# Kinematics in One Dimension (PowerPoint) by gjjur4356

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• pg 1
```									Kinematics in Two Dimensions

• Position, velocity, acceleration vectors
• Constant acceleration in 2-D
• Free fall in 2-D

Serway and Jewett : 4.1 to 4.3

Physics 1D03 - Lecture 4   1
           y
path
The Position vector r
points from the origin to
the particle.
(x,y)
yj          
r

xi                        x


The components of r are the coordinates (x,y) of the

particle: r  x i  y j

For a moving particle, r (t ), x(t), y(t) are functions of
time.
Physics 1D03 - Lecture 4       2
  
Displacement : r  rf  ri
y
final                  vavg
         
r
Average Velocity :                  rf
initial
v avg  r / t                        
                   ri
(a vector parallel to r )
x

Instantaneous Velocity :         y
 
v  dr / dt  is tangent to the                v
path of the particle

r
x
Physics 1D03 - Lecture 4          3

Acceleration is the
v (t  t )
rate of change of                
v (t )
velocity :                                   time t  t

 dv                            time t
a
dt                 path of particle

v                                     
 lim                                       v
t 0 t                                 a
    t
                 v
v (t )

v (t  t )

Physics 1D03 - Lecture 4   4
a is the rate of change of v (Recall: a derivative gives
the “rate of change” of function wrt a variable, like
time).

Velocity changes if
i) speed changes
ii) direction changes (even at constant speed)
iii) both speed and direction change

In general, acceleration is not parallel to the velocity.

Physics 1D03 - Lecture 4   5
Concept Quiz

A pendulum is released at (1) and swings across to (5).


At which positions is a  0 ?
(consider tangential a only!)
a) at 3 only
b) at 1 and 5 only

c) at 1, 3, and 5

1                  5     d) none of the above
2    3    4

Physics 1D03 - Lecture 4   6
Components: Each vector relation implies 3
separate relations for the 3 Cartesian components.


r  x i y jzk         (i, j, k, are unit vectors)

We get velocity components by differentiation:

 dr
v
dt
 dx  i   dy  j   dz  k
     
the unit vectors are
constants
 dt   dt   dt 
 v xi  v y j  v z k

Physics 1D03 - Lecture 4   7
Each component of the velocity vector looks like the
1-D “velocity” we saw earlier. Similarly for
acceleration:

 dv  dvx   dv y   dvz 
a       i      j   k
dt  dt   dt   dt 

dx           dvx d 2 x
vx     ,    ax        2
dt            dt dt
dy           dv y d 2 y
vy  ,      ay        2
dt            dt   dt
dz          dvz d 2 z
vz     ,   az        2
dt           dt dt

Physics 1D03 - Lecture 4   8
Common Notation – for time derivatives only, a dot
is often used:


 dr  
v    r
dt

 dv   
 r
a    v
dt
Physics 1D03 - Lecture 4   9
Constant Acceleration + Projectile Motion

If a is constant (magnitude and direction), then:
             
v (t )  v o  a t
                     
r (t )  ro  v o t  2
1 a t2

 
Where ro , v o are the initial values at t = 0.

In 2-D, each vector equation is equivalent to a pair of
component equations:
x(t )  xo  vox t  12 a x t 2
y (t )  yo  voy t  12 a y t 2

 
Example: Free fall : a  g  9.8 m/s [down]
2

Physics 1D03 - Lecture 4   10
Shooting the Gorilla

Tarzan has a new AK-47. George the gorilla hangs
from a tree branch, and bets that Tarzan can’t hit him.
Tarzan aims at George, and as soon as he shoots his
gun George lets go of the branch and begins to fall.

Where should Tarzan be aiming his gun as he fires it?

A) above the gorilla
B) at the gorilla
C) below the gorilla

Physics 1D03 - Lecture 4   11
a=g
v0t

v0                           (1/2)gt2

r0
r(t) =r0+v0t +(1/2)gt2

Physics 1D03 - Lecture 4   12
Concept quiz
Your summer job at an historical site includes firing a
cannon to amuse tourists. Unfortunately, the cannon isn’t
properly attached, and as the cannonball shoots forward
(horizontally) the cannon slides backwards off the wall.
If the cannon hits the ground 2 seconds later, the
cannonball will hit the ground:

a)   2 seconds after firing              2 m/s                        100 m/s
b) 100 seconds after firing
c)   2
100 seconds after firing
d) Other (explain)

Physics 1D03 - Lecture 4      13
Example Problem
A stone is thrown upwards from the top of a 45.0 m high
building with a 30º angle above the horizontal. If the
initial velocity of the stone is 20.0 m/s, how long is the
stone in the air, and how far from the base of the
building does it land ?

Physics 1D03 - Lecture 4   14
Example Problem: Cannon on a slope.

100 m/s

30°

20°

How long is the cannonball in the air, and how far from
the cannon does it hit?
Try to do it two different ways: once using horizontal and vertical
axes, once using axes tilted at 20o.

Physics 1D03 - Lecture 4   15
Show that for:
d
vo   θ
Φ

2v cos  sin(   )
2
d    o
g cos 2

Physics 1D03 - Lecture 4   16
Summary

• position vector r points from origin to a particle

 dr
• velocity vector v 
dt           
 dv      v            
• acceleration vector a         [    , as t , v go to zero]
dt    t
• for constant acceleration, we can apply 1-D formulae
to each component separately
