ACCELLERATION
-Acceleration shows how fast velocity changes
-Acceleration is the “velocity of velocity”
x
v
t
v
a
t
x dx
v  lim

t 0 t
dt
a  lim
v dv

t  0 t
dt
Uniform acceleration
Uniform motion
x
t
v
v
a
t
x  x  x0
t  t  t0
v  v  v0
t  t  t 0
v  v0  at  t 0 
x  x0  vt  t 0 
v(t)
x(t)
v0
t
t0
Δx
Δt
t
Δt
t
t
v0  v
v  v0  at
 t  0
 t
2
2
at 2
at 2
x  v0  t 
 x  x0  v0  t 
2
2
at 2
if t 0  0  t  t  x  x0  v0 t 
2
x 
v(t)
Δx
v
For any motion:
x  v t
For uniform acceleration:
v0  v
Δx 
Δt
2
Velocity
dx
vt  
dt
t
x  x0   vt dt
Acceleration
dv
dt
a
t
vt   v 0   a t dt
t0
t0
Uniform motion
v  const
x  x0  vt  t 0 
v0  v
v
2
Uniform acceleration
a  const
v  v0  at  t 0 
t
t
t0
t0
x  x0   vt dt  x0   v0  at  t 0 dt
at  t 0 
x  x0  v0 t  t 0  
2
2
Base equations for 1D uniform acceleration
at 2
x  x0  v0 t 
2
v  v0  at
2 equations
2 quantities can be found
if t 0  0  t  t
-What to remember?
-How to use?
Two useful equations that can be derived from the base equations
v0  v
1. x 
t
2
See how it was derived on previous slide
2. v  v0  at
at 2
 x  x0   v0 t 
2
2ax  x0   at 2v0  at   v  v0 2v0  v  v0   vv  v0 v  v0   v 2  v02
2a x  x0   v 2  v02
Example: A train car moves along a long straight track. The graph
shows the position as a function of time for this train. The graph shows
that the train:
1. speeds up all the time.
2. slows down all the time.
3. speeds up part of the time and
slows down part of the time.
Steepness of slope
is decreasing
4. moves at a constant velocity.
time
Positive Acceleration = a smile
Negative Acceleration = a frown


time
time
Example: The graph shows position as a function of time for two
trains running on parallel tracks. Which of the following is true?
1. At time t0, both trains have the
same velocity.
2. Both trains speed up all the time.
3. Both trains have the same
velocity at some time before t0.
4. Somewhere on the graph, both
trains have the same acceleration.
t1
t0
Same slope at t = t1
Position
dx v = slope of x(t)
dt
Acceleration
Velocity
v 
dv d 2x a = slope of v(t)
a 

or
dt
dt 2
a = curvature of x(t)
Position
x  0
Displacement =
area under v(t) curve
Velocity
x 
v  0

t1
 vdt
t0

Acceleration
Change in velocity =
area under a(t) curve
v 
v
t0
t1
t1

t0
adt
Example v(t) from a(t): Draw the velocity vs. time graph that corresponds to the
following acceleration vs. time graph. Assume that the velocity at t = 0 is zero.
a
Does your graph look like
one of these?
v
t
t
A
v
v
B
t
t
C
a
t
v
t
NB: a < 0 but object
is speeding up.
NB: a > 0 but object
is slowing up.
Free fall
-Free fall acceleration: g=9.8m/s2
Using the two base equations:
at 2
x  x0  v0 t 
2
v  v0  at
Substitute the following into the base equations:
a  g
xy
To derive the following equations:
gt 2
y  y 0  v0 t 
2
v  v0  gt
Example 1. A particle, a material point, is thrown vertically up. Find
the maximum height the particle will reach and the time it will take, if
you are given the initial height and the initial velocity.
Given:
Unknown variables:
t1  t? V
y0
0
1
y?
V
y y 
v1  ? no! 2vg 1  0
v0
g
2
0
y  y max  ?
max
0
t at y max  ?
Solution:
0  v0  gt1
y max
Answer:
v0
t1 
g
v0 gv02
 y 0  v0
 2
g 2g
y max
v02
 y0 
2g
t1 
v0
g
y max
v 02
 y0 
2g
Example 2. A particle, a material point, is thrown vertically up. Find
the velocity with which the particle returns to the point from which it
was thrown, and the time this flight will take. The initial height and
the initial velocity are given.
Given
y0
v0
y  y0
v2  ?
t2  ?
Solution:
gt 2
v0 t 2 
0
t2
 2t
2v 0
v 2  v0  g
g
2
Answer
2v 0
t2 
g
v 2  v 0

1

2v 0
t2 
g
v 2  v 0
Compare to example 1:
t2  2t1
Example 3, Two particles, material points, are thrown vertically up. One particle
is thrown before the other. Find the time at which both objects are at the same
height, and the height at which the objects’ intersection occurs.
Given:
Equations
used:
v0
Unknown
gΔΔ2
Too many
y  y 0  v0 Δt 
variables:
t0
2
variables but
t3  ?
gt 2
y1  v0 t 
2
h ?
y 2  v0 t  t 0  
Solution:
t3  ?
g t  t 0 
2
y1  ?
2
gt 32
g t 3  t 0 


v0 t 3 
 v0 t 3  t 0 
2
2
gt
g t  t 
Vt 
 V t  t  
2 gt 2
2
gt 32
gt 02
3
vV0 tt3 gt  V t  Vvt0 t3 gt vgt0 t 0t gt  gt 3t 0 
2 2
2
2 2
2
gt
t V
gt202  V t  gt t t  2  g
t 0 v0
 v0 t 0  gt 3t 0 t 3  
2
2 g
2
2
2
3
0 3
3
0
3
2
3
0 3
2
3
0 3
3 0
0
3 0
2
0
0 0
2
0
0 0
0
0
0
3
Answer:
y2  ?
y1  y2
Thus, 3 equations
and 3 unknowns.
h  y1 t 3  
2
v  g t
v 
t
 v 0  0  0    0  0  
 2 g  2 2 g 
v0 t 0 v02 gt 02 gt 0 v0 gv02





2
g
8
2g
2g 2
t
v
t3  0  0 ;
2 g
v02 gt 02
h

2g
8
Free fall (review)
Example1: Ball #1 is thrown vertically upwards with a speed of v0 from the top of
a building and hits the ground with speed v1. Ball #2 is thrown vertically downwards
from the same place with the same speed v0 and hits the ground with speed v2.
Which one of the following three statements is true. Neglect air resistance.
A.
v1>v2
B.
v1=v2
C.
v1<v2
D.
Depends on which ball is more massive
E.
None of the above
Example2: You are throwing a ball straight up in the air.
At the highest point, the ball’s
1. velocity and acceleration are zero.
2. velocity is nonzero but its acceleration is zero.
3. acceleration is nonzero, but its velocity is zero.
4. velocity and acceleration are both nonzero.