Lecture 3
For constant acceleration:
v = v0 + at
•
•
Δx =
1
v + vo ) t
(
2
1 2
Δ
x
=
v
t
+
at
0
\
2
v2 = v02 + 2aΔx
Some points of vertical motion:
g = 9.8 m/s 2 downwards
v = v0 − gt
Δy ≡ y − y0 =
•
•
•
1
(v + vo )t
2
1
Δy = v0t − gt2
2
v2 − v02 = −2gΔy
Here we assumed t0=0 and the y axis
to be vertical.
Vectors and components of vectors
Addition and subtraction of vectors
Acceleration
•
Th average acceleration
The
l i is
i defined
d fi d by
b the
h change
h
off velocity
l i with
i h time
i :
Δv
a=
; Δv = v − v0
Δt
•
In analogy, the instantaneous acceleration can be computed using the same
equation, but in the limit of very small elapsed times.
Δv
a = lim
li Δt →0
Δt
•
Example: A speed boat starts from rest and reaches 3.2 m/s in 2s. What is its
average acceleration? Assuming its acceleration to be constant,
constant what is its
velocity after 5 s?
3.2m / s − 0
a=
2s
= 1.6m / s 2 = a
v = v0 + at
v0
0
= 0 + (1.6m / s 2 )(5s)
t1
2s
= 8 / ms
v1
3.2m/s
t2
5s
v2
?
clicker question
•
1)
Velocity getting more positive
2)
Velocity getting more negative
3)
Velocity getting more positive
4)
Velocity getting more negative
These four motion diagrams show the motion of a particle along the xaxis, where the positive direction is to the right. Some of them depict
positive accelerations, and some depict negative accelerations. Which
are which?
a) 1&2 are positive, 3&4 are negative
b) 1&3 are positive, 2&4 are negative
c) 1&4 are positive, 2&3 are negative
d)) 2&3 are positive,
p
, 1&4 are negative
g
The answer is b).
Motion under constant acceleration
•
•
Recall v=v0+at.
(t,v)
For const. acceleration, 
v=
v =
velocity
1
(v + v0 )
2
1
(v + v0 )
2
((t0,,v0)
time
•
Also: Δx = v t
and v = v0 + at
•
Putting it all together:
1
(v + v0 )t = 1 (v0 + at + v0 )t
2
2
1
 Δx = v0t + at 2
2
Δx =
For constant acceleration :
v = v0 + at
1
(v + vo )t
2
1
Δx = v0t + at 2
2
Δx =
examples
•
A speed
d bboatt starts
t t from
f
restt att t0=0
0 andd x0=0
0 andd accelerates
l t att a rate
t
of a=1.6 m/ss . How far does it go after t=5 seconds?
1
Δx = v0 t + at 2
2
1
= at 2
2
=
2
1
2
1.6m
/
s
5s
(
)( )
2
= 20m
t0
0
x0
0
a
1.6 m/s2
v0
0
t
5s
Δx
?
For constant accelerati on :
v = v0 + at
1
(v + vo )t
2
1
Δx = v0t + at 2
2
Δx =
One more equation
•
•
IIn principle,
i i l these
th
three
th equations
ti
are
enough. However there is a class of
problems where you know v, v0, a, and t.
O more equation
One
ti can make
k suchh
problems easier to solve.
Replace
p
t with v,, v0 and a :
v = v0 + at  t = (v − v0 ) / a
1
1
(
)
Δx = v + v0 t = (v + v0 )(v − v0 ) / a
2
2
 2aΔx = v 2 − v02
For constant
acceleration
on::
constant accelerati
v = v00 + at
1
1
Δxx =
= ((vv +
+ vvo ))tt
Δ
o
22
1
Δx = v0t + 1 at 22
Δx = v0t + 2 at
2
v2 = v02 + 2aΔx
Practice quiz
•
A man iis driving
d i i his
hi car att a velocity
l it off 10 m/s.
/ Suddenly,
S dd l he
h
notices that the car 10 m in front of him is stopped and he slams
on his brakes. Because it is winter, the traction is poor and his car
decelerates with an acceleration of only –3 m/s2. With what
velocity does he hit the stopped car in front of him? Select the
answer below that is closest to the answer that you calculate.
a)
b)
c)
d)
4 m/s
10 m/s
-4 m/s
6.5 m/s
v0
10 m/s
/
a
-3 m/s2
Δx 10 m
v
?
v 2 = v02 + 2aΔx
= (10m / s) + 2 ( −3m / s 2 ) (10m )
2
= 40m 2 / s 2
 v = 6.3m / s
Graphical determination of displacement
v
(m/s)
t=1,v=2
2
v
(m/s)
2
t=1,v=2
0
0
0
t (s)
0
1)What is the displacement covered in 1 second?
2)What is the area indicated by
?
Hint: Use Δx=v0t+1/2at2 or Δx=0.5*(v0+v)t
The area under the v-t curve is equal to the displacement
of the object! Note unit of area: area=2m/s*1s=2m
t (s)
Q 1.
Δx (m)
a) 1.
b) 1.
c) 2.
d) 2.
2.
area(m)
1.
2.
1.
2.
Now, let’s use the fact that the area under the a-t curve is equal to the velocity
of the object! Note unit of area: area=m/s2 *s=m/s in that case
•
Example
Imagine a car moving starting at rest at x=0 and having the following
acceleration as a function of time. What is the velocity and position at
2 s and at 4 s? For constant a: v = v + at; x − x = v t + 1 at 2 ; x − x = 1 v + v t
( o)
0
0
0
0
2
2
Can use area to get v(2) and v(4)
v(2)=aΔt=(1m/s2)⋅2s=2m/s
v(4)=v(2)+
v(4)
v(2)+ aΔt
aΔt=2m/s+(2m/s
2m/s+(2m/s2))⋅2s
2s
v(4)=6 m/s
x by more obvious” route
x=x0+v
+ 0t+1/2at
t+1/2 t2
x(2)=0+0⋅(2s)+1/2⋅(1m/s2)(2s)2=2m
x(4)=2m+2m/s⋅2s+1/2(2m/s2)(2s)2
x(4)=2m+4m+4m=10m
x by “less obvious” route
use x(t)=x
x(t) x0 Δx
Δx=00.55*(v
(v0+v)t to get x(t)
x(2)=0+0.5*(0+2m/s)⋅(2s)=2m
x(4)=2m+0.5*(2m/s+6m/s)⋅(2s)=10m
Conceptual Reading Quiz
•
If you d
drop an object
bj t iin th
the absence
b
off air
i resistance,
i t
it
accelerates downward at 9.8 m/s2. If instead you throw it
downward, its downward acceleration after release is:
– a)) lless th
than 9
9.8
8 m/s
/ 2.
– b) 9.8 m/s2.
– c) more than 9.8 m/s2.
“It ain't what you don't know that gets you into
trouble It
trouble.
It'ss what you know for sure that just ain
ain'tt so.
so ”
Mark Twain
Motion under Earth’s gravitational attraction
•
Near th
N
the E
Earth’s
th’ surface,
f
all
ll objects
bj t are accelerated
l t d by
b gravity
it
downwards with an acceleration of g=9.8 m/s2.
g = 9.8 m/s 2 downwards
v = v0 − gt
Δy ≡ y − y0 =
1
(v + vo )t
2
1
Δy = v0t − gt2
2
v2 − v02 = −2gΔy
A timed drop can provide
one measurement of g.
http://www.youtube.com/watch?v=-4_rceVPVSY&feature=related
Example 1
•
At t0=0
0L
Lucy throws
th
a ball
b ll straight
t i ht upwards
d with
ith a velocity
l it off 30 m/s.
/
It rises a distance h and then falls downward where Lucy catches it.
– 1) When does the ball stop?
g = 9.8 m/s 2 downwards
– 2) How high does it go?
v = v0 − gt
v = vo − gt
v0
g
t
v=0
gt
t
30m / d
30
9.8m / s 2
v 2 = v02 − 2g
2 Δy
2gh
h=
v
2
0
v02
(
2
2 9.8m / s
2
)
Δy ≡ y − y0 =
3.06s
= 46m
1
(v + vo )t
2
1
Δy = v0t − gt2
2
v2 − v02 = −2gΔy
2 h
2gh
v02
2g
h
(30m / s)
v0
v0 30 m/s
v
0
Example 2
•
At t0 Lucy
L
throws
th
a ball
b ll straight
t i ht upwards
d with
ith a velocity
l it off 30 m/s.
/ It
rises a distance h and then falls downward where Lucy catches it.
– 1) When does the ball return to Lucy’s hand?
g = 9.8
9 8 m/s 2 downwards
– 2) What is its velocity when Lucy catches it?
v = v0 − gt
• a) 30 m/s
1
Δy ≡ y − y0 = (v + vo )t
• b) 60 m/s
2
• c) – 30 m/s
1
Δy = v0t − gt2
• d) – 60 m/s
2
v2 − v02 = −2gΔy
v0
30 m/s
Δy 0
Symmetries of trajectory
Trajectory of Lucy's ball
velocity of Lucy's ball
50
40
30
30
20
Δt
Δt
0
-1
0
1
2
3
4
Δt
10
0
-10
-20
tmax
10
Δt
-30
5
6
7
-40
-1
1
Δy = v0 t − gt 2
2
y(t
( max+Δt)
Δ ) = y(t
( max-Δt)
Δ)
It takes the same time to go up
as down.
0
1
2
3
4
5
6
7
t (s)
t (s)
•
•
tmax
20
v (m/s)
y (m)
40
v = v0 − gt
•
•
( max+Δt)
Δ ) = -v(t
( max--Δt)
Δ)
v(t
Speed at the end is the same as
in the beginning.
The acceleration remains –g because
the slope remains –g everywhere!!!
Clicker quiz
• Where does the
acceleration vanish?
50
40
y (m
m)
a)) att 0 s
b) at 6 s
c) at 0 s and 6 s.
s
d) at 3 s
e)) never
Trajectory of Lucy's ball
30
20
Δt
Δt
tmax
10
0
-1
0
1
2
3
t (s)
4
5
6
7