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Mechanics Kinematics and Newton's Laws

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Mechanics Kinematics and Newton's Laws Powered By Docstoc
					Kinematics
Distance versus Displacement
      distance = 7 blocksblocks east
        displacement= 3 blocks
           distance = 7
            distance 10
   displacement = 3=blocks south

   A            7 blocks         B
             23

                                  3 blocks
          7.6 blocks

                                 C

         displacement = 7.6 blocks, 23 south of east
Definition
The displacement is a vector that
points from an object’s initial
position toward its final position
and has a magnitude that equals
the shortest distance between
the two positions.
SI Unit - meter (m)
Speed versus Velocity


   A              7 blocks          B

  leave:                       arrive:
  9:20 AM                      9:31 AM
                              displacement
         average velocity 
       average speed 
                         distance
                       elapsedelapsed time
                               time
                               7
                        7 blocks blocks
                                        east
                              11 minutes
                       11 minutes
                      0.64  0.64 blocks/minute, east
                            blocks/min ute
                              0.64 blocks/min
Definition
The average velocity is a vector.
Its magnitude is the
displacement divided by the
elapsed time.
                    x x0
                 v
                     t  t0
SI Unit - meter per second (m/s)
Example: You are driving your car at 27 m/s. You
look away from the road for 2.0 s to tune in a
decent radio station. How far does your car go
during that time?


          x
       v   x  vt  (27 m/s)(2.0 s) = 54 m
          t
Example: You are driving your car along a straight,
level road for 8.4 km at 70.0 km/h, when your car
stops, having run out of gas. Over the next 30
minutes, you walk another 2.0 km to a gas station.
What is your average velocity from the beginning of
your drive to the time when you arrive at the gas
station?


Reasoning: Considering the definition of average
velocity, we’ll need to know the total displacement
and the elapsed time. There are two segments in
which you are moving at two different speeds.
We’ll need to consider each segment separately
and add the two together.
Displacement:
Car: This is given as 8.4 km
Walk: We are given that you walk 2.0 km
Since both of these are in the same direction, we
can just add them together…
              x = 8.4 + 2.0 = +10.4 km
Elapsed Time:
Car: The car is traveling at 70.0 km/h for a distance
of 8.4 km. We can use this to figure out how long
you were driving.            x
                         v
                              t
                                  x   8.4 km
                        t                   0.12 h
                                  v 70.0 km/h

Walk: We are told that you walk for 30 minutes or
0.5 h. So the total elapsed time is 0.62 h.
  x 10.4 km
v          17 km/h
  t  0.62 h
Instantaneous
Velocity
 This refers to how
 fast something is
 moving at a
 particular instant in
 time.
Definition
The average acceleration is a vector. Its
magnitude is the change in velocity divided
by the elapsed time.
                        v v 0
                       a
                           t  t0
SI Unit - meter per second squared (m/s2)
 At time t = 0 s, the plane starts from rest.




At time t = 29 s, the plane’s velocity is 260 km/h and
it takes off.

     
  v  v o 260 km h  0 km h        km h
 a                          9.0
    t  to      29 s  0 s            s
Acceleration and Decreasing Velocity
                    
                 v  v o 13 m s  28 m s
                a                        5.0 m s 2
                   t  to    12 s  9 s
Equations of Kinematics




                     Five Variables
                    x - displacement
         v0 - initial velocity at time t0 = 0 s
               v - final velocity at time t
              a - acceleration (constant)
            t - elapsed time since t0 = 0 s
Reasoning Strategy
1. Make a drawing.

2. Decide which directions are to be called positive (+) and
negative (-).

3. Write down the values that are given for any of the five
kinematic variables.

4. Verify that the information contains values for at least three
of the five kinematic variables. Select the appropriate equation.

5. When the motion is divided into segments, remember that
the final velocity of one segment is the initial velocity for the next.

6. Keep in mind that there may be two possible answers to a
kinematics problem.
       Speed Trap
The state police placed speed sensors
at two locations, 201 m apart, on an
interstate highway. A motorcycle
passes the first sensor at time t0 = 0 s
and passes the second sensor 7.25 s
later. Unaware of the sensors, the
motorcyclist increases his speed by
maintaining a constant acceleration of
1.88 m/s2 between them.
(a) Discuss the direction of each of
    the following vectors:
    displacement, velocity, and
    acceleration.

(b) What is the speed of the
motorcycle as it passes each sensor?
In which direction does each of the following vectors for
the motorcycle point:
       displacement, velocity, and acceleration?
  Answer: Let’s assume that the direction the
  motorcycle is initially moving is the positive x
  direction.

  The motorcycle continues moving in this
  direction; therefore, the velocity and displacement
  are in the positive x direction.

  Finally, since the speed is increasing, the
  acceleration must also point in the positive x
  direction.
Answer: Three of the five kinematic variables are given in the
statement of the problem:
       motorcycle data         x          a       v v0       t
                              201 m    +1.88 m/s2   ?   ?   7.25 s

By knowing three of the variables, the other two can be found.


                1 2               1 2
      x  v 0t  at            x  at
                2         v0     2
                                  t
                                        1
                               (201 m) (1.88 m/s2 )(7.25 s) 2
                                       2
                                          7.25 s
                              20.9 m/s
Once the initial speed is known, we can find the final speed.

       2       2
   v        v 0  2ax

                 2
           v   v0    2ax
                             2                2
             (20.9 m/s)  2(1.88 m/s )(201 m)
             34.5 m/s
A car will appear at the bottom that travels at a constant velocity.
The car is leaking oil at a regular time interval.
At the instant the police car will accelerate to catch up.
By a freakish quirk of fate, the police car also leaks oil at the
same rate.
When the cars are side by side, are the cars traveling at
the same velocity?
A car, starting from rest, accelerates in a straight line
path at a constant rate of 2.5 m/s2. How far will the
car travel in 10 seconds?
x=?

                  x  vot  at
vo = 0 m/s                          1   2
v=?                                 2
a = 2.5 m/s2
t = 10 s
                  x  at 
                       1
                       2
                           2    1
                                2    2.5 m/s  10 s
                                             2           2



                     125 m
A car starts from rest and accelerates at a
constant rate in a straight line. In the first
second, the car moves 2.0 m. How fast will the
car be moving at the end of the second
second?
Part 1
x = 2.0 m
             x  vot  at 1
                          2
                              2

vo = 0 m/s
v=?          x  at
                 1
                 2
                      2

a=?
               2 x 2(2.0 m)
t = 1.0 s
             a 2              4.0 m/s 2

                    1.0 s 
                             2
                t
Part 2
x=?
vo = 0 m/s
v=?
a = 4.0 m/s2
t = 2.0 s

          v  v o  at
                                   
                 0 m/s   4.0 m/s2 (2.0 s) = 8.0 m/s
Motion in Two Dimensions




vx  vox  a xt        x   1
                            2
                                vox  vx  t
x  voxt  a x t
           1
           2
                   2
                       v  v  2a x x
                        2
                        x
                                  2
                                  ox
v y  voy  a y t

y  voyt  a yt    1
                   2
                        2



y   1
     2
         v   oy    vy t

v  v  2ay y
 2
 y
         2
         oy
Galileo’s Experiments
Galileo’s Experiments
The x part of the motion occurs exactly as it would if the
y part did not occur at all, and vice versa.
Drop the Ball

				
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