Motion with Constant Acceleration in Two and Three Dimensions
SummaryMotion with Constant Acceleration in Two and Three Dimensions
We have already seen that motion in more than one dimension that undergoes
constant acceleration is given by the vector equation:
x(t) = at2 + v0t + x0,
where a, v0 and x0 are constant vectors
denoting the acceleration, intitial velocity, and initial position,
respectively. Our next task will be to analyze special cases of this equation
that describe important examples of two- and three-dimensional motion with
constant acceleration: mainly, we will study projectile motion.
Projectile Motion
Simply stated, projectile motion is just the motion of an object near the
earth's surface which experiences acceleration only due to the earth's
gravitational pull. In the section on one-dimensional motion with constant
acceleration, we learned that this
acceleration is given by g = 9.8 m/s2. Using a three-dimensional
coordinate system, with the z-axis pointing upwards to the sky, the
corresponding acceleration vector becomes a = (0, 0, - g). This turns
out to be the only piece of information we need to write down the general vector
equation for projectile motion.
x(t) = (0, 0, - g)t2 + v0t + x0
As an example, consider a creature fired out of a canon with speed v at an angle
θ from the earth's surface. How far away will the creature be when it
falls back down to earth?
Figure %: Diagram of a creature fired out of a canon at an angle θ.
To answer this question we must first determine the position function,
x(t), which means we must find v0 and x0.
We can choose the x-axis to point in the direction of the creature's
horizontal movement across the earth. This means that the creature's movement
will be constrained to the x-z plane, and so we can completely ignore the
y-direction, effectively reducing our problem to two dimensions. (In fact,
using this kind of trick we can always reduce projectile motion problems to two
dimensions!) From the initial speed and angle of projection, we can determine
that v0 = (v cosθ, 0, v sinθ). Since the canon is fired
from the surface of the earth, we can set x0 = 0 (where
0 = (0, 0, 0), the zero-vector). This leaves us with the position
function:
x(t) = (0, 0, - g)t2 + (v cosθ, 0, v sinθ)t
The y-equation is pretty much useless. If we break this up into x- and
z-components we get:
x(t)
=
v cosθt
z(t)
=
v sinθt - gt2
The next step is to find that time at which the creature will hit the ground.
Setting z(t) = 0 and solving for t we find that the time at which the creature
will hit the ground is tf = . Finally, we need to plug
this time into the equation for the x-position, to see how far the creature
has traveled horizontally in this time.
x(tf) =
Using the trig identity sin(2θ) = 2 sinθcosθ we find that when
the creature hits the ground its distance from the canon will be:
at2 + v0t + x0,
