Chapter 4 Two–Dimensional Kinematics
Section 4.1
Constant velocity motion A body moving with constant velocity in two dimensions
travels equal distances for equal time intervals (not necessarily the same distance along
each dimension, i.e. ∆x is not generally the same as ∆y).
On a graph of x versus time, constant velocity still appears as a straight line where: positive
velocity motion has a positive slope (points up), zero velocity motion is a horizontal line,
and negative velocity motion has a negative slope (points down). The slope of a straight
line on the graph of x versus time is equal to the x component of velocity.
Replacing the "x" in the previous paragraph with "y" would hold for the y component of
velocity as well.
Constant acceleration motion A body moving with constant vector acceleration travels
different distances for equal time intervals. In two dimensions, there are three possibilities:
1) only an x component of acceleration, 2) only a y component of acceleration, or 3)
acceleration components along both x and y directions. (The case where acceleration is zero
in both x and y directions is just the same as the constant velocity case above)
Graphs can be drawn of velocity versus time for the two dimensional case, but separate
graphs would be needed for vx and vy versus time in order to calculate the x and y
displacements (∆x and ∆y) from the area under the graph line.
Kinematic equations in two dimensions Motion in two dimensions (x and y) under
constant acceleration is described using the following two sets of equations (Note that the
set of x components is used to describe one-dimensional motion.):
x components
y components
vx = v0x + ax t
vy = v0y + ay t
1
1
x – x0 = 2 (v0x + vx ) t
y – y0 = 2 (v0y + vy ) t
1
1
x – x0 = v0x t + 2 axt 2
y – y0 = v0y t + 2 ayt 2
vx2 = v0x2 + 2 ax (x – x0)
vy2 = v0y2 + 2 ay (y – y0)
Meaning of symbols above (for y components, replace x below with y):
x – x0 = displacement along x-axis (from origin)
vx = final x-component of velocity
v0 x = initial x-component of velocity
ax = x-component of acceleration
t = time that has passed (from t = 0)
Displacement and velocity can also be represented using the unit vector notation, e.g.:




→
1
1
∆r =  v0x t + 2 axt2  ^
x +  v0 y t + 2 ay t2  y^
→
v = ( v0 x + ax t) ^
x + ( v0 y + ay t) y^
Variables in equations above for specific situations:
Two-dimensional projectile motion(vertical versus horizontal): Either v0x or v0y may be nonzero. vx will not change over time, but vy ≠ v0y. For accelerations, ax = 0 and ay = –g =
–9.81 m/s2 .
Section 4.3
Case 1 (Zero Launch Angle): v0y = 0, v0x ≠ 0. The path traced out by the object will be half a
downward parabola (Fig 4-5). vx = v0x over the course of the path (ax = 0) while vy = –gt
and becomes increasing larger in the negative y direction.
Section 4.4,5
Case 2 (General Launch Angle, where ∆y = 0): v0y ≠ 0, v0x ≠ 0. The path is a symmetric
parabola opening down and centered around the maximum height (Fig 4-7,8, 11). If given
the initial velocity magnitude and launch angle, the time of flight, maximum height, range,
and final velocity can be calculated. In the absence of air friction, the maximum horizontal
range occurs when the launch angle is 45°. The velocity vector has the same magnitude
going up through a value of y as it does coming back down through the same value of y
(Fig 4-10). The velocity takes on its minimum magnitude at the maximum height (= v0x).
Case 3 (General Launch Angle, where ∆y ≠ 0): v0y ≠ 0, v0x ≠ 0. The path is no longer a
symmetric parabola, but is still parabolic in nature (Fig 4-20). Problems of this sort that ask
for time will generate a quadratic equation in time that must be solved.
Some additional hints for problem solving (some of these are repeated from Chapter 2):
• Make sure that you understand whether the motion is in two dimensions or in one
dimension. If it is in two dimensions, you will need to consider both x and y
component equations. Make sure that you correctly identify what each of the physical
quantities (sx, vfy, ax, etc..) are in the problem you are working – this includes
understanding which values you are given and which you need to solve for in the
problem.
• Make sure that you understand exactly what is being asked for BEFORE you start
looking for an equation to apply. For example: Are initial values given or being asked
to solve for? Are there values implied but not stated? ( i.e. starting from rest = initial
velocity of zero, or object in free fall means that y-acceleration is –9.81 m/s2).
• Make sure you understand whether the motion is free fall or whether the object
experiences some other acceleration (in either x or y direction or both).
• The component equations above have in common the variable of time (= t). Thus, the
same time (t) must pass for simultaneous x and y motion.
• If acceleration takes on different values over different parts of an object's motion, the
problem must be divided into individual parts for each of the different accelerations.