Calculus in Physics II x 0 = 0 x axis Knowing the car’s displacement history, find its velocity and acceleration. 1

Given a displacement history, calculate the velocity history. Average Velocity = Δ Displacement Total time V = ΔxΔx ΔtΔt x 0 = 0 x axis 2

Given a displacement history, calculate the velocity history. V = ΔxΔx ΔtΔt x 0 = 0 x axis = Slope 3

Given a displacement history, calculate the velocity history. V = ΔxΔx ΔtΔt x 0 = 0 x axis = Slope 4

Given a displacement history, calculate the velocity history. V = ΔxΔx ΔtΔt x 0 = 0 x axis = Slope 5

Given a displacement history, calculate the velocity history. V = ΔxΔx ΔtΔt x 0 = 0 x axis = Slope 6

Given a displacement history, calculate the velocity history. V = ΔxΔx ΔtΔt x 0 = 0 x axis = Slope 7

Given a displacement history, calculate the velocity history. V = ΔxΔx ΔtΔt x = 0 x axis ≈ Slope (if Δt small enough) x 0 = 15 Slope constantly changing, V changes in time! Find slope at time = 2.8 sec approximately 8

Given a displacement history, calculate the velocity history. x = 0 x axis x 0 = 15 Find slope at time = 2.8 sec V = ΔxΔx ΔtΔt ≈ Slope 9

Given a displacement history, calculate the velocity history. V = ΔxΔx ΔtΔt x = 0 x axis ≈ Slope x 0 = 15 Find slope at time = 2.8 sec 10

Given a displacement history, calculate the velocity history. x = 0 x axis x 0 = 15 Find slope at time = 2.8 sec Exact answer = -1.6 V = ΔxΔx ΔtΔt ≈ Slope 11

Given a displacement history, calculate the velocity history. x = 0 x axis x 0 = 15 Find slope at time = 2.8 sec Exact answer = -1.6 Differential Calculus V = ΔxΔx ΔtΔt = Slope =lim Δt→0 dx dt V = ΔxΔx ΔtΔt ≈ Slope 12

Differential Calculus for Physics Power Rule Given: x = a t b ; where a and b are constants. = ab t b-1 dx dt Given: y = a x b ; where a and b are constants. = ab x b-1 dy dx 13

Find the derivative (dx/dt) ≡ Find the Velocity x = 10 t 2 x = 5 t 3 x = 3 x = 3 t t x = 0 x axis 14

y = 10 t 2 y = 3 t 3 y = 6 t 5 y = 2 t t y = 0 y axis Find the derivative (dy/dt) ≡ Find the Velocity 15

Find the derivative (dy/dx): y = 4 x 2 y = 2 x 3 + 2x +33 y = 6 x 4 y = 8 x x

x = a t b x = - t 2 +4 t + 15 = v = -2 t +4 dx dt (t = 2.8) = v = -2(2.8)+4 dx dt (t = 2.8) = -1.6 dx dt ft sec Use calculus to find the velocity at 2.8 sec. 17

x 0 = 0 x axis Slope constantly changing, V changes in time! V = ΔxΔx ΔtΔt = Slope =lim Δt→0 dx dt Use calculus to find the velocity at 10. sec. 18

Given a displacement history, calculate the velocity history. x 0 = 0 x axis Find velocity at 10. sec V = ΔxΔx ΔtΔt = Slope =lim Δt→0 dx dt x = 0.1 t 2 = v = 0.2 t dx dt (t = 10) = 2 dx dt ft sec 19

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 Velocity Total time a = ΔvΔv ΔtΔt ΔvΔv ΔtΔt = Slope of velocity =lim Δt→0 dv dt 20

a = ΔvΔv ΔtΔt = Slope of velocity =lim Δt→0 dv dt v = 0.2 t = 0.2 dv dt ft sec 2 a = Given a velocity history, calculate the acceleration history. 21

Find the derivative (dv/dt) ≡ Find the Acceleration v = 20 t v = 15 t 2 v = 0 v = 9 t t x = 0 x axis 22

Given a displacement history, calculate the acceleration. Find acceleration at 10. sec x = 0.1 t 2 = v = 0.2 t dx dt = 0.2 dv dt ft sec 2 a = = d2xd2x dt 2 dx dt = d Find acceleration at 20. sec 23

Find second derivative (d 2 x/dt 2 ) (i.e., acceleration): x = 10 t 2 x = 5 t 3 x = 3 x = 3 t t

y = 10 t 2 y = 3 t 3 y = 6 t 5 y = 2 t t Find second derivative (d 2 y/dt 2 ) (i.e., acceleration): 25

y = 4 x 2 y = 2 x 3 + 2x +33 y = 6 x 4 y = 8 x x - 27 Find second derivative (d 2 y/dx 2 ): 26

Summary of Definitions v = dx dt dv dt a = dv dt a = = d2xd2x dt 2 dx dt = d 27

x 0 = 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. 28

Given a velocity history, calculate the displacement history. Average Velocity = Δ Displacement Total time v = ΔxΔx ΔtΔt x 0 = 0 x axis Δx = vΔtΔt Know: Therefore: 29

Check out these possible solutions. x 0 = 0 x axis V = ΔxΔx ΔtΔt = Slope =lim Δt→0 dx dt Which curve satisfies the above definition of velocity to produce a velocity = 1? All of them! 30

Given a velocity history, calculate the displacement at t = 2 sec. V = ΔxΔx ΔtΔt x 0 = 0 x axis Δx = vΔtΔt v = (1 ft/s) (2 s) = 2 ft The area under the curve! 31

Possible solutions. x 0 = 0 x axis Which curve has a displacement of 2 ft at 2 sec? x = t 32

Given a velocity history, calculate the displacement at t = 2 sec. Area under velocity curve is the displacement. x 0 = 0 x axis Δx = vΔtΔt Sum all the areas is integral calculus. Δt = dt dx = vdt x = 33

Integral Calculus for Physics Reverse Power Rule at (b+1) (b+1) + C ax (b+1) (b+1) + C 34

Find the integrals of the following ≡ Find the displacement v = 20 t v = 15 t 2 v = 0 v = 9 t t x = 0 x axis x = 10 t 2 + C x = 5 t 3 + C x = C x = 3 t t 2 + C v = 2x = 2t + C 35

Given a velocity history, displacement = 0 at start, calculate the displacement history. x 0 = 0 x axis Δt = dt x = Given: v = 1 x = t + x 0 The initial displacement at t = 0 was zero; therefore, x 0 = 0 x = t 36

Given a velocity history, and knowing that the initial displacement was 0 (X 0 = 0), calculate the displacement history. x 0 = 0 x axis x = Given: v = 0.2 t x = 0.1 t 2 + x 0 The initial displacement (x 0 ) at t = 0 was zero; therefore, x 0 = 0 x = 0.1 t 2 37

Verify that the area under the velocity curve is the displacement. x = x (6 sec) = 0.5 (6 sec) (1.2 ft/sec) X (6 sec) = 3.6 ft (6 sec, 3.6 ft) 38

Given an acceleration history, find the velocity history. v 0 = 0 x axis v = Average Acceleration= Δ Velocity Total time a = ΔvΔv ΔtΔt Δv = aΔtΔt Define: Therefore: dv = a dt 39

Given an acceleration history, and knowing that the initial velocity was 0 (v 0 = 0), calculate the velocity history. v 0 = 0 x axis v = Given: a = 0.2 v = = 0.2 t + v 0 At t = 0, v = 0; therefore, v 0 = 0. v = 0.2 t 40

Summary of Definitions v = dx dt dv dt a = x = v = dv dt a = = d2xd2x dt 2 dx dt = d x = 41

Constant acceleration (gravity) v = = -g d2xd2x dt 2 x = v 0 = v a = -g dx dt = -g t + v 0 -g t v 0 t + x 0 x 0 = 0 The hardest problem will be to solve two simultaneous equations. 42