# solved problems on elementary kinematics by lindash

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```									Solved Problems on Elementary 1-D Kinematics

Useful Formulae :
While solving a problem in kinematics a student first carefully decide about the type
the type of motion.

Motion

A                                                    B
Uniform                                           Variable
Acceleration                                     Acceleration

A1                       A2              A3         B1               B2                   B3
Straight line   Motion in plane   Motion in 3     Straight line     Plane motion     Motion in 3
motion          Eg .Projectile    dimensions      motion            e.g., circular   dimensions
motion                                              motion

Once the nature of motion is decided, the appropriate formula/formula should be
employed to solve the problem. For certain cases, a list of formulas (not exhaustive) has
been given to help you. But remember, in case of any complication the last resort is
always the basic definitions which, in case are :
dr                                                       r
(i)      v                                            (ii)    v  
dt                                                       t
dv                                                       v
(iii)    a                                            (iv)    a 
dt                                                       t
and      (v)     Average speed = distance traveled /time
Solved Problems :

Q.1    A wind is blowing with a constant velocity v in the direction denoted by the arrow
in the figure. Two airplanes start out from a point A and fly with a constant speed
c. One flies against the wind to a point B and then returns to point A, while the
other flies in the direction perpendicular to the wind to a point C and then returns
to point A. The distances A B and A C are the same. Which plane will return to
point A first and what will be the ratio of the flight times of the two planes?

C


v

A                       B

Sol.1 If AB  AC  l, then the times of flight from A to B and form B to A are,
respectively,    l /( c  v) and l /( c  v) .

t    1      2lc
The entire flight time is t 1                2     .
c  c   c  v2

For the second airplane to fly form A, to C, its velocity must be directed at an
angle to the direction of the wind in such a manner that the resulting velocity
directed to ward C is equal to (c 2  v 2 ) 1 / 2 in magnitude. The entire flight time of
2l
this airplane will be     t2                 .
2
c  v2
The second airplane will arrive before the first, and the flight time ratio is
t 2 /t 1  1  v 2 /c 2 .
Q.2    A boat is moving across a river whose waters flow with a velocity u. The velocity
of the boat with respect to the current, v 0 , is directed at an angle  to the line
perpendicular to the current. What will be the angle  at which the boat moves
with respect to this line?


                  u
v0


What will be the velocity v of the boat with respect to the river banks? What
should be the angle at which the boat moves -directly across the current with
given u and v?

Sol.2 The figure shows that
 sin   u                u
tan   0            tan             .
 0 cos              0 cos 
Velocity v can be found form the equation
2
( v 0 sin  u) 2   0 cos 2    2 ,
which yields
2
u           u    
v  v0 1  2       sin   
v      .

v0          0    
The boat will travel directly across the river if   0. Under this condition,
sin   u / 0 . Obviously, the boat can travel at right angles to the current only if
 0 is greater then u.
Q.3    From a point A on a bank of a channel with still waters a person must get to a
point B on the opposite bank. All the distances are shown in the figure. The
person uses a boat to travel across the channel and then walks along the bank to
point B. The velocity of the boat is v 1 and the velocity of the walking person is
v 2 . Prove that the fastest way for the person to get from A to B is to select the
angles  1 and  2 in such a manner that (sin  1 /(sin  2 ) = v1 /v 2 .

A

1                        a
c

x              2              b
d
B

Sol.3 The time of travel by boat form A to C is
t 1  x 2  a 2 / 1 .
The total time of travel is
t 2  ( d  x) 2  b 2 / 2 .

The total time of travel is

x2  a2              (d  x) 2  b 2
t  t1  t2                                              .
1                    2

The extremum condition is dt / dx  0 , or
dt              x                        dz
                                                      0.
dx           x2  a2             2 (d  x) 2  b 2
1
Since
x                                       dx
 sin  1 and                               sin 2 ,
x2  a2                               (d  x) 2  b 2
sin  1 v1
we can write sin  1 / 1  sin  2 2 , whence                              .
sin  2 v2

We can easily see that the extremum corresponds to the minimum of time of travel.
Q.4    An object slides without friction down an inclined plane from a point B to a point
C that is distant a from a point A.
B

h

C                                    A

a

At what height h (or at what angle  ) is the sliding time minimal?

Sol.4 The time of travel along straight line BC is determined by the length S of segment
BC and the acceleration w. The figure shows that
h
S  a2  h2 , w                              g.
2           2
a h
2
Since S  wt / 2 , we can write
g         h
a2  h2                             t2 ,
2     2
a h      2

whence
2 (a 2  h 2 )
t                    .
e      h
Nullifying the derivative (the extremum condition),
dt      h2  a2
                       0,
dh   2 gh 3 ( a 2  h 2 )
yields h = 0.

The same result is obtained if we express S and w in terms of  :
S  a / cos  , w  g sin  ,
2       a
t                     .
g sin  . cos 
Nullifying the derivative dt / d , we find that   45.
Q.5     The time dependence of the lengths. of the paths of two bodies moving in a
straight line is given by curves a and b, respectively.
s
a

b

t
What curve corresponds to accelerated motion and what curve to decelerated
motion?

Sol.5   The acceleration in rectilinear motion is the second derivative of the distance
traveled with respect to time. For a concave cure the second derivative is
positive, while for a convex curve the second derivative is negative, whereby
curve (a) corresponds to decelerated motion and curve (b) to accelerated motion.

Q.6     A material particle is moving along a straight line in such a manner that its
velocity varies as shown in the figure.

v

1   2   3       t1   t2 4    t

At which moment in time numbered successively on the time axis will the
acceleration of the particle be maximal? How should one use the graph to
determine the average velocity of motion over the time interval from t 1 to t 2 ?

Sol.6 By definition, acceleration is the time derivative of velocity, w  dv /dt. For
rectilinear motion the vector equation can be written in scalar form. The
acceleration is the highest when the derivative is the greatest, that is, when the
curvature of the curve is maximal. The curvature is determined by the slope of
the tangent line to the particular point on the curve. This corresponds to moment
2 on the time axis.
Note that for curvilinear motion the question contains an ambiguity, since to
determine the acceleration we must known the radius of the trajectory at every
moment in the course of the motion in addition to the magnitude of the velocity.
To find the average velocity, we must known the distance traveled by the particle
in the course of a definite time interval. In terms of the velocity vs. time graph,
the distance traveled is the area of the figure bounded by the curve, the time axis,
and the vertical straight lines passing through the initial and final moments of
time on the time axis. Analytically the distance is calculated via the integral :
t0


S  v dt,
t1
t2

  dt
t1
when the average velocity is                        .
t2  t1

Q.7    The velocity of a particle moving in a straight line varies with time in such a
manner that the v vs. t curve is represented by one half of an ellipse.
v
vm

0                           t

The maximal velocity is v m and the total time of motion is t. What is the path
traversed by the particle and the average velocity over t? Can such motion
actually occur?

Sol.7 In terms of the velocity vs. time graph, the distance traveled is determined by the
area bounded by the curve and the curve and the time axis. This area is

S   mt.
4
The average velocity is
S 
   m .
t   4
Such motion cannot be realized in practical terms since at the initial and final
moments of the motion the acceleration, which is d /dt , is inifinitely large in
absolute value.
Q.8    The velocity of a particle decreases in relation to the path traversed according to
the linear law v  v0  ax .
v
A

B
0
xm                 x

After what time will the particle get to a point B that lies on the axis of abscissas
distant x m from the origin of coordinates?

Sol.8 The particle will never get to point B will approach it without bound. Indeed, form
the equation    0  ax we get
dx
 dt.
 0  ax
Integration of this expression yields
 x  0 / a 
   / a    at ,
In             
     0      
Whence
0         at
x       (1  e   ).
a
The limit value x m   0 / a can be attained only at t  . The dependence of x on
t defined by Eq. is represented by the shown in the figure.
Q.9     The velocity of a particle moving in a straight line increases according to the
linear law v  v0  kx .
V

V0

0                             X

How does the acceleration change in the course of such motion? Does it increase
or decrease or stay constant?

Sol.9 The acceleration
dw dv dx
w              k ( v0  kx)
dt dx dt
increases with x. The same result can be obtained form the following line of
reasoning: at constant acceleration the relationship between the velocity and the
distance traveled is given by the formula
2
 2   0  2 wx ,
so that the velocity increases in proportion to the square root of the distance.
Hence, for the velocity to increase linearly with x, the acceleration must increase.

Q.10 The figure shows the "timetable" of a train, the dependence of the speed of the
train on the distance traveled.
V

0                                  X
How can this graph be used to determine the average speed over the time interval
it took the train to travel the entire distance?

Sol.10 The train covers the distance dx in the course of dt  dx / ( x ), where  (x ) is the
speed with which it travels over dx. The total time of motion is
S
dx
t     (x) .
0
The average speed is determined by dividing the distance covered by the train by
the time t.
Q.11 A rod of length l leans by its upper end against a smooth vertical wall, while its
other end leans against the floor.
y

l
Vy
Vx

0                                   x
The end that leans against the wall moves uniformly downward. Will the other
end move uniformly, too?

Sol.11 The speed with which the lower end of the rod moves,  x  dx /dt , can be written
in the form
dy dx
x          .
dt dy
Since x  l 2  y 2 , we can write
dx       y
         ,
dy    l2  y2
whence
y       dy   y|v y |
x                               .
l2  y2   dt   l2  y2
Thus, the speed of the lower end gets smaller and smaller and vanishes at y = 0.
Q.12 An object is thrown upward with an initial velocity v0 . The drag on the object is
assumed to be proportional to the velocity. What time will it take the object to
move upward and what maximal altitude will it reach?

Sol.12 Since the drag is proportional to the velocity of the object, so is the acceleration
caused by this force (with a minus sign). Hence, by Newton’s second law,
dv
  g  rv,
dt
where r is the proportionality factor. Whence
                           t
d

0
  g /r
 r dt.      
0
Integration yields*
            g  rl g
   0          e  .
            r      r
For   0 this yields
1         r 
t m  ln  1  0 .
        
r                    g 
To find the maximal altitude, we rewrite in the form
dh       g        g
  0   e rt  .
dt       h        r
Integrating this equation up to t, we find that
     g1              g
h   0   (1  e rt )  t.
     rr              r
Bearing in mind that at the point of greatest ascent  = dh/dt = 0 and combining
this result with we get
     g          g
 0   e rt m  .
     r          r
Combining with yields
 0  gtm
h                      .
r
Substituting t m form we arrive at the final result
1       
g        r 0
h                 .
 0  ln  1 
 
r
     r 
      g
When drag is extremely low, or r 0 g  1, we can employ the expansion
2
    r  r   1  r          r 0   
ln  1  0   0   0
       
         .
     g    g  2 g

 g



This results in the well - known formula
2
0
h            .
2g
* The section of the curve that lies below the curve lies below the t axis (see the
figure) corresponds to the descent of the object after the objects has reached the
maximal altitude. The rate of descent asymptotically approaches the value the
force of gravity is balanced by the drag.

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