AccelerationVelocity Models

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Acceleration-Velocity Models Runge-Kutta Method

• 10-12-2006

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Acceleration-Velocity Models

• Newtons 2nd Law

The rate of change of the momentum of a body is
directly proportional to the net force acting on
it, and the direction of the change in momentum
takes place in the direction of the net force.
F m a
force
mass
acceleration
1726 by John Venderbank
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Gravity

• dv/dt v(t) a
• m v F
• The gravity acceleration g9.8m/s
• The force of gravity Fg -m g
• m v Fg -m g
• v -g
• General Solution v(t) -g t C , C is
  constant
• Given v0 v(0), we get the particular solution
• v(t) -g t v0

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Example 1

• A crossbow bolt is shot straight upward from the
  ground (h(0)h00) with initial velocity v049
  (m/s).
• v -g -9.8
• v(t) -9.8 t v0 -9.8 t 49.
• The bolts height h(t) v(t).
• h(t) ? v(t) dt ? (-9.8 t 49) dt
• -4.9t249th0 -4.9t249t -4.9
  t(t-10)
• Q1 0V(T)-9.8 T 49, so T5 (s)
• ? After 5 seconds, the velocity is zero.
• Q2 hmax h(T) h(5) -4.95(5-10)122.5 (m)
• ? The maximum height is 122.5 meter.
• Q3 0y(t)-4.9 t(t-10), so t 10 (s)
• ? After 10 seconds, the bolt returns to the
  ground.

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Matlab

• t1 00.110
• h1 -4.9t1.2 49t1
• plot(t,h1)

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Resistance 1

• Resistance proportional to velocity Fr - k v
• F Fg Fr - m g k v
• mv-mg-kv
• So, v - g p v, where p k/m is constant.
• This equation is a separable first-order DE, the
  solution is
• v(t) (v0 g/p) exp(-pt) g/p
• When t -gt 8, v -g/p.
• Therefore, there is a terminal speed V g/p.
• The height h(t) v(t), so the general
  solution
• h(t) -( v0 g / p) exp(-pt) / p - g t / p
  C
• h0h(0) gives us C h0 ( v0 g / p )/ p
• Thus, h(t) ( v0 g / p) (1 - exp(-pt) ) / p -
  g t / p h0

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Example 2

• Same problem as example 1, but we take the air
  resistance into account, with p0.04.
• Q1 Maximum height ?
• Q2 When it will return to the ground?
• H00, v049, p0.04, so
• v(t) 294 exp(-4/25) 245.
• h(t) 7350 - 245 t - 7350 exp(-t/25).
• 0v(t) 294 exp(-4/25) 245, so t25
  ln(294/245)4.558 (s)
• Maximum height hmax(tm) 108.28 (m)
• 0y(t) 7350 - 245 t - 7350 exp(-t/25)
• MATLAB eval(solve('7350-245t-7350exp(-t/25)0'
  ,'t'))
• Result t 9.411 (s)

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Matlab

• t2 00.110
• h27350 - 245.t2 - 7350 exp(-t2./25)
• plot(t2,yh)

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Example 1 and 2 together

• t 00.110
• h1 -4.9t.2 49t
• h27350 - 245.t - 7350 exp(-t./25)
• plot(t,h1,r,t,h2,b)

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Resistance 2

• Resistance proportional to the square of the
  velocity
• Fr k v2 - k v v
• Then, m v Fg Fr - m g k v v
• v(t) -g p v v, p k/m
• Upward
• v(t) - g p v2 , h(t) v(t)
• Solutions for velocity and height?
• Downward
• v(t) -g p v2 , h(t) v(t)
• Solutions for velocity and height?

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Example 3

• Same as example 1, but we assume the air
  resistance proportional to the square of the
  velocity, with p 0.0011.
• Upward
• dsolve('Dv-g-pv2','v(0)v0','t')
• OR dsolve('Dv - 9.8 -0.0011v2','v(0)49','t'
  )
• fuinline(??','t',h')
• t3u,h3uode45(fu,0,?,0)
• Downward
• dsolve('Dv-gpv2','v(0)v0','t')
• OR dsolve('Dv - 9.8 0.0011v2','v(0)49','t'
  )
• fdinline(??','t',h')
• t3u,h3uode45(fd,?,?,?)

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Examples 1, 2 and 3 together

• plot(t,h1,'r',t,h2,'m',t3u,h3u,'k',t3d,h3d,'b')

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Runge-Kutta Method

• dy/dx f(x,y), with IC y(x0) y0
• K1 f(Xn,Yn)
• K2 f(Xn0.5h, Yn0.5hk1)
• K3 f(Xn0.5h, Yn0.5hk2)
• K4 f(Xn1, Ynhk3)
• Yn1 Yn (k12k22k3k4)h/6.

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Exercises

• dy/dx x y, y(0) 1
• dy/dx x2 y2, y(0) 1
• dy/dx 5y 6 exp(-x), y(0) 1
• Compare with Eulers method?

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• Project 2 on the web (DUE next Thurs, Oct.19)
• http//math.asu.edu/jliu/MAT275/