velocity and acceleration

figure showing r(t), v(t) and a(t)
figure 1: r(t), with v(t) and a(t) shown.

Let's look at how position, velocity and acceleration are related and can be written.

In the figure to the right, we show a space curve, which is described by a position vector r(t). Then v(t) gives the velocity at any time t. The velocity, which is r'(t), is also tangent to the curve, as shown. The acceleration, which is r''(t) is the same as v'(t), which is the rate of change of velocity. So it tells how fast the the velocity (change of position) is changing.

Ok, now let's think just about v and a. We saw in class (and in our book) that a can be broken into exactly two pieces: one part in the direction of v, which tells how much we're speeding up or slowing down as we move along the space curve (that is, that part of acceleration isn't trying to change the direction of motion - just speed it up or slow it down), and the part perpendicular to that, which is changing the actual direction of motion.

The first of these is the projection of a onto v: projva - as shown in the lower diagram in the figure. And the perpendicular is what's left: a - projva.

So what? Ok, here's our main point:

We can break the acceleration of an object into exactly two pieces, one in the direction of the velocity, which we call the tangential component of the acceleration, and the other perpendicular to this, which we call the normal component. With T(t), the unit tangent to the curve ( = r '(t)/|r '(t)| ) and N(t), the principle unit normal ( = T '(t)/|T '(t)| ), we can therefore write
a(t) = aT T(t) + aN N(t).
And, not only that, aT is just the length of the projection of a onto v (and aN N(t) can be found by subraction if we know aT).

An Example:
Let's see how this works in practice. Suppose that r = <2t2, t, 1>. Find v, a, and aT and aN.

The velocity and acceleration are easy to find:

v = r' = <4t, 1, 0>
a = v' = r'' = <2, 0, 0>.

Then, to find the tangential and normal components of acceleration we need T and N, the unit tangent and principal unit normal vectors. From their definitions, we have

T = v / |v| = <4t, 1, 0> / sqrt(16t2 + 1),

and N = T' / |T'|, which isn't all that nice to calculate. This is why we usually try and avoid calculating it... Before doing that, notice that we can easily now find aT. This is just the magnitude of the projection of a on T (see the projections page):

aT = |a| cos(theta)
   = a . T,

because a . T = |a| |T| cos(theta) and |T| = 1, so

aT = <2, 0, 0> . <4t, 1, 0> / sqrt(16t2 + 1)
   = 8 t / sqrt(16t2 + 1)

Then we can easily find aN N by subtracting:

a = aT T + aN N, so
aN N = a - aT T, which is
aN N = <2, 0, 0> - ( 8 t / sqrt(16t2 + 1) ) ( <4t, 1, 0> / sqrt(16t2 + 1) )
   = <2, 0, 0> - <32t2, 8t, 0> / (16t2 + 1).

Got all that? It's just a bunch of pushing vectors around. Go back and make sure that it makes sense. We can carry out the subtraction, too:

aN N = <2, 0, 0> - <32t2, 8t, 0> / (16t2 + 1)
   = <2, 0, 0> - <32t2 / (16t2 + 1), 8t / (16t2 + 1), 0>
   = < 2 - 32t2 / (16t2 + 1), -8t / (16t2 + 1), 0>
   = < 2 / (16t2 + 1), -8t / (16t2 + 1), 0>
   = < 2, -8t, 0> / (16t2 + 1)

Finally, if we wanted to find aN and N, the easiest way to do it at this point would probably to find aN by finding the magnitude of the vector aN N we found above, and then find N by finding a unit vector in the direction of the vector aN N.

But I'm not going to do that here.    :)


velocity and acceleration
Last modified: Tue Sep 28 08:55:47 EDT 2004
©2004 Gavin LaRose, UM Math Dept.