Rectilinear Motion Revisited

Objective: We will look at rectilinear motion using Integration

Position and Velocity

• Remember the relationships between position and velocity and velocity and acceleration:

)()(/ tvts )()(/ tatv

Position and Velocity

• Remember the relationships between position and velocity and velocity and acceleration:

• We will now look at these relationships using Integration.

)()(/ tvts )()(/ tatv

)()( tsdttv )()( tvdtta

Position and Velocity

• If we know the velocity function of a particle in rectilinear motion, then by Integrating we can produce a family of position functions from that velocity function. In addition, if we know the position of the particle at any time t, then we are able to find the constant of Integration and find a unique position function.

Velocity and Acceleration

• Similarly, if we know the acceleration function of the particle, then by Integrating we can produce a family of velocity functions. Also, if know the velocity at any time, then we can solve for the constant of Integration to produce a unique velocity function.

Example 1

• Suppose that a particle moves with velocity v(t) = cost along a coordinate line. Assuming that the particle has coordinate s = 4 at time t = 0, find its position function.

Example 1

• Suppose that a particle moves with velocity v(t) = cost along a coordinate line. Assuming that the particle has coordinate s = 4 at time t = 0, find its position function.

• The position function is Cttdtdttvts

sin1

cos)()(

Example 1

• Suppose that a particle moves with velocity v(t) = cost along a coordinate line. Assuming that the particle has coordinate s = 4 at time t = 0, find its position function.

• The position function is

• Since s = 4 when t = 0, we have

Cttdtdttvts

sin1

cos)()(

C )0(sin1

4 4)(sin

1)( tts

Displacement

• Recall that displacement over a time interval of a particle in rectilinear motion is its final coordinate minus its initial coordinate. In other words, it is the difference of where it started from where it finished.

• Remember, Integration gives the amount of change.

1

0

)()()( 01

t

t

tstsdttvntdisplaceme

Distance Traveled

• In contrast, to find the distance traveled by the particle over the time interval [t0 , t1 ], we must Integrate the absolute value of the velocity function.

1

0

|)(|.t

t

dttvtraveleddist

Displacement vs. Distance Traveled

• Since the absolute value of velocity is speed, we can summarize the last two equations by the following:

• Integrating velocity over a time interval produces displacement, and integrating speed over a time interval produces distance traveled.

Example 2

• Suppose that a particle moves on a coordinate line so that its velocity at time t is v(t) = t2 – 2t m/s.

a) Find the displacement of the particle during the time interval 0 < t < 3.

b) Find the distance traveled by the particle during the time interval 0 < t < 3.

Example 2

• Suppose that a particle moves on a coordinate line so that its velocity at time t is v(t) = t2 – 2t m/s.

a) Find the displacement of the particle during the time interval 0 < t < 3.

• Displacement is the value of the derivative.

03

)2()(3

0

3

0

23

23

0

t

tdtttdttv

Example 2

• Suppose that a particle moves on a coordinate line so that its velocity at time t is v(t) = t2 – 2t m/s.

b) Find the distance traveled by the particle during the time interval 0 < t < 3.

• To find the distance traveled, we need to find the zeros of the velocity function and solve two different integrals.

• v(t) = t(t – 2)• |__-__|__+__

3

0

3

2

2

0

)()(|)(| dttvdttvdttv

0 2

Example 2

• Suppose that a particle moves on a coordinate line so that its velocity at time t is v(t) = t2 – 2t m/s.

b) Find the distance traveled by the particle during the time interval 0 < t < 3.

• Distance Traveled is: 2

0

3

2

22 )2()2( dtttdttt

3

8

3

4

3

4

33

3

2

232

0

23

t

tt

t

Velocity vs. Time Curve

• 6.7.1 FINDING DISPLACEMENT AND DISTANCE TRAVELED FROM THE VELOCITY VS. TIME CURVE.

• For a particle in rectilinear motion, the net signed area between the velocity versus time curve and the interval [t0

, t1] on the t-axis represents the displacement of the particle over that time interval, and the total area between the velocity versus time curve and the interval [t0 , t1] on the t-axis represents the distance traveled by the particle over that time interval.

Example 3

• Look at the following velocity vs. time curves for a particle in rectilinear motion along a horizontal line with the positive direction to the right. For each graph, find the displacement and distance traveled for the particle over the interval 0 < t < 4.

Example 3

• Look at the following velocity vs. time curves for a particle in rectilinear motion along a horizontal line with the positive direction to the right. For each graph, find the displacement and distance traveled for the particle over the interval 0 < t < 4.

• Disp. = 2 Disp. = -2 Disp. = 0• D.T. = 2 D. T. = 2 D. T. = 2

Uniformly Accelerated Motion

• An important case of rectilinear motion occurs when a particle has constant acceleration. We call this uniformly accelerated motion. We will now see how to find the equations necessary to solve these problems.

Uniformly Accelerated Motion

• Suppose the particle has acceleration a(t) = a and the initial conditions

when when

• We will integrate acceleration to find velocity.

• We know that the v0 = 0 at t = 0, so C1 = v0.

0

0

t

t

0

0

vv

ss

1Catadt

Uniformly Accelerated Motion

• Suppose the particle has acceleration a(t) = a and the initial conditions

when when

• We will now integrate velocity to find position.

• We know that the s0 = 0 at t = 0, so C2 = s0.

0

0

t

t

0

0

vv

ss

20

2

0 2)( Ctv

atdtvat

Uniformly Accelerated Motion

• 6.7.2 Uniformly Accelerated Motion• If a particle moves with constant acceleration a along

an s-axis, and if the position and velocity at time t = 0 are s0 and v0 , respectively, then the position and velocity functions of the particle are

200 2

1)( attvsts

atvtv 0)(

Example 4

• Suppose that an intergalactic spacecraft uses a sail and the “solar wind” to produce a constant acceleration of .032 m/s2. Assuming that the spacecraft has a velocity of 10,000 m/s when the sail is first raised, how far will the spacecraft travel in 1 hour, and what will its velocity be at the end of this hour?

032.

/000,10

0

0

0

a

smv

s

Example 4

• Suppose that an intergalactic spacecraft uses a sail and the “solar wind” to produce a constant acceleration of .032 m/s2. Assuming that the spacecraft has a velocity of 10,000 m/s when the sail is first raised, how far will the spacecraft travel in 1 hour, and what will its velocity be at the end of this hour?

032.

/000,10

0

0

0

a

smv

s 2)032(.2

1)(000,100)( ttts

ttv )032(.000,10)(

Example 4

• Suppose that an intergalactic spacecraft uses a sail and the “solar wind” to produce a constant acceleration of .032 m/s2. Assuming that the spacecraft has a velocity of 10,000 m/s when the sail is first raised, how far will the spacecraft travel in 1 hour, and what will its velocity be at the end of this hour?

032.

/000,10

0

0

0

a

smv

s ms 000,200,36)3600)(032(.2

1)3600(000,100)3600( 2

smv /100,10)3600)(032(.000,10)3600(

Example 5

• A bus has stopped to pick up riders, and a woman is running at a constant velocity of 5 m/s to catch it. When she is 11m behind the front door the bus pulls away with a constant acceleration of 1 m/s2. From that point in time, how long will it take for the woman to reach the front door of the bus if she keeps running with a velocity of 5 m/s?

1

0

11

0

0

b

b

b

a

v

s

0

5

00

w

w

w

a

v

s

Example 5

• A bus has stopped to pick up riders, and a woman is running at a constant velocity of 5 m/s to catch it. When she is 11m behind the front door the bus pulls away with a constant acceleration of 1 m/s2. From that point in time, how long will it take for the woman to reach the front door of the bus if she keeps running with a velocity of 5 m/s?

1

0

11

0

0

b

b

b

a

v

s

0

5

00

w

w

w

a

v

stsw 50

212

1011 ttsb

Example 5

• A bus has stopped to pick up riders, and a woman is running at a constant velocity of 5 m/s to catch it. When she is 11m behind the front door the bus pulls away with a constant acceleration of 1 m/s2. From that point in time, how long will it take for the woman to reach the front door of the bus if she keeps running with a velocity of 5 m/s?

st

st

t

7.6

3.3

35

02210

512

111

2

2

tt

tt

Free-Fall Model

• Motion that occurs along a vertical line is called free-fall motion. We will assume that the only force acting on the object is gravity, and since the object stays sufficiently close to the Earth, the force is constant.

• It is a fact of physics that a particle with free-fall motion has constant acceleration. This acceleration is

2

2

/32

/8.9

sft

sm

Free-Fall Model

• The equations for free –fall are a little different than motion along a line.

200

0

2

1)(

)(

)(

gttvsts

gtvtv

gta

Example 6

• An object is launched from a height of 7 ft with an upward velocity of 100 ft/s. How high does the object go and how long does it take to reach this peak? How long does it take to return to the launchpad?

2

0

0

/32

/100

7

sfta

sftv

s

Example 6

• An object is launched from a height of 7 ft with an upward velocity of 100 ft/s. How high does the object go and how long does it take to reach this peak? How long does it take to return to the launchpad?

ttv

ttts

32100)(

161007)( 2

2

0

0

/32

/100

7

sfta

sftv

s

Example 6

• An object is launched from a height of 7 ft with an upward velocity of 100 ft/s. How high does the object go and how long does it take to reach this peak? How long does it take to return to the launchpad?

• The object will reach its peak when the velocity is zero.

st

t

8/25

321000

fts 25.163)8/25(16)8/25(1007)8/25( 2

Example 6

• An object is launched from a height of 7 ft with an upward velocity of 100 ft/s. How high does the object go and how long does it take to reach this peak? How long does it take to return to the launchpad?

• The object will return to the launchpad when its position is 7.

• The time it takes to go up is always equal to the time it takes to come down.

st

tt

8/50

1610077 2

Example 7

• A penny is released from rest near the top of the Empire State Building at a point that is 1250 ft above the ground. Assuming that the free-fall model applies, how long does it take for the penny to hit the ground, and what is its speed at the time of impact?

sfta

v

fts

/32

0

1250

0

0

Example 7

• A penny is released from rest near the top of the Empire State Building at a point that is 1250 ft above the ground. Assuming that the free-fall model applies, how long does it take for the penny to hit the ground, and what is its speed at the time of impact?

sfta

v

fts

/32

0

1250

0

0

2)32(

2

101250)( ttts

Example 7

• A penny is released from rest near the top of the Empire State Building at a point that is 1250 ft above the ground. Assuming that the free-fall model applies, how long does it take for the penny to hit the ground, and what is its speed at the time of impact?

sfta

v

fts

/32

0

1250

0

0

2)32(

2

101250)( ttts

21612500 t st 8.8

8.8t

Example 7

• A penny is released from rest near the top of the Empire State Building at a point that is 1250 ft above the ground. Assuming that the free-fall model applies, how long does it take for the penny to hit the ground, and what is its speed at the time of impact?

sfta

v

fts

/32

0

1250

0

0

sfttspeed /8.282|)8.8(32||320|

|| velocityspeed

Homework

• Pages 416-418

• 1-17 odd

• 29, 35, 37, 39, 41