Announcements
• You may attend any of the instructors’ office hours. List maintained on Blackboard under “Staff / Office Hours”.
• HW #1 due tonight at 11:59 p m• HW #1 due tonight at 11:59 p.m.
• HW #2 now available, due Thurs Sept 9 at 11:59 p.m.
• Solutions to HW #1 will be available on Blackboard (under “Solutions”) after the due date/time.) /
• Quiz grades will be posted to Blackboard after your recitation instructor has graded the quizzes (also returned next week).
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• Last Time: Motion in One Dimension
Di l V l i A l iDisplacement, Velocity, Acceleration
• Today:
One‐Dimensional Motion with Constant AccelerationOne Dimensional Motion with Constant Acceleration
Freely Falling Objects (under gravity)
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Constant/Uniform Acceleration• Constant/Uniform Acceleration: magnitude and direction of
the acceleration does not change.
• Constant acceleration is important, because it applies to many natural phenomena.
g
p
• One such example is objects in “free fall” near the surface of h E h N l i i i ll bj (i d d
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the Earth. Neglecting air resistance, all objects (independent of mass) “fall” with the same downward acceleration !
Constant Acceleration• Key Point:
Under constant acceleration the instantaneousUnder constant acceleration, the instantaneous acceleration at any point in a time interval is equal to the average acceleration over the entire time interval.a e age acce e a o o e e e e e e a
velocityThis Means:l
y
if
ttvv
aa
slope = a
vif tt
Let: ti = 0, vi = v0, tf = t, and vf = vat
i , i 0, f , f
vva 0 atvv
v0v0
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timeta atvv 0
t(for constant a only)
Constant Accelerationvelocity velocity
a > 0 a < 0a > 0
vv
time time
v0
Δt
A l it i ti i t l Δt iAverage velocity in any time interval Δt is :
0 vv Just the average of the velocities at the
20 vvv
the velocities at the start/finish of the time interval !
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(for constant a only)
Constant Acceleration
Recall: The displacement Δx average velocity vand
0xxxxx if txv
Set:t
If ti = 0 :i
tvvtvvtvx
00
21
2(for constant a only)
022
Using + atUsing v = v0 + at :
1 21
(for constant
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tatvvx 0022
0 2attvx (for constant
a only)
Constant Accelerationvelocity
slope = a
For constant acceleration, the displacement Δx is just
v
v0
at the displacement Δx is just equal to the area under the velocity‐vs‐time graph !
time
v0
t
1 20 2
1 attvx Recall: Area of Triangle = ½ (base) (height)
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2
Constant Acceleration
Finally: Recall
tvvx 021
and atvv 0vvt 0
02 0 a
Plugging in for t gives us:
22 avv
avvvvx
221 2
02
00
xavv 222(for constant a only)
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xavv 20(for constant a only)
“Equations of Motion” for l iConstant Acceleration
Equation Information
atvv 0 Velocity as a function of time
20 2
1 attvx Displacement as a function of time
xavv 220
2 Velocity as a function of displacement
These are for motion along a single axis (x‐axis) with acceleration along that direction.• If acceleration and motion along y axis just change x y
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• If acceleration and motion along y‐axis, just change x y.• Assumes that at t = 0, v = v0.• Recall: Δx = x – x0
Example
A car starting from rest undergoes a constant acceleration of a = 2.0 m/s2.
(a) What is the velocity of the car after it has traveled 100 m?
(b) How much time does it take to travel the 100 m?
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Example: Problem 2.27A car traveling east at 40.0 m/s (= 89.48 mph !) passes a trooper hiding at roadside The driverpasses a trooper hiding at roadside. The driver uniformly reduces his speed to 25.0 m/s in 3.50 s.
(a) What is the magnitude and direction of the car’s acceleration as it slows down?
(b) How far does the car travel during the 3.50‐s time period?
v0 = 40 m/s
+xEastWest
‐x
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Example: Problem 2.29A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final velocity of 2 80 m/sslowing down to a final velocity of 2.80 m/s.
(a) Find the truck’s original speed.( ) g p(b) Find its acceleration.
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Freely Falling Objects• Key Point:
h l bl ll b f llWhen air resistance is negligible, all objects falling under the influence of gravity near the Earth’s surface fall at the same constant accelerationfall at the same constant acceleration.
Pl tAristotle (384 – 322 B.C.)
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Plato (428 – 348 B.C.) Heavier objects fall faster than lighter objects.
Galileo Galilei (1564‐1642)
• Italian physicist and philosopher
• Professor at the University of P dPadua
• Invented the thermometer andInvented the thermometer and the pendulum clock
“Galileo, perhaps more than any other single person, wasother single person, was responsible for the birth of modern science.”
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modern science.– Stephen Hawking
Galileo Galilei (1564‐1642)
• First to observe the heavens with a telescope
• Defended the (disturbing) idea of Copernicus (1473‐1543) that the Earth was not the center of the universe1543) that the Earth was not the center of the universe• Tried for heresy by Catholic Church (1992: Vatican found him not guilty)
U d t k i f i t l t di t d ib• Undertook series of experimental studies to describe the motion of bodies: mechanics
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Freely Falling Objects• What is a freely falling object ?
Any object moving freely under the influence of gravity aloneAny object moving freely under the influence of gravity alone, regardless of its initial motion.
• Denote magnitude of free fall acceleration as:
If
• Denote magnitude of free‐fall acceleration as:
g = 9.80 m/s2 (magnitude on the surface of the Earth)
• If we:
• Neglect air resistance
l d ’ h l d ( h• Assume acceleration doesn’t vary with altitude (over short vertical distances)
“up” = +y
Motion of freely falling object is the same as one‐dimensional motion
up = +y
2m/s9 80 ga
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same as one dimensional motion with constant acceleration !
m/s9.80 ga
“Equations of Motion” for l lli bj
+y = “up”
Freely Falling Objects 2m/s 9.80 ga
Equation Informationq
gtvv 0 Velocity as a function of timeg0
21 tt
y
Di l t f ti f ti20 2
gttvy Displacement as a function of time
ygvv 220
2 Velocity as a function of displacement
18Recall: Δy = y – y0
Objects Rising Against Gravity
If object’s initial (upward) velocity is v0 , object s t a (up a d) e oc ty s 0 ,how high will it rise?
yy
y = 0
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y = 0
Objects Rising Against Gravity
If object’s initial (upward) velocity is v0 , object s t a (up a d) e oc ty s 0 ,how high will it rise?
y
First, solve for the time to reach the highest point.
At highest point, v = 0 !y g p
Second, calculate the maximum height after solving for the time to reach the highest point.
Alternative method …
y = 0
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y = 0
Example: Multiple Choice #10 (p. 48)
A student at the top of a building throws a red ball upward with speed v0, and then throws a green ball downwards with the same initial speed v0. Immediately before the balls hit the ground …
l h l i i f
v
y True or False: The velocities of the two balls are equal.
ygvv 220
2
v0
v0y = h
h
21y = 0
Example: Multiple Choice #10 (p. 48)
A student at the top of a building throws a red ball upward with speed v0, and then throws a green ball downwards with the same initial speed v0. Immediately before the balls hit the ground …
l h d f
v
y True or False: The speed of each ball is greater than v0.
ygvv 220
2
v0
v0y = h
h
22y = 0
Example: Multiple Choice #10 (p. 48)
A student at the top of a building throws a red ball upward with speed v0, and then throws a green ball downwards with the same initial speed v0. Immediately before the balls hit the ground …
l h l i f
v
y True or False: The acceleration of the green ball is greater than h f h d b llv0
v0y = h
that of the red ball.
h
23y = 0
Example: Problem 2.50
A small mailbag is released from a helicopter that is descending steadily at 1.50 m/s. After 2.00 s:
(a) What is the speed of the mailbag? ; and(b) How far is it below the helicopter ?
(c) How do (a) and (b) change if the helicopter is rising steadily at 1.50 m/s?
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