Lecture 3

Motion in one-dimension (aka Kinematics)
 Examine systems with non-zero acceleration (often
constant)
 Solve 1D problems with zero and constant acceleration
(including free-fall and motion on an incline)
Physics 201: Lecture 3, Pg 1
Position, instantaneous velocity & acceleration
x(t )  xi  function (t f  ti )
x
x  x(t f )  x(ti )
xi
t
dx
vx 
dt
vx
vi
t
2
dvx d x
ax 
 2
dt
dt
ax
ti
tf
Physics 201: Lecture 3, Pg 2
t
Example problem

A car moves to the right first for 2.0 sec at 1.0 m/s and then
4.0 seconds at 2.0 m/s.

What was the average velocity?

Two legs with constant velocity but ….
vx, avg

vx
Average velocity:
v1  v2

2
vx, avg 
change in position
change in time
Physics 201: Lecture 3, Pg 4
t
Example problem

A particle moves to the right first for 2.0 seconds at 1.0 m/s
and then 4.0 seconds at 2.0 m/s.

What was the average velocity?
vx

Two legs with constant velocity

Find the total displacement (x2 –x0)

Constant velocity
x(Δt)=x0+vx Δt
x1 = x0 + v0 (t1-t0)


t
x2 = x1 + v1 (t2-t1)
x2 - x0 = (x2 - x1) + (x1 – x0) = v1 (t2-t1) + v0 (t1-t0)
x2 –x0 = 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6.0 s or 1.7 m/s
Physics 201: Lecture 3, Pg 5
Another special case, constant acceleration
 Particle motion with constant acceleration
 The velocity vector changes
a
dv x
ax 
 constant
dt
a x dt  dv x
t
0
t
ti
tf
v = area under curve = a t
a x t  v x  v x  v x
f
i
Physics 201: Lecture 3, Pg 6
Constant acceleration in 1D

“Particle” motion with constant acceleration

A car, starting at rest, with smoothly increasing velocity (to the right):
vx (t )  vx  a x t  0  a x t
i
a
vx  a x t
v0
v1
v2
v3
v4
v5
vx
ax
0
t
0
t
Physics 201: Lecture 3, Pg 7
If constant acceleration we can integrate twice
ax
a x  const
t
vx (t )  vx  a x t
vx
i
vxi
t
x  xi  vx t  a x t
i
1
2
2
x
xi
Physics 201: Lecture 3, Pg 8
t
If constant acceleration then we also get:
v  v  2a x ( x f  xi )
2
xf
2
xi
vx ,avg  (vx  vx )
1
2
i
f
Physics 201: Lecture 3, Pg 9
vx,avg  (vx  vx )
1
2
i
f
x f  xi  vx t  a x t
2
x f  xi  vx t  a x t
2
1
2
i
1
2
i
x f  xi
1
 vx  2 a x t
t
i
vx , avg  vx  a x t  vx  (vx  vx )
1
2
i
1
2
f
i
Physics 201: Lecture 3, Pg 10
Displacement with constant acceleration
x  xi  vx t  a x t
i
1
2
2
A particle starting at rest & moving along a line
with constant acceleration has a displacement
whose magnitude is proportional to t2
( x  xi )   a x t
1
2
2
1. This can be tested
2. This is a potentially useful result
Physics 201: Lecture 3, Pg 11
Free Fall

When any object is let go it falls toward the ground !!
The force that causes the objects to fall is called
gravity.

This acceleration on the Earth’s surface, caused by
gravity, is typically written as “little” g

Any object, be it a baseball or an elephant,
experiences the same acceleration (g) when it is
dropped, thrown, spit, or hurled, i.e. g is a constant.
a y  -g
y(t )  y0  v y t  g t
0
1
2
2
Physics 201: Lecture 3, Pg 12
Gravity facts:

g does not depend on the nature of the
material !
 Galileo (1564-1642) figured this out
without fancy clocks & rulers!

Feather & penny behave just the same in
vacuum

Nominally,
g = 9.81 m/s2
At the equator g = 9.78 m/s2
At the North pole
g = 9.83 m/s2
Physics 201: Lecture 3, Pg 13
Exercise 1
Motion in One Dimension
When throwing a ball straight up, which of the following is
true about its velocity v and its acceleration a at the highest
point in its path?
A.
B.
C.
D.
Both v = 0 and a = 0
v  0, but a = 0
v = 0, but a  0
None of the above
y
Physics 207: Lecture 3, Pg 14
Throwing a ball up
You throw a ball up at 9.8 m/s, how high does it goes
and how long does it take?
 initial velocity +9.8 m/s
final velocity 0 m/s

v y (t )  v y  a y t
i
0 m/s  9.8 m/s  9.8 m/s t
2
t  1.0 s
y (t )  y0  v y t  12 g t 2
i
 9.8 m/s (1s)  4.9 m/s 2 (1s) 2
y  4.9 m
Physics 207: Lecture 3, Pg 15
Throwing a ball up
You throw a ball up at 9.8 m/s,
 Ignoring air resistance, how fast is it travelling when it
falls past you?

y (t )  y0  v y t  12 g t 2
i
0 m  9.8 m/s (t )  4.9 m/s 2 (t ) 2
t  0.0 s , t  2.0 s
v y (t )  v y  a y t
i
 9.8 m/s  9.8 m/s (2.0 s)
2
 9.8 m/s
Physics 207: Lecture 3, Pg 16
Exercise 1D Freefall

Alice and Bill are standing at the top of a cliff of
height H. Both throw a ball with initial speed v0,
Alice straight down and Bill straight up. The speed
of the balls when they hit the ground are vA and vB
respectively.
A.
v A < vB
Alice
B.
v A = vB
v0
Bill
v0
C.
v A > vB
H
vA
vB
Physics 207: Lecture 3, Pg 18
Exercise 1D Freefall : Graphical solution

Alice and Bill are standing at the top of a cliff of
height H. Both throw a ball with initial speed v0,
Alice straight down and Bill straight up.
cliff
v0
vx
turnaround
point
back
at
cliff
v= -g t
t
identical displacements
(one + and one -)
-v0
vground
ground
ground
Physics 207: Lecture 3, Pg 19
Exercise 2,1D Freefall
The graph at right shows the y
velocity versus time graph for a
ball. Gravity is acting downward
in the -y direction and the x-axis
is along the horizontal.
Which explanation best fits the
motion of the ball as shown by
the velocity-time graph below?
A.
B.
C.
D.
E.
The ball is falling straight down, is caught, and is then thrown straight
down with greater velocity.
The ball is rolling horizontally, stops, and then continues rolling.
The ball is rising straight up, hits the ceiling, bounces, and then falls
straight down.
The ball is falling straight down, hits the floor, and then bounces
straight up.
The ball is rising straight up, is caught and held for awhile, and then is
thrown straight down.
Physics 207: Lecture 3, Pg 20
Problem Solution Method:
Five Steps:
1)
Focus the Problem
-
2)
Describe the physics
-
3)
what are the relevant physics equations
Execute the plan
-
5)
what physics ideas are applicable
what are the relevant variables known and unknown
Plan the solution
-
4)
draw a picture – what are we asking for?
solve in terms of variables
solve in terms of numbers
Evaluate the answer
-
are the dimensions and units correct?
do the numbers make sense?
Physics 207: Lecture 3, Pg 21
A science project
You drop a bus off the Willis Tower (442 m above the
side walk). It so happens that Superman flies by at
the same instant you release the bus. Superman is
flying down at 35 m/s.
 How fast is the bus going when it catches up to
Superman?

Physics 207: Lecture 3, Pg 22
A “science” project
You drop a bus off the Willis Tower (442 m above the
side walk). It so happens that Superman flies by at
the same instant you release the car. Superman is
flying down at 35 m/s.
 How fast is the bus going when it catches up to
Superman?

yi

Draw a picture
y
0
t
Physics 207: Lecture 3, Pg 23
A “science” project
yi
Draw a picture
 Curves intersect at
two points

y
0
g 2
y   t  vSuperman t
2
g
2
 t  vSuperman
t   vSuperman
2
g
2
v bus   gt  g vSuperman  2vSuperman
g
Physics 207: Lecture 3, Pg 24
t