Showing position vector follows parabolic path

[physicspupil]

1. The problem statement, all variables and given/known data

Hi, got my HW back from prof. There were a few problems... I want to understand these before the test.

Show that r(t) = (at^2)/2 + vt + r lies in a plane and that if a and v are not parallel, then r(t) traces out a parabola. Note a, v, and r are constant vectors here (i.e. acceleration, velocity, and position and t is time)


2. Relevant equations

1. y = Ax + Bx^2 (prof said to show this)



3. The attempt at a solution

I wasn't sure about this. I put that since v(r) = dr/dt = at + v, there is no 3rd direction in which particle will move (since r disappears). Also, we can pass a plane through any two vectors...

In regards to showing y = Ax + Bx^2, is it as simple as letting x = t and y = r(t) -r? This would make it fit the form since A = a/2 = constant, right? Or do I have to do something by decompose these vectors such as

x(t) = (axt^2)/2 + vxt + x, y(t) = (ayt^2)/2 + vyt + y, etc. I'm just not sure what prof. wants???

Thanks!

[Dick]

You'll want to make a selection of basis vectors for the plane. I would suggest to take y1 to be a unit vector parallel to a. Then take x1 to be a unit vector perpendicular to a, but in the same plane as a and v. Now write x(t)=x1.(r(t)-r) and y(t)=y1.(r(t)-r) ('.' being the dot product). Since x1 is perpendicular to a, x(t) is a linear function of t, and x(t) is a quadratic function of t. So it should be easy to write it in the form y=Ax+Bx^2.

[physicspupil]

x(t) = x1 * [r(t) – r] = x1 * (½ at^2 + vt)
= x1 * ½ at^2 + x1 * vt (since a not || v, then v not perpendicular to x)
= 0 + |x1||v|cos(theta) = vtcos(theta) where v is now a scalar

Similarly

y(t) = (½ at^2 + vt)cos(theta)

So now do I do this… ½ at^2 + vt = B (vt)^2 + Avt (i.e. y = Bx^2 + Ax) and find that A = 1 and B = a/(2v^2) always works? So is the answer just to pick this A and B, and then r(t) will fit the form of a parabola? Sorry, I’m still kind of shaky on these type of proof/showing problems. Thanks again for the help!

[Dick]

Well, you are on the right track. x(t)=(x1.v)*t. x1.v is just a constant - no need to say what it is. y(t)=(y1.a)*t^2/2+(y1.v)*t. This is a constant times t^2 plus a constant time t. No need to say more. Solve the x(t) equation for t and substitute into the y(t) equation and you'll get a quadratic equation for y(t) in terms of x(t) which is what you are after. Note that y(t) and x(t) are just coordinates of the path in the plane defined by the direction of vectors a and v with origin at r.