Transcript Uniform Acceleration

Uniform Acceleration
Motion in One Dimension
Sects. 5-7: Outline
1. Kinematics of one dimensional motion with uniform acceleration:
v f  vi  at
x f  xi  vi t  at
v f  vi  2ax
x f  xi  vt
2
2
1
2
2. Special case: Freely falling objects  a = -g = - 9.8 m/s2
3. Special case: a = 0  motion with uniform velocity v = constant
2
Uniform Acceleration
• The derivation is in the text twice, once using algebra &
once using calculus. Read on your own!
• Notation (drop subscripts x on v & a, for motion along x only!)
t  0 = time when the problem begins
xi  initial position (at t = 0, often xi = 0)
vi  initial velocity (at t1 = 0)
t  time when we wish to know other quantities
xf  position at time t
vf  velocity at time t
a  acceleration = constant
(Average & instantaneous accelerations are equal)
Uniform Acceleration Equations
• Results (one dimensional motion only!):
vf = vi + at
(1)
xf = xi + vi t + (½)a t2
(2)
(vf)2 = (vi)2 + 2a(x - x0) (3)
vavg = (½)(vf + vi)
(4)
NOT VALID UNLESS a = CONSTANT!!!
Often xi = 0. Sometimes vi = 0
Physics and Equations
• IMPORTANT!!!
– Even though these equations & their applications
are important, Physics is not a collection of
formulas to memorize & blindly apply!
– Physics is a set of PHYSICAL PRINCIPLES.
– Blindly searching for the “equation which will
work for this problem” is DANGEROUS!!!!
– On exams, you get to have an 3´´  5´´ index card
with anything written on it (both sides) you wish.
On quizzes, I will give you relevant formulas.
Example: Acceleration of Car
Known: x0 = 0, x = 30 m, v0 = 0, a = 2.0 m/s2
Wanted: t
Use: x = (½)a t2

t = (2x/a)½ = 5.48 s
Example: Estimate Breaking
Distances

v = v0 = constant = 14 m/s
t = 0.50 s
a=0
x = v0t = 7 m

a = - 6.0 m/s2
v decreases from 14 m/s to zero
x0 = 7 m, v0 = 14 m/s, v = 0
v2 = (v0)2 + 2a(x – x0)
 x = x0 + [v2 - (v0)2]/(2a)
= 7 m + 16 m = 23 m
Example: Fastball
Known: x0 = 0, x = 3.5 m, v0 = 0, v = 44 m/s
Wanted: a
Use: v2 = (v0)2 + 2a (x - x0)
 a = (½)[v2 - (v0)2]/(x - x0) = 280 m/s2 !
Example 2.7: Carrier Landing, p. 35
Problem: A jet lands on an aircraft carrier at 140 mi/h (63 m/s).
a) Calculate the acceleration (assumed constant) if it stops in t = 2.0 s
due to the arresting cable that snags the airplane & stops it.
b) If the plane touches down at position xi = 0, calculate it’s final
position.
Example 2.8, Watch Out for the Speed Limit! p. 39
Problem: A car traveling at a constant velocity of magnitude 41.4 m/s passes a trooper
hidden behind a billboard. One second after the speeding car passes the billboard, the
trooper sets out from the billboard to catch it, accelerating at a constant rate of 3.90
m/s2. How long does it take her to overtake the speeding car?
Freely Falling Objects
Freely Falling Objects
• Important & common special case of uniformly
accelerated motion:
“FREE FALL”
Objects falling in Earth’s gravity. Neglect air resistance.
Use one dimensional uniform acceleration equations
(with some changes in notation, as we will see)
A COMMON MISCONCEPTION!
• Experiment finds that the acceleration of
falling objects (neglecting air resistance) is
always (approximately) the same, no matter
how light or heavy the object.
• Acceleration due to gravity, a  g
g = 9.8 m/s2
(approximately!)
• Acceleration of falling objects is always the
same, no matter how light or heavy.
• Acceleration due to gravity, g = 9.8 m/s2
• First proven by Galileo Galilei
Legend: Dropped
objects off of the
leaning tower of Pisa.
• Acceleration due to gravity
g = 9.8 m/s2
(approximately)
– Depends on location on Earth, latitude, & altitude:
• Note: My treatment is slightly different than the book’s, but it
is, of course, equivalent!
• To treat motion of falling objects, use the same
equations we already have, but change notation
slightly:
Replace a by g = 9.8 m/s2
– But in the equations it could have a + or a - sign
in front of it! Discuss this next!
• Usually, we consider vertical motion to be in the y
direction, so replace xf by yf and xi by yi (often y0i = 0)
NOTE!!!
Whenever I write the symbol g, it
ALWAYS means the POSITIVE
numerical value 9.8 m/s2! It NEVER
is negative!!! The sign (+ or -) of the
gravitational acceleration is taken
into account in the equations we now
discuss!
Sign of g in 1d Equations
• Magnitude (size) of g = 9.8 m/s2 (POSITIVE!)
– But, acceleration is a vector (1 dimensional), with 2
possible directions.
– Call these + and -.
– However, which way is + and which way is - is
ARBITRARY & UP TO US!
– May seem “natural” for “up” to be + y and “down”
to be - y, but we could also choose (we sometimes
will!) “down” to be + y and “up” to be - y
– So, in equations g could have a + or a - sign in
front of it, depending on our choice!
Directions of Velocity & Acceleration
• Objects in free fall ALWAYS have downward
acceleration.
• Still use the same equations for objects thrown
upward with some initial velocity vi
• An object goes up until it stops at some point
& then it falls back down. Acceleration is
always g in the downward direction. For the
first half of flight, the velocity is UPWARD.
 For the first part of the flight, velocity &
acceleration are in opposite directions!
VELOCITY & ACCELERATION
ARE NOT NECESSARILY IN
THE SAME DIRECTION!
Equations for Bodies in Free Fall
• Written taking “up” as + y!
v f = vi - g t
yf = yi + vit – (½)gt2
(vf)2 = (vi)2 - 2g(yf - yi)
vavg = (½)(vf + vi)
g = 9.8 m/s2
Often, yi = 0. Sometimes vi = 0
(1)
(2)
(3)
(4)
Equations for Bodies in Free Fall
• Written taking “down” as + y!
v f = vi + g t
yf = yi + vit + (½)gt2
(vf)2 = (vi)2 + 2g(yf - yi)
vavg = (½)(vf + vi)
g = 9.8 m/s2
Often, yi = 0. Sometimes vi = 0
(1)
(2)
(3)
(4)
Example
v1 = (9.8)(1)
= 9.8 m/s
v2 = (9.8)(2)
= 19.6 m/s
v3 = (9.8)(3)
= 29.4 m/s
Note: y is positive
DOWNWARD!
v = at
y = (½) at2
a = g = 9.8 m/s2
Example
Example 2.12: Not a bad throw for a rookie! p. 38
Problem: A stone is thrown at point (A) from the top
of a building with an initial velocity of vi = 19.2 m/s
straight upward. The building is H = 49.8 m high, and
the stone just misses the edge of the roof on its way
down, as in the figure. Answer these questions:
a) Calculate the time at which the stone reaches
its maximum height.
b) Calculate the maximum height of the stone above
the rooftop.
c) Calculate the time at which the stone returns to the
level of the thrower
d) Calculate the velocity of the stone at this instant.
e) Calculate the velocity & position of the stone at time t = 5 s.
Motion in One Dimension
Sects. 5-7: Summary
1. Kinematics of one dimensional motion with uniform acceleration:
v f  vi  at
x f  xi  vi t  at
v f  vi  2ax
x f  xi  vt
2
2
1
2
2. Special case: Freely falling objects  a = -g = - 9.8 m/s2
3. Special case: a = 0  motion with uniform velocity v = constant
2