires A freely falling object requires 1.50 s1.50 s to travel the last to travel the last 30.0 m30.0 m before it hits the ground. From what height above the ground before it hits the ground. From what height above the ground did it fall?did it fall?


Consider the last 30 m30 m of fall. We find its speed 30 m30 m above the ground:

2

22

12

10 30 m 1.5 s 9.8 m s 1.5 s

230 m 11.0 m

12.6 m s.1.5 s

fi yi y

yi

yi

y y v t a t

v

v


Now consider the portion of its fall above the Now consider the portion of its fall above the 30 m30 m point. We point. We assume it starts from restassume it starts from rest

2 2

2 2

2 2

2

2

12.6 m s 0 2 9.8 m s

160 m s8.16 m.

19.6 m s

yf yi y fiv v a y y

y

y

Its original height was then: Its original height was then: 30 m 8.16 m 38.2 m

Motion Equations from CalculusMotion Equations from Calculus

Displacement Displacement equals the area equals the area under the velocity – under the velocity – time curvetime curve

The limit of the sum The limit of the sum is a definite integralis a definite integral

0lim ( )

f

in

t

xn n xttn

v t v t dt

Kinematic Equations – General Kinematic Equations – General Calculus FormCalculus Form

0

0

xx

t

xf xi x

x

t

f i x

dva

dt

v v a dt

dxv

dt

x x v dt

Kinematic Equations – Calculus Form Kinematic Equations – Calculus Form with Constant Accelerationwith Constant Acceleration

The integration form of The integration form of vvff – v – vii gives gives

The integration form of The integration form of xxff – x – xii gives gives

xf xi xv v a t

21

2f i xi xx x v t a t

The height of a helicopter above the ground is given by The height of a helicopter above the ground is given by h = 3.00th = 3.00t33, where , where hh is in meters and is in meters and tt is in seconds. After is in seconds. After 2.00 s2.00 s, the helicopter releases a small mailbag. How long , the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?after its release does the mailbag reach the ground?


33.00y t 2.00 st 33.00 2.00 24.0 my

29.00 36.0 m sydy

v tdt


The equation of motion of the mailbag is: The equation of motion of the mailbag is:

2 21 124.0 36.0 9.80

2 2b bi iy y v t gt t t 0by 20 24.0 36.0 4.90t t 7.96 st

Automotive engineers refer to the time rate of change of Automotive engineers refer to the time rate of change of acceleration as the "acceleration as the "jerkjerk." If an object moves in one ." If an object moves in one dimension such that its jerk dimension such that its jerk JJ is constant, (a) determine is constant, (a) determine expressions for its acceleration expressions for its acceleration aaxx(t)(t), velocity , velocity vvxx(t)(t), and , and

position position x(t)x(t), given that its initial acceleration, speed, and , given that its initial acceleration, speed, and position are position are aaxixi , , vvxixi, and , and xxii , respectively. (b) Show that, respectively. (b) Show that

)(222xixxix vvjaa

Automotive engineers refer to the time rate of change of acceleration Automotive engineers refer to the time rate of change of acceleration as the "as the "jerkjerk." If an object moves in one dimension such that its jerk ." If an object moves in one dimension such that its jerk JJ is constant, (a) determine expressions for its acceleration is constant, (a) determine expressions for its acceleration aaxx(t)(t), velocity , velocity

vvxx(t)(t), and position , and position x(t)x(t), given that its initial acceleration, speed, and , given that its initial acceleration, speed, and

position are position are aaxixi , , vvxixi, and , and xxii , respectively. (b) Show that , respectively. (b) Show that

)(222

xixxix vvjaa

(a) daJ

dt constant da Jdt

1a J dt Jt c ia a when 0t 1 ic a

ia Jt a

Automotive engineers refer to the time rate of change of acceleration as the "Automotive engineers refer to the time rate of change of acceleration as the " jerkjerk." If an ." If an object moves in one dimension such that its jerk object moves in one dimension such that its jerk JJ is constant, (a) determine is constant, (a) determine expressions for its acceleration expressions for its acceleration aaxx(t)(t), velocity , velocity vvxx(t)(t), and position , and position x(t)x(t), given that its initial , given that its initial

acceleration, speed, and position are acceleration, speed, and position are aaxixi , , vvxixi, and , and xxii , respectively. (b) Show that, respectively. (b) Show that

)(222xixxix vvjaa

(a)

22

12i i

dva

dtdv adt

v adt Jt a dt Jt at c

iv v when 0t 2 ic v 21

2 i iv Jt at v

Automotive engineers refer to the time rate of change of acceleration as the Automotive engineers refer to the time rate of change of acceleration as the ""jerkjerk." If an object moves in one dimension such that its jerk ." If an object moves in one dimension such that its jerk JJ is constant, is constant, (a) determine expressions for its acceleration (a) determine expressions for its acceleration aaxx(t)(t), velocity , velocity vvxx(t)(t), and , and position position x(t)x(t), given that its initial acceleration, speed, and position are , given that its initial acceleration, speed, and position are aaxixi , , vvxixi, and , and xxii , respectively. (b) Show that , respectively. (b) Show that

)(222xixxix vvjaa

(a)

2

3 23

12

1 16 2

i i

i i

i

dxv

dtdx vdt

x vdt Jt at v dt

x Jt at v t c

x x

when 0t 3 ic x 3 21 16 2 i i ix Jt at v t x

(b) 22 2 2 2 2i i ia Jt a J t a Jat

2 2 2 2 2i ia a J t Jat 2 2 21

22i ia a J Jt at

Recall the expression for v : 212 i iv Jt at v

212i iv v Jt at

2 2 2i ia a J v v

The acceleration of a marble in a certain fluid is The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is proportional to the speed of the marble squared, and is given (in given (in SI SI units) by units) by a = –3.00 va = –3.00 v22 for for vv > 0 > 0. If the marble . If the marble enters this fluid with a speed of enters this fluid with a speed of 1.50 m/s1.50 m/s, how long will it , how long will it take before the marble's speed is reduced to half of its take before the marble's speed is reduced to half of its initial value?initial value?

The acceleration of a marble in a certain fluid is proportional to the The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in speed of the marble squared, and is given (in SISI units) by units) by a = –3.00 va = –3.00 v22 for for vv > 0 > 0. If the marble enters this fluid with a speed of . If the marble enters this fluid with a speed of 1.50 m/s1.50 m/s, how , how long will it take before the marble's speed is reduced to half of its initial long will it take before the marble's speed is reduced to half of its initial value?value?

23.00dv

a vdt

1.50 m siv

23.00dv

vdt

2

0

3.00

1 1 1 13.00 or 3.00 .

i

v t

v v t

i i

v dv dt

t tv v v v

The acceleration of a marble in a certain fluid is proportional to the The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by speed of the marble squared, and is given (in SI units) by a = –3.00 va = –3.00 v22 for for vv > 0 > 0. If the marble enters this fluid with a speed of . If the marble enters this fluid with a speed of 1.50 m/s1.50 m/s, how , how long will it take before the marble's speed is reduced to half of its initial long will it take before the marble's speed is reduced to half of its initial value?value?

23.00dv

a vdt

1.50 m siv

2ivv

10.222 s

3.00 i

tv

A test rocket is fired vertically upward from a well. A A test rocket is fired vertically upward from a well. A catapult gives it initial velocity catapult gives it initial velocity 80.0 m/s80.0 m/s at ground level. Its at ground level. Its engines then fire and it accelerates upward at engines then fire and it accelerates upward at 4.00 m/s4.00 m/s22 until it reaches an altitude of until it reaches an altitude of 1 000 m1 000 m. At that point its . At that point its engines fail and the rocket goes into free fall, with an engines fail and the rocket goes into free fall, with an acceleration of acceleration of –9.80 m/s–9.80 m/s22. (a) How long is the rocket in . (a) How long is the rocket in motion above the ground? (b) What is its maximum motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with altitude? (c) What is its velocity just before it collides with the Earth? the Earth?

A test rocket is fired vertically A test rocket is fired vertically upward from a well. A catapult upward from a well. A catapult gives it initial velocity gives it initial velocity 80.0 80.0 m/sm/s at ground level. Its at ground level. Its engines then fire and it engines then fire and it accelerates upward at accelerates upward at 4.00 4.00 m/sm/s22 until it reaches an until it reaches an altitude of altitude of 1 000 m1 000 m. At that . At that point its engines fail and the point its engines fail and the rocket goes into free fall, with rocket goes into free fall, with an acceleration of an acceleration of –9.80 m/s–9.80 m/s22. . (a) How long is the rocket in (a) How long is the rocket in motion above the ground? (b) motion above the ground? (b) What is its maximum altitude? What is its maximum altitude? (c) What is its velocity just (c) What is its velocity just before it collides with the before it collides with the Earth? Earth?

Let point 0 be at ground level and Let point 0 be at ground level and point 1 be at the end of the engine point 1 be at the end of the engine burn. Let point 2 be the highest burn. Let point 2 be the highest point the rocket reaches and point 3 point the rocket reaches and point 3 be just before impact. The data in be just before impact. The data in the table are found for each phase the table are found for each phase of the rocket’s motion. of the rocket’s motion. (0 to 1):(0 to 1):

so so

22 80.0 2 4.00 1000fv 120 m sfv

120 80.0 4.00 t 10.0 st

A test rocket is fired vertically A test rocket is fired vertically upward from a well. A catapult upward from a well. A catapult gives it initial velocity gives it initial velocity 80.0 80.0 m/sm/s at ground level. Its at ground level. Its engines then fire and it engines then fire and it accelerates upward at accelerates upward at 4.00 4.00 m/sm/s22 until it reaches an until it reaches an altitude of altitude of 1 000 m1 000 m. At that . At that point its engines fail and the point its engines fail and the rocket goes into free fall, with rocket goes into free fall, with an acceleration of an acceleration of –9.80 m/s–9.80 m/s22. . (a) How long is the rocket in (a) How long is the rocket in motion above the ground? (b) motion above the ground? (b) What is its maximum altitude? What is its maximum altitude? (c) What is its velocity just (c) What is its velocity just before it collides with the before it collides with the Earth? Earth?

(1 to 2): 20 120 2 9.80 fix x 735 mfix x

0 120 9.80t 12.2 stThis is the time of maximum height of This is the time of maximum height of the rocket. the rocket.

A test rocket is fired A test rocket is fired vertically upward from a vertically upward from a well. A catapult gives it well. A catapult gives it initial velocity initial velocity 80.0 m/s80.0 m/s at at ground level. Its engines ground level. Its engines then fire and it accelerates then fire and it accelerates upward at upward at 4.00 m/s4.00 m/s22 until it until it reaches an altitude of reaches an altitude of 1 1 000 m000 m. At that point its . At that point its engines fail and the rocket engines fail and the rocket goes into free fall, with an goes into free fall, with an acceleration of acceleration of –9.80 m/s–9.80 m/s22. . (a) How long is the rocket in (a) How long is the rocket in motion above the ground? motion above the ground? (b) What is its maximum (b) What is its maximum altitude? (c) What is its altitude? (c) What is its velocity just before it velocity just before it collides with the Earth? collides with the Earth?

(2 to 3): (2 to 3): 2 0 2 9.80 1735fv smv f /184

184 9.80fv t 18.8 st

(a): (a):

(b):(b):

total 10 12.2 18.8 41.0 st

total

1.73 kmfix x

(c):(c):final 184 m sv

t x v a

0 Launch 0.0 0 80 +4.00

#1 End Thrust

10.0 1 000 120 +4.00

#2 Rise Upwards

22.2 1 735 0 –9.80

#3 Fall to Earth

41.0 0 –184 –9.80

An inquisitive physics student and mountain climber An inquisitive physics student and mountain climber climbs a climbs a 50.0-m50.0-m cliff that overhangs a calm pool of water. cliff that overhangs a calm pool of water. He throws two stones vertically downward, He throws two stones vertically downward, 1.00 s1.00 s apart, apart, and observes that they cause a single splash. The first and observes that they cause a single splash. The first stone has an initial speed of stone has an initial speed of 2.00 m/s2.00 m/s. (a) How long after . (a) How long after release of the first stone do the two stones hit the water? release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if (b) What initial velocity must the second stone have if they are to hit simultaneously? (c) What is the speed of they are to hit simultaneously? (c) What is the speed of each stone at the instant the two hit the water?each stone at the instant the two hit the water?

An An inquisitive physics inquisitive physics student and mountain student and mountain climber climbs a climber climbs a 50.0-m50.0-m cliff that overhangs a calm cliff that overhangs a calm pool of water. He throws pool of water. He throws two stones vertically two stones vertically downward, downward, 1.00 s1.00 s apart, apart, and observes that they and observes that they cause a single splash. The cause a single splash. The first stone has an initial first stone has an initial speed of speed of 2.00 m/s2.00 m/s. (a) . (a) How long after release of How long after release of the first stone do the two the first stone do the two stones hit the water? (b) stones hit the water? (b) What initial velocity must What initial velocity must the second stone have if the second stone have if they are to hit they are to hit simultaneously? (c) What simultaneously? (c) What is the speed of each stone is the speed of each stone at the instant the two hit at the instant the two hit the water?the water?

(a)(a)

Only the positive root is physically Only the positive root is physically meaningful:meaningful:

after the first stone is thrown.after the first stone is thrown.

2 21

1 150.0 2.00 9.80

2 2fiy v t at t t

24.90 2.00 50.0 0t t

22.00 2.00 4 4.90 50.0

2 4.90t

3.00 st

(b)(b)2

212fiy v t at

3.00 1.00 2.00 st

221

50.0 2.00 9.80 2.002iv

2 15.3 m siv downward

(c)(c)

1 1 2.00 9.80 3.00 31.4 m sfiv v at downward

2 2 15.3 9.80 2.00 34.8 m sfiv v at downward

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Chapter 2