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When considering morion in one dimension, that is, when the motion is restricted to a straight line, it is easy enough to directly apply the formulas derived earlier to determine instantaneous velocity, acceleration, and displacement. However, in two dimensions, with morion possible in any direction in a given plane, you must consider the kinematic properties of velocity, acceleration, and displacement as vectors.
Using rectangular coordinates in the standard Cartesian coordinate system, you must account for the x- and y-components of displacement, velocity, and acceleration. Essentially, you can treat the x- and y-components separately and then superimpose these components to define the corresponding vector quantities.
To help keep track of these x- and y-components, let i and j be unit vectors in the x- and y-directions, respectively. Now you can write the kinematic property vectors in terms of their components as follows:
vdt = ds, where v = vi/(l + vifet)
Note that in this equation In is the natural logarithm operator.
This example demonstrates the relative complexity of nonconscant acceleration problems versus constant acceleration problems. Itís a fairly simple example in which you are
2D Particle Kinematics
V = vxi + vyj a = axi + ayj
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If x is the displacement in the xdirection and y is the displacement in the ydirection, then the displacement vector is
ē ? V'
S = XI + yj !
It then follows that
v = ds/dt = dx/dt i 4- dy/dt j a ó dv/dt = d2s/dt = d2x/dt i + d2y/dtj
Consider a simple example in which youíre writing a hunting game and you need to figure out the vertical drop in a fired bullet from its aim point to the point at which it actually hits the target. In this example, assume that there is no wind and no drag on the bullet as it flies through the air (Iíll deal with wind and drag on projectiles in Chapter 6). These assumptions reduce the problem to one of constant acceleration, which in this case is that due to gravity It is this gravitational acceleration that is responsible for the drop in the bullet as it travels from the rifle to the target. Figure 2-2 illustrates the problem.
Figure 2-2. 2D Kinematics Example Problem
Let the origin of the 2D coordinate system be at the end of the rifle, with the x-axis pointing toward the target and the y-axis pointing upward. Positive displacements along the x-axis are toward the target, and positive displacements along the y-axis are upward. This implies that the gravitational acceleration will act in the negative y-direction.
Treating the x- and y-components separately allows you to break the problem up into small, easy-to-manage pieces. Looking at the x-component first, you know that the bullet will leave the rifle with an initial muzzle velocity vm in the x-direction, and since we are neglecting drag, this speed will be constant. Thus,
aK = 0
Vy = Vm X = vxt = Vmt
Now looking at the y-component, you know that the initial speed in the y-direction,
32 | Chapter 2: Kinematic
as the bullet leaves the rifle, is zero, but the y acceleration is óg (due to gravity). Thus,
ay = óg ó dvy/dt Vy --- Oyt = -gt
y = (l/2)ayt2 = -iVDgt2 The displacement, velocity, and acceleration vectors can now be written as follows:
s = (vmt)i - (l/2gt2)j v = (vji - (gt)j
These equations give the'instantaneous displacement, velocity, and acceleration for any given time instant between the time the bullet leaves the rifle and the time it hits the target. The magnitudes of these vectors give the total displacement, velocity, and acceleration at a given time. For example,
s = V(vmt)2 + (l/2gt2)2 V = \/W)Z + (gt)2 a = y/rSI = S
To calculate the bulletís vertical drop at the instant the bullet hits the target, you must first calculate the time required to reach the target, and then you can use that time to calculate the y-component of displacement, which is the vertical drop. Here are the formulas to use:
fhit = ^-hir/fm ó ti/vm d = )hit = -(l/2)g(fhit)2
where n is the distance from the rifle to the target and d is the vertical drop of the bullet at the target.
If the distance to the target, n, equals 500 meters (m) and the muzzle velocity, vm, equals 800 m/s, then the equations for thk and d give
hut = 0-625 s d = 1.9 m
These results tell you that to hit the intended target at that range, youíll need to aim for a point about 2 m above it.
3D Particle Kinematics
Extending the kinematic property vectors to three dimensions is not very difficult. It simply involves the addition of one more component to the vector representations
3D Particle Kinematics 33
shown in the previous section on 2D kinematics. Introducing k as the unit vector in the z-direction, you can now write
s = xi + yj + zk !
v = ds/dt = dx/dt i + dy/dt j + dz/dtk a = d2s/dt ó d2x/dt i + d2y/dtj + d2z/dt k
Instead of treating two components separately and then superimposing them, you now treat three components separately and superimpose these. This is best illustrated by an example.