v = = i-Clicker Question

Physics 123 Lecture 2 1 Dimensional Motion Correction:
Review:
• Displacement: Dx = x2 - x1
(If Dx < 0, the displacement vector points to the left.)
• Average velocity:
vav 
(Not the same as
average speed)
x2 x1
t 2 t1

x1
Dx
Dt
x2
Calculus is a co-requisite for this course, not a
prerequisite.
•
If you have not taken calculus before, you will learn
about derivatives this semester in Analytical Physics
and in Calculus.
Quiz in recitation next week:
slope = vav
• Based on last week’s homework.
x-t diagram:
• Understand how to solve HW problems.
v(t2)
x = x(t)
•
v(t2) > v(t1)
• Be able to solve problems fairly quickly (there are only
10 minutes for the quiz).
v(t)
• Instantaneous velocity:
v  lim DDxt 
Dt 0
dx
dt
v(t) is slope of tangent to x-t plot at time t.
v is not constant in time
v = v(t)  acceleration
i-Clicker Question
How many beans are in the 900 ml beaker?
• acceleration: time rate of change of velocity
A. Fewer than 1000
B. 1000-1500
C. 1500-2000
D. 2000-2500
• average acceleration:
aav 
v 2 v1
t 2 t1

• instantaneous acceleration: a  lim
E. More than 2500
(slope of line tangent to v(t) at time t )
a
dv d 2 x

dt dt 2
Dt 0
Dv
Dt
Dv
Dt

dv
dt
a is slope of v(t) graph.
a is curvature of x(t) graph.
In v-t plot above, v(t) is a straight line  constant acceleration
Show estimation game!
i.e.: v(t) = [const]t  dv/dt = [const]
Not always true!
Suppose:
v(t) = Ct3
Then: a = a(t) = dv/dt = 3Ct2  constant in time!
Finding acceleration on a v-t graph
Acceleration from x-t plot:
The (t) graph may be used to find the
instantaneous acceleration and the average
acceleration.
>0
> 0 (> vA)
> 0 (> vB)
> 0 (< vC)
=0
<0
<0
>0
>0
=0
<0
<0
=0
>0
Slope of v-t plot
gives instantaneous
acceleration
Copyright © 2012 Pearson Education Inc.
i-Clicker Question
Constant acceleration is an important
special case!
Deserves special attention!!
differentiate
a* (t-ta) v(t) = v(ta) + a* (t-ta)
vx(ta)
Let ta = 0  v(t) = vo + at
differentiate
½at2
Show motion sensor!
vo t
xo
x(t) = xo + vot + ½at2
KINEMATIC EQUATIONS in 1D
1
Lets put these equations to work!
Drag race: Constant acceleration along
v(t) = vo + at
constant
acceleration
2 x(t) = xo + vo t + ½ at2
400 m track.
v = 150 m/s at end.
Other helpful relationships:
vav 
3
x xo
t
;
vav 
4
vo  v
2
const.
acc.
only
Known: (x – xo) = 400 m; v = 150 m/s; vo = 0
Need: a = ?
ALGEBRA:
 x – xo = vav t
3
subst.
vavx
 v  vo 
( x xo )  
t
 2 
from 4
t = (v - vo) / a
rewrite 1 
• What is the acceleration?
( x xo ) 
5
6
plug 6 into 5
 v  vo   v vo 
( x xo )  


 2  a 
( x xo ) 
v 2 vo2
2a
a
v 2 vo2
a
2a
v 2 vo2
0
2( x xo )
(150 m/s ) 2
 28 m/s 2
2(400 m)
• How long does the race take?
Known: (x – xo) , v, vo, and a
Need: t = ?
x(t) = xo + vot + ½at2
0
x - xo = vot + ½at2
400 m = ½ (28 m/s2) t2 
t = 5.3 s
Yellow Light
iClicker
A motorcycle traveling along the x
axis is
accelerating at a rate of a = 4m/s2.
• Driving at 30 m/s
• Light turns yellow when you are 30 m from int.
• Decelerate at 10 m/s2.
• Will
you stop before intersection? No!
Known: vo = 30 m/s; a = -10 m/s2 ; vf = 0 m/s;
Need: (xf - xo) = ?
(x f
xo ) 
v 2f
2
vo
2a

0 (30 m/s) 2
 45 m
2(10 m/s 2 )
• What should a be?
Known: (xf - xo) = 30 m; vo = 30 m/s; vf = 0 m/s
Need: a = ?
a
v 2f
2( x f
2
vo
xo )

0 (30 m/s) 2
 15 m/s 2
2(30 m)
a. The motorcycle is speeding up.
b. The motorcycle is slowing down.
c. The motorcycle is neither speeding up nor
slowing down.
d. The motorcycle is both speeding up and
slowing down.
e. The motorcycle may be slowing down or
speeding up.
v
Slowing down
a
v
• If a = -30 m/s2, where will I stop?
(xf - xo) ~ 1/a so (xf - xo) = 15 m
Speeding up
a
Freely falling bodies
FREE FALL
Motion in 1-D under the influence of gravity.
• Free fall is the
motion of an
object under the
influence of only
gravity.
• acceleration due to gravity is constant
(at Earth’s surface)
a = -g where g = 9.80 m/s2
• gravity acts vertically downward
(choose y-axis as vertical)
• In the figure, a
strobe light
flashes with equal
time intervals
between flashes.
 Same equations of motion…
BUT: a is replaced with –g!
v(t) = vo - gt
• The velocity
change is the
same in each time
interval, so the
acceleration is
constant.
y(t) = yo + vot - ½gt2
( y yo ) 
v 2 vo2
2g
Copyright © 2012 Pearson Education Inc.
EXAMPLE: REACTION TIME
(red rulers)
EXAMPLE:
Drop a penny from top of the
Empire State Building !
(DO NOT TRY THIS!)
Observe: The penny takes 8.1 s to hit ground
• How tall is building?
Known: vo = 0 m/s; a = -g; t = 8.1 s; yo = 0
Need: y - yo
Known: yo = 0 m; vo = 0 m/s ; a = -g ; yf = - 0.10 m
Need:
0
t = ??
0
y = yo + vot - ½gt2
0
0
y = yo + vot - ½gt2
yf = - ½gt2
2y
2(0.10 m)
t

2
g
9.8 m/s
t  0.02 s 2  0.14 s
y = - ½gt2 = -(½)(9.8 m/s2)(8.1 s)2
y = - 320 m
• What’s the velocity of the penny just before
it hits the ground?
Known: vo = 0 m/s; a = -g; t = 8.1 s; and (y - yo )= -320 m
v = -gt = - (9.8 m/s2)(8.1 s) = -79 m/s
What if I first throw coin upward
with speed of 67 mi/hr (=30 m/s)?
• When will coin reach max height?
(above starting point)
Known: vo = +30 m/s; a = -g
Need:
t when v = 0
v = vo - gt
Things you always wanted to know but were afraid to ask…
1. Can a penny dropped from the Empire
State Building kill a person or embed itself
in the sidewalk?
2. Is it OK to neglect air resistance?
0 m = 30 m/s – (9.8 m/s2)t
t
vo
30 m/s

 3s
g 9.8 m/s 2
Ask the Mythbusters!
• When will it pass me on the way down?
y = yo + vot - ½gt2 but y = yo = 0
0 = vot - ½gt2 = t (vo - ½gt)
t = 0 or t = 6 s
• What is velocity just before hitting ground?
( y yo ) 
v 2 vo2
v = - 85 m/s
2g
18
EXAMPLE:
Drop a penny from top of the
Empire State Building !
(DO NOT TRY THIS!)
Observe: The penny takes 8.1 s to hit ground
• How tall is building?
Known: vyo = 0 m/s; a = -g; t = 8.1 s; yo = 0
Need: y - yo
0
0
y = yo + voyt - ½gt2
y = - ½gt2 = -(½)(9.8 m/s2)(8.1 s)2
y = - 320 m
• What’s the velocity of the penny just before
it hits the ground?
Known: vyo = 0 m/s; a = -g; t = 8.1 s; and (y - yo )= -320 m
v = -gt = - (9.8 m/s2)(8.1 s) = -79 m/s
BUT: Terminal velocity = -29 m/s !!!

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