Calculus in Physics II

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Calculus in Physics II

x0 = 0

x axis

Knowing the car’s displacement history, find its velocity and acceleration.

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Given a displacement history, calculate the velocity history.

Average Velocity = Δ DisplacementTotal time

V = ΔxΔt

x0 = 0x axis

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V = ΔxΔt

x0 = 0x axis

= Slope

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V = ΔxΔt

x0 = 0x axis

= Slope

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V = ΔxΔt

x0 = 0x axis

= Slope

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V = ΔxΔt

x0 = 0x axis

= Slope

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V = ΔxΔt

x0 = 0x axis

= Slope

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V = ΔxΔt

x = 0x axis

≈ Slope (if Δt small enough)

x0 = 15

Slope constantly changing,V changes in time!

Find slope at time = 2.8 sec

approximately

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x = 0x axis

x0 = 15


Time x2.8 18.36 m = -4.65.8 4.56

V = ΔxΔt

≈ Slope

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V = ΔxΔt

x = 0x axis

≈ Slope

x0 = 15


Time x2.8 18.36 m = -2.63.8 15.76

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x = 0x axis

x0 = 15


Time x2.8 18.36 m = -1.72.9 18.19

Exact answer = -1.6

V = ΔxΔt

≈ Slope

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x = 0x axis

x0 = 15


Exact answer = -1.6

Differential Calculus

V = ΔxΔt

= Slope = limΔt→0

dxdt

V = ΔxΔt

≈ Slope

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Differential Calculus for PhysicsPower Rule

Given: x = a tb; where a and b are constants.

= ab tb-1dxdt

Given: y = a xb; where a and b are constants.

= ab xb-1dydx

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Find the derivative (dx/dt) ≡ Find the Velocity

x = 10 t2

x = 5 t3

x = 3

x = 3 t3 - 2 t2 + 27

x = 0

x axis

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y = 10 t2

y = 3 t3

y = 6 t5

y = 2 t3 + 8 t2 - 27

y = 0

y axis

Find the derivative (dy/dt) ≡ Find the Velocity

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Find the derivative (dy/dx):y = 4 x2

y = 2 x3+ 2x +33

y = 6 x4

y = 8 x2 + 8 x - 27

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x = a tb

x = - t2 +4 t + 15

= v = -2 t +4dxdt

(t = 2.8) = v = -2(2.8)+4dxdt

(t = 2.8) = -1.6dxdt

ftsec

Use calculus to find the velocity at 2.8 sec.

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x0 = 0x axis

Slope constantly changing,V changes in time!

V = ΔxΔt


dxdt

Use calculus to find the velocity at 10. sec.

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x0 = 0x axis

Find velocity at 10. sec

V = ΔxΔt


dxdt

x = 0.1 t2

= v = 0.2 tdxdt

(t = 10) = 2dxdt

ftsec

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If the displacement is a parabola, the velocity is linear.

If the velocity is increasing what does that say about the acceleration?

Average acceleration= Δ VelocityTotal time

a = ΔvΔt

a = ΔvΔt

= Slope of velocity = limΔt→0

dvdt

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a = ΔvΔt

= Slope of velocity = limΔt→0

dvdt

v = 0.2 t

= 0.2dvdt

ftsec2a =

Given a velocity history, calculate the acceleration history.

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Find the derivative (dv/dt) ≡ Find the Acceleration

v = 20 t

v = 15 t2

v = 0

v = 9 t2 - 4 t

x = 0

x axis

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Given a displacement history, calculate the acceleration.

Find acceleration at 10. sec

x = 0.1 t2

= v = 0.2 tdxdt

= 0.2dvdt

ftsec2a = = d2x

dt2dxdt

= d dt

Find acceleration at 20. sec

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Find second derivative (d2x/dt2) (i.e., acceleration):

x = 10 t2

x = 5 t3

x = 3

x = 3 t3 - 2 t2 + 27

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y = 10 t2

y = 3 t3

y = 6 t5

y = 2 t3 + 8 t2 - 27

Find second derivative (d2y/dt2) (i.e., acceleration):

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y = 4 x2

y = 2 x3+ 2x +33

y = 6 x4

y = 8 x2 + 8 x - 27

Find second derivative (d2y/dx2):

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Summary of Definitions

v = dxdt

dvdt

a =

dvdt

a = = d2xdt2

dxdt

= d dt

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x0 = 0

x axis

Knowing the car’s velocity history, find its displacement.

Calculus in Physics II

We will be working the earlier problems in reverse; integration is the inverse operation of

differentiation.

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Given a velocity history, calculate the displacement history.

Average Velocity = Δ DisplacementTotal time

v = ΔxΔt

x0 = 0x axis

Δx = v Δt

Know:

Therefore:

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Check out these possible solutions.

x0 = 0x axis

V = ΔxΔt


dxdt

Which curve satisfies the above definition of velocity to produce a velocity = 1?

All of them!

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Given a velocity history, calculate the displacement at t = 2 sec.

V = ΔxΔt

x0 = 0x axis

Δx = v Δt

v = (1 ft/s) (2 s) = 2 ftThe area under the curve!

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Possible solutions.

x0 = 0x axis

Which curve has a displacement of 2 ft at 2 sec?

x = t

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Given a velocity history, calculate the displacement at t = 2 sec.

Area under velocity curve is the displacement.

x0 = 0x axis

Δx = v Δt

Sum all the areas is integral calculus.

Δt = dt

dx = v dt

∫𝑑𝑥=¿∫𝑣 𝑑𝑡

∫𝑣 𝑑𝑡x =

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Integral Calculus for PhysicsReverse Power Rule

= at(b+1)

(b+1) + C

= ax(b+1)

(b+1) + C

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Find the integrals of the following ≡ Find the displacement

v = 20 t

v = 15 t2

v = 0

v = 9 t2 - 4 t

x = 0

x axis

x = 10 t2 + C

x = 5 t3 + C

x = C

x = 3 t3 - 2 t2 + C

v = 2 x = 2t + C

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Given a velocity history, displacement = 0 at start, calculate the displacement history.

x0 = 0x axis

Δt = dt

∫𝑣 𝑑𝑡x =

Given: v = 1

x = t + x0

The initial displacement at t = 0 was zero; therefore, x0 = 0

x = t

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Given a velocity history, and knowing that the initial displacement was 0 (X0 = 0), calculate the displacement history.

x0 = 0x axis

∫𝑣 𝑑𝑡x =

Given: v = 0.2 t

x = 0.1 t2 + x0

The initial displacement (x0) at t = 0 was zero; therefore, x0 = 0

x = 0.1 t2

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Verify that the area under the velocity curve is the displacement.

∫𝑣 𝑑𝑡x =

x (6 sec) = 0.5 (6 sec) (1.2 ft/sec)X (6 sec) = 3.6 ft

(6 sec, 3.6 ft)

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Given an acceleration history, find the velocity history.

v0 = 0x axis

∫𝑎𝑑𝑡v =

Average Acceleration= Δ VelocityTotal time

a = ΔvΔt

Δv = aΔt

Define:

Therefore:

dv = a dt

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Given an acceleration history, and knowing that the initial velocity was 0 (v0 = 0), calculate the velocity history.

v0 = 0x axis

∫𝑎𝑑𝑡v =

Given: a = 0.2

∫ 0.2𝑑𝑡v = = 0.2 t + v0

At t = 0, v = 0; therefore, v0 = 0.

v = 0.2 t

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Summary of Definitions

v = dxdt

dvdt

a =

∫𝒗𝒅𝒕x =

∫𝒂𝒅𝒕v =

dvdt

a = = d2xdt2

dxdt

= d dt x = ∬ 𝒂𝒅𝒕𝟐

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Constant acceleration (gravity)

v =

= -gd2xdt2

x =

v0 = v

a = -g

dxdt

= -g t + v0

-g t2

2 + v0t + x0

x0 = 0

The hardest problem will be to solve two simultaneous equations.

Calculus in Physics II

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