i-Clicker Question
How many beans are in the 900 ml beaker?
A. Fewer than 1000
B. 1000-1500
C. 1500-2000
D. 2000-2500
E. More than 2500
Show estimation game!
Physics 123 Lecture 2
1 Dimensional Motion
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
slope = vav
x-t diagram:
x = x(t)
v(t2)
v(t2) > v(t1)
v(t)
• Instantaneous velocity:
v  lim
Dt 0
Dx
Dt
v(t) is slope of tangent to x-t plot at time t.

dx
dt
v is not constant in time
v = v(t)  acceleration
• acceleration: time rate of change of velocity
• average acceleration:
aav 
v 2  v1
t 2 t1

• instantaneous acceleration: a  lim
(slope of line tangent to v(t) at time t )
dv d 2 x
a

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
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
The (t) graph may be used to find the
instantaneous acceleration and the average
acceleration.
Copyright © 2012 Pearson Education Inc.
Acceleration from x-t plot:
>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
i-Clicker Question
Show motion sensor!
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
vo t
xo
x(t) = xo + vot + ½at2
KINEMATIC EQUATIONS in 1D
1
v(t) = vo + at
constant
acceleration
2 x(t) = xo + vo t + ½ at2
Other helpful relationships:
vav 
3
x  xo
t
;
4
vav 
vo  v
2
const.
acc.
only
ALGEBRA:
3  x – xo = vav t
subst.
vavx
 v  vo 
( x  xo )  
t
 2 
from 4
rewrite 1 
t = (v - vo) / a
plug 6 into 5
 v  vo   v  vo 
( x  xo )  


2
a



( x  xo ) 
v 2  vo2
2a
6
5
Lets put these equations to work!
Drag race: Constant acceleration along
400 m track.
v = 150 m/s at end.
• What is the acceleration?
Known: (x – xo) = 400 m; v = 150 m/s; vo = 0
Need: a = ?
v v
2
( x  xo ) 
2
o
a
2a
v 2  vo2
0
2( x  xo )
(150 m/s ) 2
a
 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
• 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  vo2
2a
0  (30 m/s) 2

 45 m
2
2(10 m/s )
• What should a be?
Known: (xf - xo) = 30 m; vo = 30 m/s; vf = 0 m/s
Need: a = ?
v 2f  vo2
0  (30 m/s) 2
a

 15 m/s 2
2( x f  xo )
2(30 m)
• If a = -30 m/s2, where will I stop?
(xf - xo) ~ 1/a so (xf - xo) = 15 m
iClicker
A motorcycle traveling along the x-axis is
accelerating at a rate of a = -4m/s2.
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
Speeding up
a
Freely falling bodies
• Free fall is the
motion of an
object under the
influence of only
gravity.
• In the figure, a
strobe light
flashes with equal
time intervals
between flashes.
• The velocity
change is the
same in each time
interval, so the
acceleration is
constant.
Copyright © 2012 Pearson Education Inc.
FREE FALL
Motion in 1-D under the influence of 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)
 Same equations of motion…
BUT: a is replaced with –g!
v(t) = vo - gt
y(t) = yo + vot - ½gt2
v v
2
( y  yo ) 
 2g
2
o
EXAMPLE: REACTION TIME
(red rulers)
Known: yo = 0 m; vo = 0 m/s ; a = -g ; yf = - 0.10 m
Need:
t = ??
0
0
y = yo + vot - ½gt2
yf = - ½gt2
2y
2(0.10 m)
t

g
 9.8 m/s 2
t  0.02 s 2  0.14 s
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
0
0
y = yo + vot - 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: 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
0 = 30 m/s – (9.8 m/s2)t
vo
30 m/s
t 
 3s
2
g 9.8 m/s
• 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
 2g
v = - 85 m/s
Things you always wanted to know but were afraid to ask…
1. Can a penny dropped from the Empire
State Building embed itself in the
sidewalk (or a person’s skull)?
2. Is it OK to neglect air resistance (drag)?
Ask the Mythbusters!
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 !!!