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ONLINE: MATHEMATICS EXTENSION 2
Topic 6
6.4
MECHANICS
RESISTIVE MOTION IN THE
HORIZONTAL DIRECTION
Many problems in the mathematical analysis of particles moving under the influence of
resistive forces, you start with the equation of motion. Then to find from the forces or
acceleration you must then integrate the equation of motion to find velocities as
functions of time and /or displacement and displacements as functions of time and /or
velocity.
To be successful in this topic you need to be very familiar with the exponential and log
functions and be competent in integrating functions.
BASIC KNOWLEDGE (need to know this information very well)
log e  y   ln  y 
log e  A1 A2   log e  A1   log e  A2 
log e  A1 / A2   log e  A1   log e  A2 
log e  A  n log e  A
n
y  A1e A2 x
log e  y   ln  y 
log e  y   log e  A1e A2 x   log e  A1   A2 x
dx
1
 a  b x  b log  a  b x   C
e

f '( x ) dx
 log e  f ( x )   C
f ( x)
dx
1
1
x
x
 tan 1    C  a tan    C
2
x
a
a
a
a
dx
1
ax
 a 2  x 2  2a loge  a  x   C
a
2
dx (t )
x (t )   v (t ) dt
dt
dv (t )
a (t ) 
v (t )   a (t ) dt
dt
dv (t )
dv ( x )
v
v
a (v ) 
v
dx 
dv
x
dv
dt
dx
a (v )
a (v )
dv (t ) d 1 2
a( x) 
 2 v 
d  12 v 2    a ( x ) dx
dt
dx
v (t ) 
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tan  A  B  
tan A  tan B
1  tan A tan B
tan  A  B  
tan A  tan B
1  tan A tan B
tan( A)  z A= atan  z   tan 1  z  tan 1  tan( A)   A
MOTION UNDER THE INFLUENCE OF A RESISTIVE FORCE
In many mathematics textbook capital letters are used for forces, such as T for tension,
G for gravity. In physics terms this not the best notation to use. The best use of symbols
for force is the use F with a subscript to identify the type of force. For example
gravitational force (weight) FG
string tension FT
resistive force FR
Consider a particle of mass m acted upon by an applied force FA and a resistive force FR.
The resistive force FR is one that opposes the motion, i.e., the direction of the resistive
force FR is opposite to the velocity v of the particle. For one-dimensional motion, the
direction of a vector is indicted as a positive or negative number with respect to a
chosen frame of reference.
Fig. 1. Forces acting on a particles of mass m are an applied force FA and a resistive
force FR opposing the motion of the object. The frame of reference is the + X axis
pointing to the right.
The equation of motion of the particle can be derived from Newton’s Second Law
(1)
(2)
a
a
1
m
F
i
Newton’s Second Law
i
FA  FR
m
Usually the resistive force FR is a function of velocity v. Two very important examples
are
(3)
FR    v
(4)
FR   v 2
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Most HSC exam questions on resistive motion focus on “lots” of algebraic
manipulations and not on physical interpretations of the motions. Your best approach to
this Topic is to do many exercises to get experience in solving HSC style questions.
Example (HSC 2012 Question 13a)
An object on the surface of the water of a liquid is released at time t = 0 and
immediately sinks. Let x be its displacement in metres in a downward direction from the
surface of the liquid at time t seconds.
The equation of motion is given by
dv
v2
a
 10 
dt
40
where v is the velocity of the object.
Show that
v
20  et  1
e 1
t
and
 400 
x  20log e 
2 
 400  v 
How far does the object sink in the first 4 s?
Solution
We can solve the problem in two different ways (1) by substitution / differentiation and
(2) integration
Approach 1 substitution
(1)
v
2
20  et  1
et  1
e
 400
e
t
 1
2
(2)
v
(3)
dv
v 2 400  v 2
 10 

dt
40
40
t
 1
2
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Differentiate (1) w.r.t. t
v
20  et  1
 20  et  1 et  1
e 1
t
1
1
2
dv
  20   et  et  1   20   et  1 et  et  1
dt
1
1
dv
  20   et  et  1 1   et  1 et  1
dt
t
t
1   e  1   e  1 
dv
t
t
  20   e  e  1 


dt
 et  1 



t
t
1   e  1   e  1 
dv
t
t
  20   e  e  1 

t


dt
e

1




(4)
dv
40 et

dt  et  12
Substitute (2) into (3)
10
dv

dt  et  12
e  1  e  1 
t
2
t
2
10 e2t  2et  1  e2t  2et  1
dv



dt  et  12
(5)
t
dv  40   e 

dt  et  12
same result as equation (4)
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Approach 2 integration
dv
v 2 202  v 2
 10 

dt
40
40
dt
dv
 2
40 20  v 2
t dt
v
dv

0 40 0 202  v 2
a
2
dx
1
ax

log e 

2
x
2a
ax
t  1 
 20  v 
   log e 

40  40 
 20  v 
 20  v 
et  

 20  v 
et  20  v    20  v 
v  et  1  20  et  1
v
20  et  1
e
t
 1
QED
Find the equation for x
dv
dv
v 2 400  v 2
a
v
 10 

dt
dx
40
40
dx
v dv

40 400  v 2
x dx
v
v dv

0 40 0 400  v 2
v
x  1 

   log e  400  v 2 
40  2 
0
x  20 log e  400  v 2   log e  400  
 400 
x  20 log e 
2 
 400  v 
QED
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v and x at time t = 4 s
v
20  et  1
et  1

20  e4  1
e4  1
 400 
x  20log e 
2 
 400  v 
2
  e 4  12 
 e4  1 

400  v  400  400  4
  400 1  4
2
e

1




  e  1 
2
 e8  2e 4  1  e 8  2e 4  1 
 4e 4 



400  v  400
 400 
2
2
4
4




 e  1


  e  1 
2
4
 400    e  1 



2 
2
 400  v   2e 
  e 4  1 
x  20 log e 

 2e 2 


2
2
 e4  1 
x  40log e 
2 
 2e 
QED
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EXAMPLE (Syllabus)
A particle of unit mass and velocity v moves in a straight line against a resistive force
FR where
FR    v  v 3 
Initially the particle is at the origin and travelling with a velocity v0 in a positive
direction.
(a)
Show that the displacement x and velocity v are related by the equation
 v v 
x  atan  0

 1  v0 v 
(b)
atan  tan 1
Show that the time t which has elapsed when the particle is travelling with
velocity v is given by
 v02 1  v 2  
t  loge  2

2
 v 1  v0  
1
2
(c)
Find v2 as a function of t.
(d)
Finding the limiting values of v and x as t  .
Solution
(a)
Equation of motion
a
dv
dv
v
  v  v3 
dt
dx
We can find x as a function of v by integrating the equation of motion. The integration
limits are determined by the initial conditions and final values of x and v.
v dv
dv

3
v  v 1  v2
x
v
dv
  dx  
0
v0 1  v 2
dx 
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Using the standard integral
a
2
dx
1
1
x
 x
 tan 1    C  a tan    C
2
x
a
a
a
a
a=1
 x  atan  v   v
v
0
x  atan  v0   atan  v 
Let A  atan  v0  and B  atan  v  and then take the tangent of x and using the facts that
tan( A  B ) 
tan A  tan B
1  tan A tan B
tan( A)  z A= atan  z   tan 1  z  tan 1  tan( A)   A
tan  x  
v0  v
1  v0 v
 v v 
x  atan  0

 1  v0 v 
QED
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(b)
Equation of motion
a
dv
  v  v3 
dt
We can find v as a function of t by integrating the equation of motion. The integration
limits are determined by the initial conditions and final values of t and v.
dv
a
  v  v3 
dt
dv
dv
v 
1
dt 

  
dv
3
2 
2
vv
v 1  v 
 v 1 v 
v
v
1
dt

0 v0  1  v 2  v  dv
t
t   12 log e 1  v 2   log e  v  
v
v0
t  12 log e 1  v 2   2 log e  v  
v0
v
v

 1  v 2 
t  log e  2  
 v   v0

1
2

 1  v0 2  
 1  v2 
t  12 log e  2   log e 

2
 v 
 v0  

 v0 2 1  v 2  
t  log e  2

 v 1  v0 2  


1
2
QED
(c)
The can rearrange the expression for t as a function v2 to find v2
 v0 2 1  v 2  
t  12 log e  2

 v 1  v0 2  


e 
2t
v0 2 1  v 2 
v 2 1  v0 2 
e 2 t v 2 1  v0 2   v0 2 1  v 2 


v2 
v0 2
e 2 t 1  v0 2   v0 2
v 2 e 2 t 1  v0 2   v0 2  v0 2
QED
(d)
t   e2t   v  0 x  atan  v0 
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