Kinematics - The Study of Motion

1. Draw a simplified picture of what happens during the problem.
   1. Define the coordinate system you will use for assigning magnitudes and signs for displacement, velocity, and acceleration
   2. The picture should be able to directly correlate to equations, so assign variable names according to the picture rather than starting with equations.

2. Assign variables as known or unknown and establish whether acceleration is constant throughout the problem.

3. If the acceleration is constant, then begin using equations for motion which contribute to the solution. If it is not constant, then break the problem into pieces for which the acceleration is constant.

4. Reread the problem to be sure you have not overlooked what is required and to re-establish what you are supposed to find as a solution.

Let's try the technique on a few examples:

1. A bus slows with constant deceleration from 24 m/s to 16 m/s and moves 50 m in the process.
   1. How much further does it travel before coming to a stop?
   2. How long does it take to stop from 24 m/s?

   Solution:
   First, we can find the value of the deceleration a. The figure below describes what happens as a picture.

   

   The equation for motion that relates acceleration, velocity, and distance is
   

   v2	= v02 - 2a(50 m) ==>
   a	= (v2 - v02)/[(-2)(50 m)]
   	= ((16 m/s)2 - (24 m/s)2)/[(-2)(50 m)]
   a	= 3.2 m/s2

   1. This acceleration determines how much further the bus travels.
      

      v02 = 0	= v2 - 2ax ==>
      x	= v2/ 2a
      	= (16 m/s)2/ 2(3.2 m/s2)
      x	= 40 m

   2. The time needed to come to a stop is
      

      v = 0	= v0 - at ==>
      t	= v0/a
      	= (24 m/s)/(3.2 m/s2)
      t	= 7.5 s

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2. In the figure below you see a frictionless track which has an inclined section and a horizontally flat section (also frictionless). The blue ball, initially held at rest, is released and slides down the incline.
   
   As it slides, at some time the ball passes between points a and b. Still later, the ball passes between points c and g. Both points a and b and c and g are separated by a distance d. Neglecting any air resistance, is the time it takes for the ball to go from a to b
   1. greater than the time needed to go from c to g
   2. less than the time needed to go from c to g
   3. the same as the time needed to go from c to g
   4. impossible to compare to the time needed to go from c to g without a measurement
   5. dependent on the weight of the ball

   You can play around with this type of setup interactively using this Java applet.

   Solution:
   We have to make use of one of the equations for motion. The simplest to use in this case is that the distance d must satisfy the equation
   

   d = ½ (v_a + v_b)t₁

   where v_a is the instantaneous velocity of the ball at point a and v_b is its velocity at point b. Since the ball is traveling downward, its acceleration due to gravity increases the speed as it goes down and v_a < v_b. The same effect demands that v_b < v_c. Since c and g are on horizontal ground, v_c = v_g and the distance d satisfies the equation
   

   d = v_ct₂

   Therefore, ½ (v_a + v_b)t₁ = v_ct₂.

   We want to know whether t₂ is greater than, less than, or equal to t₁. Mathematics gives us the answer. Since, for any d > 0 we must have v_a < v_b, then ½ (v_a + v_b) < v_b < v_c, so t₂ must be less than t₁. The answer is a.

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3. A disgruntled gardener, disappointed with his plant progeny, drops a flower pot from the roof. The pot begins from rest a distance h above the top floor window as shown in the figure below. The windows on each floor have a length equal to h and are spaced a distance h apart along the vertical direction.
   

   If the pot takes a time t to pass from the top to the bottom of the top floor window, the time it will take to drop from the top to the bottom of the next window down is

   1. 3/2 t
   2. 2/3 t
   3. 1/4 t
   4. 1/3 t
   5. 1/9 t
   6. The answer I like

While we could find the answer to the first question easily, the second and third were uncertain until we made use of what we understood about the equations for motion. Certainly the flower pot problem is impossible to intuit even though we can visualize very well what happens to the flower pot. The lack of a firm understanding of how to calculate this kind of motion led to continual attempts to better understand motion. When Galileo first proposed a new philosophy for science, that of experimentation as the only means of discovering the way nature works, he took it to heart by completely re-examining the concept of motion. To be precise in a scientific sense you must have definitions or descriptions of physical phenomena which are practical (i.e. useful in describing many physical phenomena) and readily understood in terms of mathematics.

The logical starting point for describing motion is to define first what we mean by saying that an object "has moved" and then to relate the amount of movement to the time over which it takes place. Therefore, we described "has moved" as a change in position or a displacement. Symbolically, we relate the position of object relative to an origin of a coordinate system. Along one dimension, for example, the initial position of an object might be listed as x₀. If we later observe the object at a different position x₁ relative to the same origin as before, then x₁ - x₀ is the displacement. The time over which this displacement takes place is usually referred to as just t since the starting time of the observation (at least for motion) is never important. So, Galileo quantifies the amount of motion that occurs as

where v_avg stands for the average velocity. Since we often find that the "motion" of an object also changes with time, it is practical to also define the acceleration as

Here we must be careful to distinguish between v_avg and the v used in the acceleration definition. The instantaneous velocity is defined as the derivative of displacement with respect to time. More about that later.

These definitions of acceleration, velocity, and displacement are connected through mathematics in what are called the equations for motion:

One thing you can not forget is that these equations are only valid if the acceleration, a, is constant for the entire time interval for which you use them.

To make use of these equations to answer the questions above, let's look at what these equations do best: predict the motion or displacement for the future based on what we have determined about what influences the motion. In many problems the main "influencer" of motion is gravity. Gravity provides a constant acceleration of magnitude g = 9.8 m/s² directed towards the center of the earth (downward in almost every case we will consider until near the end of the course). However, any case of constant acceleration can be handled by these equations.

Consider the case of the flowerpot problem. Here is the solution.

This is a tough problem but it's indicative of several skills you need to adopt to approach problems on motion. First, we note that the flower pot is accelerating as it goes down so we expect that the time to pass the top window will be greater than the time to pass the bottom window since the window distances are the same (namely, h) while the velocity of the plot during the whole time it passes the bottom window is higher than any velocity it reaches while passing the top window. Therefore, answer a is excluded before we even begin.

To find a value for t, the time necessary to fall past the top window, we use one of the equations for motion as applied to the picture below

where t₁ = t as described in the problem, y₁ is the vertical position of the bottom of the top window, y₀₁ is the vertical position of the top of the top window, v₀₁ is the speed of the pot when it reaches the top of the top window, and g is the acceleration due to gravity. To find v₀₁, we use an equation for motion which relates the velocity at the roof to the velocity at the top of the top window.

Our original equation is a quadratic equation so its solution is

Notice that we chose the minus sign in choice between plus and minus in the numerator since only a positive time makes physical sense for this problem.

We use the same approach for the bottom window. The only difference is that the initial velocity at the top of the bottom window (which we will call v₀₂) is greater than v₀₁. However, we find it in the same manner, by relating it to the initial velocity of zero from the roof

The time to fall past the second window is

The ratio of t₂ and t₁ is the answer we originally set out to determine, so

This ratio is roughly 2/3, so answer b is closest.

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