ONLINE: MATHEMATICS EXTENSION 2
Topic 6
MECHANICS
EXERCISE p6502
Consider an object of mass m falling due to gravity. The object was released with an
initially velocity v0. The resistive force due to the medium the object falls through is of
the form FR    v and directed in the opposite direction to the motion. Derive the
following results
vT 
mg

 
a     vT  v0  e    / m  t
m
v  vT   v0  vT  e    / m t
m
x  vT t     vT  v0  e    / m  t
 
 v  v0  
 m 
x      v0  v   vT log e  T

 b 
 vT  v  
Comment on the acceleration a, velocity v and displacement x as t   ?
Sketch graphs for acceleration a, velocity v and displacement x time graphs for
v0 > vT v0 = 0 and v0 < vT where v0 is the initial velocity.
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Solution
The forces acting on the object are the
gravitational force FG (weight) and the
resistive force FR. In our frame of
reference, we will take down as the
positive direction.
The equation of motion of the object is
determined from Newton’s Second Law.
ma  m
dv
 FG  FR  m g   v
dt
where a is the acceleration of the object at any instance.
The initial conditions are
t0
v  v0
 
x  0 a  g    v0
m
When a = 0, the velocity is constant v = vT where vT is the terminal velocity
0  m g   vT
vT 
mg

physics.usyd.edu.au/teach_res/hsp/math/math.htm
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We start with the equation of motion then integrate this equation where the limits of the
integration are determined by the initial conditions (t = 0 and v = v0) and final
conditions (t and v)
a
dv
mg 
 
  
 g   v   v 
dt
 
m
 m 
uv
mg

du
 
    dt  
0
u
0
u
m
t
mg

 v 

 v0  m g



v
du
 
   dt 
u
m
du  dv
u
 
   t  log e  u   m g
v0 
m

v
mg
mg

 v 
 log e 
 v0  m g











  / m t
 e




m g    / m t
  v0 
e
 
 
mg
v  vT   v0  vT  e
vT 
mg

  / m t
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
v0  0  v  vT 1  e    / m  t

v0  vT  v  vT
v0  vT  v increases to vT
v0  vT  v decreases to vT
In every case, the velocity v tends towards the limiting value vT.
Plots of the velocity v as a function of time t
m = 2.00 kg
 = 5.00 kg.s-1
g = 9.80 m.s-2
vT = 3.92 m.s-1
Initial values for velocity v0 [m.s-1]
blue: 10 red: vT
magenta: 2
cyan: 0
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The acceleration a as a function of time t is
v  vT   v0  vT  e    / m  t

dv d
  / m t

vT   v0  vT  e 
dt dt
      / m t
a   v0  vT  
e
 m 
a

 
a   vT  v0    e    / m  t
m
  vT
v0  0  a  
 m
   / m t
e

 a  g e   / m t
v0  vT  a  0
v0  vT  a  0 and decreases to 0
v0  vT  a  0 and a increases to 0
t a0
Initial values for velocity v0 [m.s-1]
blue: 10 red: vT
magenta: 2
cyan: 0
physics.usyd.edu.au/teach_res/hsp/math/math.htm
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We can now calculate the displacement x as a function of velocity t
v  vT   v0  vT  e    / m  t
v

x
0
dx
dt
dx  v dt


dx   vT   v0  vT  e    / m  t dt
t
0
t


m
x   vT t     v0  vT  e    / m  t 
 

0
m
m
x  vT t     v0  vT  e    / m  t     v0  vT 
 
 
m
x  vT t     v0  vT  1  e    / m  t
 
t
x  vT


 m

v0  0  x  vT  t    e    / m t  1 
  



Initial values for velocity v0 [m.s-1]
blue: 10 red: vT
magenta: 2
cyan: 0
physics.usyd.edu.au/teach_res/hsp/math/math.htm
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So far we have only considered the case where the initial velocity was either zero or a
positive quantity ( v0  0 ), i.e., the object was released from rest or projected downward.
We will now consider the case where the object was project vertically upward (v0 < 0).
Note: in our frame of reference, the origin is taken as x = 0, the position of the object at
time t = 0; down is the positive direction and up is the negative direction.
When the object is launched upward at time t = 0, the initial velocity has a negative
value. Let u be the magnitude of the initial velocity v0
v0  0 v0  u u  0
Therefore, the equation for the velocity v as a function of time t can be expressed as
v  vT   v0  vT  e    / m  t
v  vT   u  vT  e 
  / m t
We can now find the time tup it takes for the object to rise to its maximum height xup
above the origin (remember: up is negative). At the highest point v = 0, therefore,
0  vT   u  vT  e
   / m  tup

m
u
tup    log e  1  
 
 vT 
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The maximum height xup reached by the object in time t = tup is
m
x  vT t     v0  vT  1  e    / m  t
 



m
   / m  tup
xup  vT tup     u  vT  1  e
 

For the parameters
m = 2.00 kg  = 5.00 kg.s-1 g = 9.8 m.s-2 u = 10 m.s-1 vT = 3.92 m.s-1
The time to reach maximum height is tup = 0.507 s
The max height hup reached is hup = 2.013 m
xup = - 2.013 m
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We can find the displacement x as a function of velocity v
av
dv
 g    / m v
dx
dx 
 1 
v dv
  / m  dx  mg /   v




v dv
g    / m v
vT  mg / 
  / m  dx 
v dv
vT  v
We can integrate this equation by a substitution method or an algebraic manipulation
method.
Substitution Method
u  vT  v du  dv v  vT  u dv  du v0  vT  u0
  vT  u 
du
u
x
u  v  u
u 
v
T
0   / m  dx  u0 u du  u 0 1  uT
u0  vT  v0
  / m  dx 
  / m  x  u  vT loge  u u
u
0

 du

u
  u  u0   vT log e  
 u0 
 vT  v0 

 vT  v 
  / m  x   v0  v   vT loge 
 v  v0  
 m 
x      v0  v   vT log e  T

 b 
 vT  v  
physics.usyd.edu.au/teach_res/hsp/math/math.htm
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Algebraic manipulation
  / m  dx 
v dv
vT  v
 1
v dv
1 
 vT  

vT  v
 vT vT  v 

0   / m  dx  vT  v
x

vu
0
 1
1  
 
 dv 
 vT vT  v  
v
 v

  / m  x  vT   log e  vT  v  
 vT
 v0

 vT  v0  

 vT  v  
  / m  x    v0  v   vT log e 

 v  v0  
 m 
x      v0  v   vT log e  T

 b 
 vT  v  
QED
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For the object projected up with an initial velocity v0 = - u where u > 0, the maximum
height reached xup occurs when v = 0
 v  v0  
 m 
x      v0  v   vT log e  T

 b 
 vT  v  
 v  u 
 m 
xup      u   vT log e  T

 b 
 vT  

u
 m 
xup     vT log e  1 
 b 
 vT


  u


Note: up is negative and down is positive in our frame of reference.
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For the parameters
m = 2.00 kg  = 5.00 kg.s-1 g = 9.8 m.s-2 u = 10 m.s-1 vT = 3.92 m.s-1
The time to reach maximum height is tup = 0.507 s
The max height hup reached is hup = 2.013 m xup = - 2.013 m
physics.usyd.edu.au/teach_res/hsp/math/math.htm
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