Acceleration
• Is a change in velocity over
time.
• Is usually described as
“speeding up” or “slowing
down”, but ……
• If an objects direction changes
its direction, it is accelerating
even if it is speed is not
changing.
• Is a vector!
• If acceleration and velocity are
in same direction, the object
speeds up. Opposite directions,
the object slows down.
Definitions
Average velocity:
vavg
x x2  x1


t t2  t1
Average acceleration:
aavg
v v2  v1


t t2  t1
Formulas based on definitions:
x  x0 
v0  v f
2
t
v f  v0  at
Derived formulas:
x  x0  v0t  at
1
2
2
x  x0  v f t  at
1
2
2
2a( x  x0 )  v  v
2
f
2
0
For constant acceleration only
Review of Symbols and Units
• Displacement (x, xo); meters (m)
• Velocity (v, vo); meters per second (m/s)
• Acceleration (a); meters per s2 (m/s2)
• Time (t); seconds (s)
• Values with an
subscript means it is the
initial value. Values without it are final.
o
Use of Initial Position x0 in Problems.
0
x  x0 
v0  v f
2
0
t
x  x0  v0t  at
1
2
0
2
x  x0  v f t  at
1
2
0
If you choose the
origin of your x,y
axes at the point of
the initial position,
you can set x0 = 0,
simplifying these
equations.
2
2a( x  x0 )  v  v
v f  v0  at
2
f
2
0
The xo term is very
useful for studying
problems involving
motion of two bodies.
Problem Solving Strategy:
 Draw and label sketch of problem.
 Indicate + direction.
 List givens and state what is to be found.
Given: ____, _____, _____ (x,v,vo,a,t)
Unkowns: ____, _____
 select Equation containing one and not
the other of the unknown quantities, and
Solve for the unknown.
Example 6: A airplane flying initially at 400
ft/s lands on a carrier deck and stops in a
distance of 300 ft. What is the acceleration?
+400 ft/s
v=0
300 ft
+
vo
X0 = 0
Step 1. Draw and label sketch.
Step 2. Indicate + direction.
Example: (Cont.)
v=0
300 ft
+
Step 3. List given; find
information with signs.
List t = ?, even though
time was not asked for.
+400 ft/s
vo
X0 = 0
Given: vo = +400 ft/s
v=0
x = +300 ft
Find: a = ?; t = ?
Continued . . .
v=0
x
+400 ft/s
300 ft
+
F
vo
X0 = 0
Step 4. Select equation
that contains a and not t.
a=
-vo2
2x
=
-(400 ft/s)2
2(300 ft)
0
0
2a(x -xo) = v2 - vo2
Initial position and
final velocity are 2zero.
a = - 267 ft/s
Why isForce
the acceleration
negative?
Because
is in a negative
direction!
Acceleration Due to Gravity
• Every object on the earth
experiences a common force: the
force due to gravity.
• This force is always directed
toward the center of the earth
(downward).
• The acceleration due to gravity is
relatively constant near the
Earth’s surface.
g
W
Earth
Gravitational Acceleration
• In a vacuum, all objects fall
with same acceleration.
• Equations for constant
acceleration apply as usual.
• Near the Earth’s surface:
a = g = 9.8 m/s2 or approximately 10 m/s2
Directed downward (usually negative).
Sign Convention:
A Ball Thrown
Vertically Upward
avy==
=-0
+
avy=
==-++
ya
=+-v ==
•
UP = +
Release Point
Displacement is positive
(+) or negative (-) based
on LOCATION.
vya==
=-0-
•
yv=
=a=Negative
Negative
Tippens
Velocity is positive (+) or
negative (-) based on
direction of motion.
• Acceleration is (+) or (-)
based on direction of force
(weight).
Same Problem Solving
Strategy Except a = g:
 Draw and label sketch of problem.
 Indicate + direction and force direction.
 List givens and state what is to be found.
Given: ____, _____, a = - 9.8 m/s2
Find: ____, _____
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.
Example 7: A ball is thrown vertically upward with
an initial velocity of 30 m/s. What are its position
and velocity after 2 s, 4 s, and 7 s?
Step 1. Draw and
label a sketch.
Step 2. Indicate + direction
and force direction.
+
a=g
Step 3. Given/find info.
a =-9.8 m/s2
t = 2, 4, 7 s
vo = + 30 m/s y = ? v = ?
vo = +30 m/s
Finding Displacement:
Step 4. Select equation
that contains y and not v.
0
y  y0  v0t  at
1
2
+
a=g
2
y = (30 m/s)t + ½(-9.8 m/s2)t2
Substitution of t = 2, 4, and 7 s
will give the following values:
vo = 30 m/s
y = 40.4 m; y = 41.6 m; y = -30.1 m
Finding Velocity:
Step 5. Find v from equation
that contains v and not x:
v f  v0  at
v f  30 m/s  (9.8 m/s )t
+
a=g
2
Substitute t = 2, 4, and 7 s:
vo = 30 m/s
v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s
Example 7: (Cont.) Now find
the maximum height attained:
Displacement is a maximum
when the velocity vf is zero.
v f  30 m/s  (9.8 m/s )t  0
2
30 m/s
t
; t  3.06 s
2
9.8 m/s
To find ymax we substitute t
= 3.00 s into the general
equation for displacement.
+
a=g
vo = +96 ft/s
y = (30 m/s)t + ½(-9.8 m/s2)t2
Example 7: (Cont.) Finding the maximum height:
y = (30 m/s)t + ½(-9.8 m/s2)t2
t = 3.00 s
Omitting units, we obtain:
+
a=g
y  (30)(3.06)  (9.8)(3.06)
1
2
y = 90.0 m – 45.0 m
vo =+30 m/s
ymax = 45.0 m
2