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Unformatted text preview: 2.3: AccelerationVelocity Models Vertical motion without Air Resistance m F G Recall: F = m a = m d v dt . So, with y H L = y and v H L = v , we have Force of Gravity is F G =  m g m d v dt = F G =  m g d v dt =  g . This has solutions v H t L =  g t + C 1 v H t L =  g t + v = d y dt y H t L = g 2 t 2 + v t + C 2 y H t L = g 2 t 2 + v t + x . ª Example Assume that v = 0, y = 20. In[1]:= vNoAir = v @ t D . DSolve @8 v' @ t D 9.8, v @ D < , v @ t D , t D Out[1]= 8 9.8 t < In[2]:= yNoAir = y @ t D . DSolve @8 y' @ t D vNoAir @@ 1 DD , y @ D 20 < , y @ t D , t D Out[2]= 9 20 4.9 t 2 = In[3]:= sol1 = t . Solve @ yNoAir @@ 1 DD 0, t D Out[3]= 8 2.02031, 2.02031 < 2 Chapter2Section3.nb In[4]:= Plot @8 vNoAir, yNoAir < , 8 t, 0, sol1 @@ 2 DD< , PlotStyle fi 88 Blue, Thick, Dashed < , 8 Blue, Thick << , AxesLabel fi 8 t, y < , PlotLabel fi "Blue = No Air resistance" D Out[4]= 0.5 1.0 1.5 2.0 t 20 10 10 20 y Blue = No Air resistance Chapter2Section3.nb 3 Vertical motion with Air Resistance (proportional to velocity) m F G F R Motion is DOWN m F G F R Motion is UP Recall: F = m a = m d v dt . Force of Gravity is F G =  m g Force of Air Resistance is F R =  k v where k is a constant chosen so that the force, F R , always opposes motion. (So, in the first picture, k < 0 while in the second picture, k > 0.) The Net Force if F = F G + F R =  m g k v So, with y H L = y and v H L = v , we have m d v dt = F =  m g k v d v dt =  g Ρ v where Ρ = k m and which has solution v H t L = K v + g Ρ O eΡ t g Ρ Notice that lim t z ¥ v H t L = g Ρ = v Τ . Thus, the two solutions can be written as v H t L = K v + g Ρ O eΡ t g Ρ = H v v Τ L eΡ t + v Τ y H t L = y + v Τ t + 1 Ρ H v v Τ L I 1 eΡ t M ....
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 Fall '11
 KeithEmmert
 Force, Mass, Hyperbolic function, V1

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