physics 170 lecture 19 chapter 12 - kinematics sections 1-7mattison/courses/phys170/p170-19.pdf ·...

Phys 170 Lecture 19 1

Physics 170 Lecture 19

Chapter 12 - KinematicsSections 1-7

The Separate and Integrate Trick for a(s)More Tangent-Normal Examples


Non-Constant-Acceleration Kinematics

a t( ) = ddtv t( ) v T( ) = v T0( ) + a t( )dt

T0

T

∫v t( ) = d

dtx t( ) x T( ) = x T0( ) + v t( )dt

T0

T

∫

x T( ) = x T0( ) + v T0( ) + a t( )dtT0

T

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥dt

T0

T

∫

x T( ) = x T0( ) + v T0( ) T −T0[ ]+ a t( )dtT0

T

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥dt

T0

T

∫


PROBLEM 12-142 (page 69, 14th edition)


The car travels around the 150 m radius circular track with speed

15 m/s. On passing A, the car’s speed is increased by (s /2.5)m/s2

where s is the distance along the track from A in meters.

• Determine the car’s speed, acceleration and travel time when

it is 20 m past A.

• Determine the car’s travel distance, speed and acceleration

2 seconds after passing A.







it is 20 m past A.



Determine the car’s speed, acceleration and travel time when it is 20 m past A.

Determine the car’s travel distance, speed and acceleration 2 seconds after passing A.

Acceleration as Function of Position

(You don’t need to know this derivation, just the result of it.)

Start from the definition of acceleration, use the chain rule, and the definition of velocity.

Now make mathematicians’ heads spin by sayingand integrating both sides


a s( ) = ddtv s( ) = dv

ds⋅dsdt

=dvds

⋅v s( )

a ⋅ds = v ⋅dv

adss1

s2

∫ = vdvv s1( )

v s2( )

∫ =v2 s2( )

2−v2 s1( )

2 or ads∫ +C =

v2 s( )2








it is 20 m past A.



a s( ) = s2.5

=sA v s = 0( ) = 15 = B

ads∫ +C =v2 s( )

2sAds∫ +C =

s2

2A+C =

v2 s( )2

v = B at s = 0 →02

2A+C =

B2

2→ C =

B2

2s2

2A+B2

2=v2

2→ v = s2

A+ B2








it is 20 m past A.



Determine the car’s speed, acceleration and travel time when it is 20 m past A.A = 2.5 B = 15 s = 20

v = s2

A+ B2 =

202

2.5+152 = 19.62 m/s

atangent s( ) = sA=

202.5

= 8.00 m/s2

That's the tangent acceleration only.There's also "normal" acceleration

anormal =v2

r=

19.622

150= 2.566m/s2

But how do we get the timeto go 20 meters?

v(s) to s(t)

We can use the separate-and-integrate trick for that too.


v s( ) = dsdt

t s( ) = F s( ) +C

s t( ) = Inverse t = dsv s( )∫ +C

⎡

⎣⎢

⎤

⎦⎥

→ dt = dsv s( )

→ dt∫ =dsv s( )∫ → t = F s( ) +C

This expression gives time as a function of position:

If we need position as a function of time, we just invert it.

We still need to find C from other conditions in the problem, either before or after the inversion.


v = dsdt

=s2

A+ B2 →

dss2

A+ B2

∫ +C = dt∫ = t s( )

dss2

A+ B2

∫ = A ⋅ ln s + A ⋅s2 + AB2

A

⎛

⎝⎜

⎞

⎠⎟

= A ⋅ ln s + s2 + AB2( )t s( ) = A ⋅ ln s + s2 + AB2( ) +Cs = 0 at t = 0 → 0 = A ⋅ ln 0 + 02 + AB2( ) +C→ C = − A ⋅ ln A ⋅B( ) = − 2.5 ⋅ ln 2.5 ⋅15( ) = −5.0062


t s( ) = A ⋅ ln s + s2 + AB2( )− 5.0062

t s = 20( ) = 2.5 ⋅ ln 20 + 202 + 2.5 ⋅152( )− 5.0062

= 1.2113 seconds








it is 20 m past A.









it is 20 m past A.



Determine the car’s speed, acceleration and travel time when it is 20 m past A.

Determine the car’s travel distance, speed and acceleration 2 seconds after passing A.

We have t(s).We can solve numericallyfor distance s at t = 2 secondsThen plug that s back in above.


t s( ) = A ⋅ ln s + s2 + AB2( ) +C2 = 2.5 ⋅ ln s + s2 + 2.5 ⋅152( )− 5.0062 → s = 38.661 m

v = s2

A+ B2 =

38.6612

2.5+152 = 28.6857 m/s

atangent =sA=

38.6612.5

= 15.4644 m/s2

anormal =v2

ρ=

28.68572

150= 5.4858 m/s2


t s( ) = A ⋅ ln s + s2 + AB2( )− A ⋅ ln 0 + 02 + AB2( )= A ⋅ ln s + s2 + AB2

0 + 02 + AB2

⎛

⎝⎜

⎞

⎠⎟ = A ⋅ ln s

B A+

s2

AB2 +1⎛

⎝⎜

⎞

⎠⎟

Define u =s

B A and T =

tA

T = ln u + u2 +1( )→ eT = u + u2 +1 → eT − u = u2 +1

e2T − 2eTu + u2 = u2 +1→ u =e2T −1

2eτ=eT − e−T

2= sinhT

sB A

= sinh tA→ s t( ) = B A sinh t

A


ddzsinh z = d

dzez − e− z

2=ez + e− z

2= cosh z

ddzcosh z = d

dzez + e− z

2=ez − e− z

2= sinh z

s t( ) = B A sinh tA

v t( ) = ddts t( ) = B A d

dtsinh t

A= B A 1

Acosh t

A= Bcosh t

A

atangent t( ) = ddtv t( ) = B d

dtcosh t

A=

BAsinh t

A

Position, Velocity, Acceleration vs Time


s t( ) v t( ) a t( )

ddt

s

v dt∫

ddt

v

a dt∫

Position, Velocity, Acceleration vs Position


t s( ) v s( ) a s( )

a = v ⋅ dvds

v = dsdt

=1

dt ds

a s( )ds∫ = v2

2+C

v = 2 a s( )ds∫ −C( )t = ds

v s( )∫ +C

Phys 170 Lecture 19

s t( ) v t( ) a t( )

v = dsdt

s = vdt∫ +C

a = dvdt

v = adt∫ +C

a = v ⋅ dvds

a s( )ds∫ = v2

2+C

v = 2 a s( )ds∫ −C( )

v s( ) a s( )t s( )

t = dsv s( )∫ +C

v = dsdt

= 1dt ds

s t( )↔ t s( ) s t( )↔ t s( ) s t( )↔ t s( )

The Map


For Next Time

Homework 6 is due at 10 PM tonight.

No homework will appear at 6 PM.

Have a nice break!

physics 170 lecture 19 chapter 12 - kinematics sections 1-7mattison/courses/phys170/p170-19.pdf ·...

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