Todays Agenda

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Todays Agenda

• kinematics in 2D
• projectiles

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In 2D motion, the direction of position,
velocity, and acceleration changes.
Vectors change. a vector change
includes direction change.
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Question

• A car moves along a circle with constant speed.
  Which one of the following statements is correct.
• acceleration of the car is zero
• Motion of the car is one-dimensional
• Velocity of the car does not change
• Magnitude of the car acceleration is constant
• Acceleration is increasing with time

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Roller coaster
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Position and Velocity
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Position and Velocity
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Position and Velocity
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Instantaneous velocity
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Components of velocity
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question
A particles position is described by vector
x2t3, y3t2, in meters. What's the speed at
t1.73sec. 1) 16 2) 21 3) 41 4) 11
5) 27m/s
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Question
Particle velocity is given by
The x component of position at t16s is 4096m.
Find the x component at zero seconds. 1) 4096m
2)0 3)-8192 4)-4096 5)neither of
these
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Acceleration
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Instantaneous acceleration
Can be decomposed into parallel and
perpe-ndicular components
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Instantaneous acceleration
Can be decomposed into x and y components (more
convenient)
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Question

• A particle is moving along a circle and the
  acceleration is pointing
• Inside the circle, as discussed before. The
  particles velocity is suddenly
• reversed, and the particle traverses the circle
  backwards. The acceleration
• After the reversal is
• Inside the circle 2) Outside the circle 3)
  Either, depending
• on situation.

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Constant acceleration in 2D is similar to that in
1D
Initial position and velocity
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Dynamics in 2D
Newtons 2nd law
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Alice tapes a small 400g model rocket to a 600g
ice hockey puck. The rocket generates 20N of
thrust. She orients the puck so that the rockets
nose points in the positive y-direction, then
pushes the puck across frictionless ice in the
positive x-direction. She releases it with a
speed of 0.5 m/s at the exact instant the rocket
fires. Find an equation for the pucks
trajectory, then graph it.

• y1.67x2
• y4.2x2
• y12x2
• y40x2
• y100x2

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Question
An arrow is shot at an angle of       above the
horizontal. The arrow hits a tree a horizontal
distance          away, at the same height
above the ground as it was shot. Use          
for the magnitude of the acceleration due to
gravity.

• Find the time that the arrow spends in the air.
•  

1) 6.7 2)12.1 3)4.1 4)5.9 5)23
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Adkins Video
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The parabolic trajectory of a ball
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2-D Kinematics

• The position, velocity, and acceleration of a
  particle in 3 dimensions can be expressed as
• r x i y j
• v vx i vy j (i , j unit vectors )
• a ax i ay j
• We have already seen the 1-D kinematics equations

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2-D Kinematics

• For 2-D, we simply apply the 1-D equations to
  each of the component equations.
• Which can be combined into the vector equations
• r r(t) v dr / dt a d2r / dt2

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BIG IDEA Treat each coordinatedirection (each
vector component) independently.

• To get the complete motion in 2-D we simply use
  vector math to add the components
  (SUPERPOSITION).

EXAMPLE for constant acceleration we have a
const v v0 a t r r0 v0 t 1/2 a
t2 (where a, v, v0, r, r0, are all vectors)
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2-D Kinematics

• We will focus mainly on 2-D problems
• Vector arithmetic in 2-D is not much different
  than in 3-D.
• 3-D problems can be reduced to 2-D problems
  when acceleration is constant
• Choose y axis to be along direction of
  acceleration
• Choose x axis to be along the other direction
  of motion
• Example Throwing a baseball (neglecting air
  resistance)
• Acceleration is constant (gravity)
• Choose y axis up ay -g
• Choose x axis along the ground in the direction
  of the throw

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Motion in 2D

• Two footballs are thrown from the same point on a
  flat field. Both are thrown at an angle of 30o
  above the horizontal. Ball 2 has twice the
  initial speed of ball 1. If ball 1 is caught a
  distance D1 from the thrower, how far away from
  the thrower D2 will the receiver of ball 2 be
  when he catches it?(1) D2 2D1 (2) D2 4D1
  (3) D2 6 D1 (4) D2 8 D1

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Solution
The distance a ball will go is simply x
(horizontal speed) x (time in air) v0x t
To figure out time in air, consider the
equation for the height of the ball
When the ball is caught, y y0
(time of catch)
(time of throw)
two solutions
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Solution
x v0x t
So the time spent in the air is proportional to
v0y
Since the angles are the same, both v0y and v0x
for ball 2are twice those of ball 1.
Ball 2 is in the air twice as long as ball 1, but
it also has twice the horizontal speed, so it
will go 4 times as far!!