1. Physics 111 Lecture 03 Motion in Two Dimensions SJ 8th Ed. : Ch. 4
Position, velocity, acceleration vectors
Average & Instantaneous Velocity
Average & Instantaneous Acceleration
Two Dimensional Motion with Constant Acceleration (Kinematics)
Projectile Motion (Free Fall)
Uniform Circular Motion
Tangential and Radial Acceleration
Relative Velocity and Relative Acceleration
4.1 Position, Velocity and
Acceleration Vectors
4.2 Two Dimensional Motion with
Constant Acceleration
4.3 Projectile Motion
4.4 A particle in Uniform Circular Motion
4.5 Tangential and Radial Acceleration
4.6 Relative Velocity and Relative
Acceleration

2. Motion in two and Three Dimensions
Extend 1 dimensional kinematics to 2 D and 3 D
Kinematic quantities become 3 dimensional
Vectors needed to manipulate quantities
Motions in the 3 perpendicular directions
can be analyzed independently
Constant acceleration Kinematic Equations
hold component-wise for each dimension
Same for z
In vector notation each equation is 3 separate ones for x, y, z

3. Position and Displacement
always points from the origin to the particle’s location
A particle moves along its path as time increases
Trajectory: y = f(x)
Parameterized by time
Displacement
Positions
Average Velocity
Same direction as
Rate of change of position
Path does not show time dependence.
Slopes of (tangents to)
x(t), y(t), z(t) graphs would show
velocity components

4. Instantaneous Velocityx y
parabolic path
Free Fall Example
is tangent to the path in x,y
i.e, to a plot of y vs x t x y
y(t) is parabolic
Components of are tangent to graphs of corresponding components of the motion, viz:
vx is tangent to x(t)
vy is tangent to y(t)

5. Displacement in 3 Dimensions

6. ax affects vx and x, but not vy or y
Average Acceleration
Rate of change of velocity x y
2D Trajectory: y = f(x)
Parameterized by time
Velocity vectors are tangent to the trajectory at A & B
Move them tail to tail
is parallel to
Instantaneous Acceleration
If the solution is known find:
vx(t), vy(t) by differentiating x(t), y(t)
ax(t), ay(t) by differentiating vx(t), vy(t)
is NOT tangent to the path
ax IS tangent to a graph of vx(t)
ay IS tangent to a graph of vy(t)
ax affects vx and x, but not vy or y
Similarly for ay

7. Summary – 3D Kinematics Formulas
Definitions:
Vector Form - Motion with Constant Acceleration - 2D or 3D
PARABOLAS
Same for z
Easiest to use in Cartesian coordinates
– The x, y, z, dimensions move independently
- Choose ax, ay, az and initial conditions independently
Cartesian Scalar Form - Constant Acceleration

8. Kinematic Equations in 2D: Graphical Representationvx vy

9. No acceleration along x
Projectile Motion: Motion of a particle under constant, downward gravitation only
Assume:
Free fall along y direction (up/down) with horizontal motion as well
ay = constant = - g, ax, az = 0 Velocity in x-z direction is constant.
Trajectory (a parabola) lies in a plane. Can choose it to be x-y.
Motion is 2D, not 3D.
Usually pick initial location at origin: x0 = 0, y0 = 0 at t=0
Initial conditions:
v0 has x and y components vx0 = v0cos(q), vy0 = v0sin(q)
vx is constant, no drag or non-gravitation forces usually
If vx = 0, motion is strictly vertical, range = 0 x y
path
v0x
v0y q
Equations:
No acceleration along x
Acceleration
= - g along y

10. Trajectory of a Projectile: y as a function of x
Launch
Range R
(return to launch altitude)
vy at E = -vyi
Maximum Height vR

11. Example: Show Projectile Trajectory is a Parabola (in x)
path
vix
viy qi
Initial Conditions:
Eliminate time from Kinematics Equations:
Parabolic in x
Limiting Cases:
As qi  0, tanqi  0, cosqi  1,
As qi  90o, tanqi  infinity, cosqi  0,

12. Example: Find the Range of a Projectile (Level Ground)
The range R is the horizontal distance traveled as a projectile returns
to it’s launch height yi = 0.
Set x = R and y = 0 in the preceding trajectory formula: y x
Rearrange and put into standard quadratic form:
R factors. Roots are:
Trigonometric Identity:
“Range Formula”
Extremes of Range versus launch angle qi

13. Example: Time of Flight for a Projectile
Find the time of flight tR for range R – the time to return
to the original launch height yi = 0.
Set y = 0 in the y-displacement formula:
This is a quadratic equation in tR – factor it:
The roots are:
and:
Find speed vf at time tR - passing y = 0 on the way down
Same speed at the same altitude
vxf = vxi = vicos(qi)
(constant)
qi = qf
Find maximum height ymax reached at time tmax Set vy,max = 0 at tmax

14. Projectile Range Launch with same initial speed, vary angle
Maximum range at q = 45o
Complementary angles produce the same range, but different maximum heights and times of flight

15. Problem Solving Strategy
Text (page 43) defines a general 4 step method:
Conceptualize
Categorize
Analyze
Finalize
Additional problem solving hints for mechanics:
Choose coordinate system(s). Try to make choices that simplify
representing the problem. For example, where acceleration a is
constant try choosing an axis along the direction of a.
Make a sketch showing axes, origin, particles.
Choose names for the important quantities that will help you
remember what they mean. Be sure to distinguish quantities
of the same type, such as v, vx, vy, vi, vf, ….
List the known quantities and the given initial conditions,
like vxi, vyi, ax, ay,… Show these on your sketch.
Choose equations to use for describing the motion in the
problem. Time connects the x, y, z dimensions and is sometimes
to be eliminated via algebra.

16. What should monkey (a Physics 111 student) do
What should monkey (a Physics 111 student) do? Where should hunter aim if monkey jumps?
Assume: hunter fires at the instant monkey lets go
and aims directly at falling monkey
path for g = 0 xm ym q s

17. What should monkey (a Physics 111 student) do
What should monkey (a Physics 111 student) do? Where should hunter aim if monkey jumps?
When should hunter fire? When should monkey jump?
Coordinates, variables: x y q h s
Bullet: xb(t), yb(t)
Monkey: s, ym(t)
Initial Conditions:
Monkey at (s, h) Bullet at (0,0)
To hit monkey:
yb(t) = ym(t) at same time t1 when
bullet is at s
For bullet at s:
For monkey:
Equate positions:
Monkey should wait until hunter fires before jumping
Hunter should aim directly at monkey after he jumps
Both know Physics, so everybody waits forever

18. Example: Find the arrow’s speed
An arrow is shot horizontally by a person whose height is not known.
It strikes the ground 33 m away horizontally, making an angle of 86o with the vertical. What was the arrow’s speed as it was released?
86o
R = 33 m v0 vf
R = v0t
t = R / v0
t = flight time = time to hit ground
Final velocity components:
No acceleration along x
Vy0 = 0, t as above
Divide equations to eliminate vf:
Rearrange:

19. Uniform Circular Motion
A particle in Uniform Circular Motion moves in a circular path with a constant speed  The radius vector is rotating.
The velocity vector is always tangent to the particle’s path (as usual). It is perpendicular to the (rotating) radius vector.
The particle is accelerating, since the direction of the velocity is constantly changing.
The centripetal acceleration vector always points toward the center of the circle.
The magnitudes of all three vectors
are constant, but their directions are changing
All three vectors are rotating with the same constant angular velocity and period.
The period of revolution T is:
The frequency f is the reciprocal of the period
f = 1/T

20. Centripetal Acceleration Magnitude Formula
tail to tail
The position vectors and velocity vectors both change due to changes in their directions:
The triangles for Dr and Dv are both right isosceles, with the same angle Dq at their apexes. They are similar:
The definition of the (magnitude of) the average acceleration is:
In the limit Dt  0:
and
The magnitude of the centripetal acceleration is
It always points to the center of the circular motion

21. Example:
Find the Earth’s centripetal acceleration in its orbit about the Sun
Assume uniform circular motion
The radius of the Earth’s orbit is r = 1.5 x 1011 m
Need to find the speed v
T = 1 year = 365X24x3600 s = x 107 s
Example:
Passengers on an amusement park ride move in circles whose radius is 5 m. They complete a full circle every 4 seconds.
What acceleration do they feel, in “g”s”?
Need to find the speed v
dimensionless

22. Uniform Circular Motion Formulas via Calculus
Unit vector along r
Calculate time derivative. Constant r means dr/dt = 0
Note: r dq/dt = v = magnitude of the speed
= 2p x r / T for constant v
Unit vector tangent to circle
v is constant
Note:
Unit vector pointing inward along r