Gas station without pumps

2012 February 18

Graphical methods using average velocity

Filed under: home school — gasstationwithoutpumps @ 09:14
Tags: , , , ,

On the Quantum Progress blog, there is a nice presentation of graphical ways to do standard constant-acceleration problems in 2D: Still learning how to solve projectile motion problems after 20 years of study.

The idea is a simple one: use the fact that under constant acceleration the average velocity over an interval is just the average of the initial and final velocities.  If you draw the initial and final velocities as vectors from a point (say the origin), the average velocity is just the vector to the midpoint of the line joining the two tips the initial and final vectors.

This visualization may make it easier for students to figure out how to solve standard projectile motion problems.  He gives the example “A projectile is launched at 20m/s at 45° on horizontal ground. Find where it lands.”Because the final displacement is purely horizontal, the average velocity must be horizontal, and the and the final velocity vector symmetrically placed to make the midpoint be on the x-axis:

Initial velocity 20 m/s up 45˚, so final velocity must be down an equal amount.Picture copied from

The change in velocity can be roughly measured from the graph, or easily computed as (0, -20 \sqrt{2}) m/s.  Dividing by the acceleration of -9.8 m/s^2 gives the desired time.  The math is the same as a purely algebraic solution, but this pictorial representation may be easier for students.

I don’t know if this is worth teaching to my students—it seems cute to me, but probably takes longer to draw than just writing down the equation and solving it.  Both my students are mathematically adept enough that they might see this as a unnecessary crutch.  Little pictures can help when setting up a problem, though, and I am trying to get my students to be able to choose the appropriate models and tools for solving physics problems, so this might be worth showing them, even just for the cuteness factor.

Of course, there is a lot hidden in the picture that make it in some ways a poor teaching tool.  For example, the symmetry of v_f and v_i does not come just from the average velocity being horizontal.  There are infinitely many values for v_f that would result in the midpoint falling on the x-axis, but they all have the same y value, so that \Delta v_y is the same no matter which one you pick.  In this particular instance, the symmetric one in the picture is correct, as there is no horizontal acceleration in the model, but it is all too easy in drawing pictures to pick up symmetry even when it is not part of the original problem.

1 Comment »

  1. One can also use geometric algebra to derive a second timeless equation. Taking \vec{v}_f+\vec{v}_i = \frac{2\Delta \vec{x}}{\Delta t} and \vec{v}_f-\vec{v}_i = \vec{a} \Delta t, we get the standard timeless equation from the dot product of the equations: {v_f}^2-{v_i}^2 = 2\vec{a}\cdot\Delta \vec{x}. If we use the wedge product instead, we get \vec{v}_i\wedge\vec{v}_f=\Delta\vec{x}\wedge\vec{a}. This has a geometric interpretation as the directed area of the parallelogram from \vec{v}_i to \vec{v}_f being equal to the directed area of the parallelogram from \Delta\vec{x} to \vec{a}. It’s not normally that useful, but you can use it—exercise for the reader!—to show that for projectile motion the range on level ground is \frac{2v_{ix}v_{iy}}{a}.

    Comment by bvancil — 2012 February 19 @ 21:19 | Reply

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