Transcript Document

Physics 218: Mechanics
Instructor: Dr. Tatiana Erukhimova
Lectures 5,6
dx(t)
v(t) =
dt
dv(t)
a(t) =
dt
ò v(t)dt
v(t) = ò a(t)dt
x(t) =
or
t
If a=ac=Const:
1 2
x(t )  ac t  v(0)t  x(0)
2
v(t )  act  v(0)
v (t2 ) - v (t1) = 2ac (x(t2 ) - x(t1))
2
2
x(t) = x 0 +
ò v(t)dt
0
t
v(t) = v 0 +
ò a(t)dt
0
A “police car” problem
x2 – x1 = 3.5 km
x=0
x1
V1=30m/s
a=const
x2
V2=40m/s
V3=20m/s
a=0
ap=kt
V(t=0)=0
You start moving from rest with constant acceleration. There is a police car
hiding behind the tree. The policeman has a metric radar. He measures your
velocity to be 30 m/s. While the policeman is converting m/s to mph, you
continue accelerating. You meet another police car. This policeman measures
your velocity to be 40 m/s. You also notice the police, drop your velocity to 20
m/s and start moving with a constant velocity. However, it is too late. This
police car starts chasing you with acceleration kt (k is a constant). After some
distance he catches you.
A “police car” problem
x2 – x1 = 3.5 km
x=0
x1
V1=30m/s
a=const
x2
V2=40m/s
V3=20m/s
a=0
ap=kt
V(t=0)=0
1. What was your acceleration before you meet the second police car?
2. How long did you travel from x1 to x2?
3. Find x1
4. At which distance does the police car catch you?
5.Convert the velocity from m/s to mph
2
v




t
An object’s velocity is measured to be x
,
where =4.00 m/s and =2.00 m/s3. At t=0 the object
is at x=0.
Calculate the object’s position and acceleration as
functions of time.
Free fall
On planet Earth, if you neglect
air resistance, any body which is
dropped will experience a
constant acceleration, called g,
independent of its size or weight.
g=9.8 m/s2=32 ft/s2
g-positive!
Galileo Galilei (1564-1624),
the basic law of motion
a
a = g = const
v
for all bodies independently on their
masses
g  9.8 m / s  32 ft / s
2
2
Galileo's “Law of Falling Bodies”
distance (S) is proportional to time (T) squared
Galileo’s notes
Free fall
A person throws a ball upward into the air with an
initial velocity of 15 m/s. Calculate
a)How much time does it take for the ball to reach
the maximum height?
b)How high does it go?
c) How long is the ball in the air before it comes
back to the thrower’s hands
d)The velocity of the ball when it returns to the
thrower’s hand
e)At what time t the ball passes a point 8.00 m
above the person’s hand
A ball is thrown vertically upward with a velocity of
magnitude v1 from a window at height H. What is the
ball’s position and velocity at any time moment? How
long does it take to reach the highest point? How long
does it take to reach the ground? What is the velocity of
the ball when it hits the ground?
A ball is thrown vertically upward with a velocity of
magnitude v1 from a window at height H. In addition to
gravity acting on the ball there is another force so that
the acceleration in the up direction is –g+t where  is a
constant and t is the time. What is the ball’s position
when the acceleration is zero?
A rat is running with known constant velocity v1
when she passes you at t=0. You start from rest with
acceleration c1 where c1 is an unknown constant.
What does c1 have to be in order to catch the rat
after she has gone a distance D?
Falling with air resistance
dv
a
 g  kv
dt
Falling with air resistance
dv
2
a=
= g - kv
dt
Terminal Velocity with Coffee Filters
mg - Fr = ma
where Fr is the resistance force.
Fr
a=gm
1. A penny and a quarter dropped from a ladder land at the
same time (air resistance is negligible).
2. A coin dropped in a coffee filter from a ladder lands later
than a coin without coffee filter (the terminal velocity is
smaller for larger cross-section area).
3. A quarter dropped in a coffee filter will land faster than a
penny in a coffee filter (the terminal velocity is larger for
larger mass)
4. Two identical coins dropped in coffee filters of different
diameters land at different times (the terminal velocity is
smaller for larger cross-section area).
Resistance force: Fr = gAv
2
A – area of the projectile
For a spherical projectile in air at g = 0.25 N ´ s /m
STP:
2
4
Terminal velocity:
Fr
a=g=0
m
Fr = mg
gAv = mg
mg
vT =
gA
2
A 70-kg man with a parachute: vT ~ 5 m/s
A 70-kg man without a parachute: vT ~ 70
m/s
Have a great day!
Reading: Chapters 3,4