Kinematics in Two and Three Dimensions

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© 1998-2008 by Glenn Elert -- A Work in Progress
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Discussion

synthetic geometry

The geometry that uses algebra is called analytic geometry, the topic of the next section of this book. The geometry that doesn't use algebra should be called synthetic geometry, but I haven't seen this term used very much. Synthesis (joining separated parts into a unified entity) is the opposite of analysis (breaking something up into its constituent parts). Before the Seventeenth Century, mathematicians used geometry to solve sets of related problems from which generalized conclusions could be made. After the Seventeenth Century, mathematicians applied algebra to generalized problems from which sets of related conclusions could be drawn. The first method requires a lot of work to arrive at a few conclusions, while the second produces a lot of conclusions from a little bit of work. This is a huge oversimplification of the history of mathematics, but you get the idea. Mathematics improved dramatically when algebra and geometry were united.

Analytic geometry is usually introduced to students in college. Synthetic geometry is a large part of secondary school mathematics (even if it isn't called by that name). Analytic geometry is a useful tool in physics, but it is not necessary to learn any new mathematics in this section. A working knowledge of synthetic geometry is sufficient to solve many kinematics problems in two and three dimensions.

Kinematic problems in two and three dimensions are essentially geometry problems. To solve them you should be able to …

Displacement, velocity, and acceleration (like all vector quantities) are geometric entities. They have magnitude, which is something like length, and direction, which is just the angle they make with respect to a reference direction like "up" or "north". It is always possible to create a vector diagram or a series of diagrams from a properly worded description of a kinematic event. The challenge for those studying this topic is learning how to effectively represent motion with a static picture.

Let's review the kinematic definitions presented in this chapter. First the scalar quantities …

And then the vector quantities …

Now let's go ahead and solve some problems.

Summary

Problems

practice

  1. I went for a walk one day. I walked north 6.0 km at 6.0 km/h and then west 10 km at 5.0 km/hr. (This problem is deceptively easy, so be careful. Begin each part by reviewing the appropriate physical definition.) Determine …
    1. the total distance of the entire trip,
    2. the total displacement of the entire trip,
    3. the average speed of the entire trip,
    4. the average velocity of the entire trip, and
    5. the average acceleration of the entire trip.
    Solutions …
    1. Distance? No problem. First I walked 6.0 km and then I walked 10 km. Thus, I walked 16 km. Distance is a scalar quantity and so the individual distances add just like regular numbers to yield the overall distance …
       
      16 km
       
    2. Displacement is a bit more challenging, but not by much. Displacement is a vector and vectors have direction, so it's best to diagram this problem (a procedure that's remarkably useful in general). The resultant displacement is the vector sum of the two displacements experienced during the trip. Since they're at right angles to one another, their magnitude can be found using Pythagorean theorem and their direction can be found using tangent …
               
             
        r =  √[(6.0 km)2 + (10 km)2]
        r =  11.6619 … km
        tan θ =  10 km  
      6.0 km
        θ =  59° W of N
             
               
    3. The speed was 6.0 km/h for the first 6.0 km and 5 km/h for the last 10 km. The naive solution is to average the speeds using the add-and-divide method taught in junior high school. This method is wrong, not because the method itself is wrong, but because it doesn't apply to this situation.
         
      6.0 km/h + 5.0 km/h  = 5.5 km/h
      2
         
      The wrong method of averaging
         
      The weight of the two segments is not equal. The second segment lasted twice as long as the first (as you will soon see).

      Go back to the definition to solve this problem. Average speed is the total distance (which we've already found) divided by the total time (which we need to find). Since time is a scalar, all we need to do is find the time for each leg of the journey and add them …
                     
      Δt1 =  Δs1  =  6.0 km  = 1.0h    Δt =  Δt1 + Δt2
      v 6.0 km/h
                  Δt =  1.0 h + 2.0 h
      Δt2 =  Δs2  =  10 km  = 2.0h   Δt =  3.0 h
      v 5.0 km/h
                     
      Thus the average speed is …
               
      v =  Δs  =  16 km  = 5.3 km/h
      Δt 3.0 h
               
    4. The velocity was 6.0 km/h north over the first 6.0 km and 5 km/h west over the last 10 km. Average velocity is the total displacement divided by the total time. Both of these quantities have already been determined.
               
      v =  Δr  =  11.66 km  = 3.8873 … = 3.9 km/h 59° W of N
      Δt 3.0 h
               
    5. Acceleration in this context is relatively meaningless. It would be better to illustrate acceleration in two dimensions with a different problem (like the one below).
  2. A swimmer heads directly across a river swimming at 1.6 m/s relative to still water. She arrives at a point 40 m downstream from the point directly across the river, which is 80 m wide. Determine …
    1. the speed of the river current,
    2. the magnitude of the swimmer's resultant velocity,
    3. the direction of the swimmer's resultant velocity, and
    4. the time it takes the swimmer to cross the river.
    Since distance and velocity are directly proportional, this begins as a similar triangles problem.
     
    [magnify]
     
    1. Since speed and distance are directly proportional, the ratio of the downstream distance to the width of the river is the same as the ratio of the current speed to the swimmer's speed.
                       
      x  =  vx ⇒ 40 m  =  vcurrent ⇒ vcurrent = 0.8 m/s
      y vy 80 m 1.6 m/s
                       
    2. Determining the resultant velocity is a simple application of Pythagoras' theorem.
         
      v2 =  vx2 + vy2
      v =  √[(0.8 m/s)2 + (1.6 m/s)2]
      v =  1.8 m/s
         
    3. Direction angles are often best determined using the tangent function. This problem is no exception. The only thing open to discussion is our choice of angle. I suggest using the angle between the resultant velocity and the displacement vector that points directly across the river, but this is just my preference. Be sure to indicate that the resultant lies on a particular side of this vector for clarity.
                   
      tan θ =  x  =  40 m  =  0.8 m/s  = 0.5
      y 80 m 1.6 m/s
      θ =  27° downstream
                   
    4. This is where it gets interesting. By now you should understood that time is the ratio of displacement to velocity. This is a vector problem, so direction matters. This is why we should probably use the words displacement and velocity instead of distance and speed. The only question is which distance and which speed should we use? The simple answer is pick the pair you like the best, just be sure they point in the same direction.
                         
      t =  x    =  y    =  r    
      vx vy v
      t =  40 m  =  80 m  =  √[(40 m)2 + (80 m)2]  = 50 m
      0.8 m/s 1.6 m/s √[(0.8 m/s)2 + (1.6 m/s)2]
                         
      There's an interesting sideline to this question that astute readers might have noticed when looking at the first ratio in the chain of three shown above. The time it takes to cross a river by a swimmer swimming straight across is independent of the speed of the river. The only factors that matter are the speed of the swimmer and the width of the river. This swimmer will always cross the river in 50 s regardless of the speed of the river. 1 m/s, 10 m/s, 100 m/s, it doesn't matter. This example is a perfect illustration of an idea to be presented in the next section of this book. Motion in two dimensions can be thoroughly described with two independent one-dimensional equations. This idea is central to the field of analytical geometry.
  3. A car enters an intersection at 20 m/s where it collides with a truck. The impact rotates the car 90° and gives it a speed of 15 m/s. Determine the average acceleration of the car if it was in contact with the truck for 1.25 s.

    Solution …

    Finding the change in velocity is complicated in this problem by the change in direction. A diagram is indispensable. Let's assume that the initial direction of the car is 0° (to the right in standard position) and that the final velocity will be 90° (toward the top of the page in standard position). The difference of two vectors drawn this way would then connect the the head of the initial vector to the head of the final vector. Use Pythagorean Theorem for magnitude and tangent for direction as usual. Only after we have done all of this can we then plug numbers into the definition.

                   
      Δv =  √((20 m/s)2 + (15 m/s)2) = 25 m/s
      tan θ =  15 m/s   ⇒ θ   = 143°
    20 m/s
      a =  Δv  =  25 m/s  = 20 m/s2
    Δt 1.25 s
      a =  20 m/s2 at 143°
  4. Write something completely different.
    • Answer it.

conceptual

  1. At what direction (somewhat upstream, directly across, somewhat downstream) should you swim in order to cross a river …
    1. with the least displacement,
    2. in the least time?

numerical

  1. In the fictitious city of Metropolis, streets run east-west while avenues run north-south. The separation between centers for both the streets and the avenues is 100 m, resulting in square city blocks. Superman is located at 33rd Street and 3rd Avenue while Lois Lane is located north and east of Superman at 45th Street and 12th Avenue.
    1. How far would Lois Lane have to walk to reach Superman?
    2. How far would Superman have to fly to reach Lois Lane?
    3. In what exact direction should Superman fly to reach Lois Lane as quickly as possible?
  2. Calculate both the distance and the magnitude of the displacement of the earth after …
    1. one complete orbit,
    2. one-half of an orbit,
    3. one-fourth of an orbit, and
    4. one-eighth of an orbit around the sun.
    5. What relationship between distance and displacement does this verify?
  3. Calculate both the speed and the magnitude of the velocity of the earth after completing …
    1. one orbit,
    2. one-half an orbit,
    3. one-fourth an orbit, and
    4. one-eighth an orbit around the sun.
    5. What relationship between speed and velocity does this verify?

algebraic

  1. Given a straight river with parallel banks, flowing uniformly at speed vriver and a swimmer capable of speed vswim > vriver, at what angle (assuming that directly across is 0°) should the swimmer head in order to cross the river …
    1. with the least displacement,
    2. in the least time?

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