# 2D Motion

Physics
• Study Guide
Summary

## Motion with Constant Acceleration in Two and Three Dimensions

Summary Motion 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?

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:

x(tf) =