Download Acceleration

Definition of Acceleration
 An acceleration is the change in velocity
per unit of time. (A vector quantity.)
 A change in velocity requires the
application of a push or pull (force).
A formal treatment of force and acceleration will
be given later. For now, you should know that:
• The direction of acceleration is same as
direction of force.
• The acceleration is
proportional to the
magnitude of the force.
The Signs of Acceleration
• Acceleration is positive (+) or negative
(-) based on the direction of force.
+
F
a (-)
F
a(+)
Choose + direction first.
Then acceleration a will
have the same sign as
that of the force F —
regardless of the
direction of velocity.
Average and Instantaneous a
aavg
Dv v2  v1


Dt t2  t1
ainst
Dv

(Dt  0)
Dt
slope
v2
Dv
Dv
v1
Dt
Dt
t1
t2
time
Example 3 (No change in direction): A constant
force changes the speed of a car from 8 m/s to
20 m/s in 4 s. What is average acceleration?
+
Force
t=4s
v1 = +8 m/s
Step
Step
Step
Step
1.
2.
3.
4.
v2 = +20 m/s
Draw a rough sketch.
Choose a positive direction (right).
Label given info with + and - signs.
Indicate direction of force F.
Example 3 (Continued): What is average
acceleration of car?
+
Force
t=4s
v1 = +8 m/s
v2 = +20 m/s
20 m/s - 8 m/s
Step 5. Recall definition
a
 3 m/s
of average acceleration.
4s
aavg
Dv v2  v1


Dt t2  t1
a  3 m/s, rightward
Example 4: A wagon moving east at 20 m/s
encounters a very strong head-wind, causing it
to change directions. After 5 s, it is traveling
west at 5 m/s. What is the average
acceleration? (Be careful of signs.)
+
vf = -5 m/s
E
Force
vo = +20 m/s
Step 1. Draw a rough sketch.
Step 2. Choose the eastward direction as positive.
Step 3. Label given info with + and - signs.
Example 4 (Cont.): Wagon moving east at 20 m/s
encounters a head-wind, causing it to change
directions. Five seconds later, it is traveling west at
5 m/s. What is the average acceleration?
Choose the eastward direction as positive.
Initial velocity, vo = +20 m/s, east (+)
Final velocity, vf = -5 m/s, west (-)
The change in velocity, Dv = vf - v0
Dv = (-5 m/s) - (+20 m/s) = -25 m/s
Example 4: (Continued)
+
vf = -5 m/s
aavg =
Dv
Dt
E
vo = +20 m/s
Force
Dv = (-5 m/s) - (+20 m/s) = -25 m/s
=
vf - vo
tf - to
a = - 5 m/s2
a=
-25 m/s
5s
Acceleration is directed to
left, west (same as F).
Signs for Displacement
+
D
vf = -5 m/s
C
A
vo = +20 m/s
E
Force
B
a = - 5 m/s2
Time t = 0 at point A. What are the signs
(+ or -) of displacement at B, C, and D?
At B, x is positive, right of origin
At C, x is positive, right of origin
At D, x is negative, left of origin
+
D
vf = -5 m/s
Signs for Velocity
x=0
A
vo = +20 m/s
C
E
Force
a = - 5 m/s2
What are the signs (+ or -) of velocity at
points B, C, and D?
 At B, v is zero - no sign needed.
 At C, v is positive on way out and
negative on the way back.
 At D, v is negative, moving to left.
B
Signs for Acceleration
+
D
vf = -5 m/s
C
A
vo = +20 m/s
E
Force
B
a = - 5 m/s2
What are the signs (+ or -) of acceleration at
points B, C, and D?
 At B, C, and D, a = -5 m/s, negative
at all points.
 The force is constant and always directed
to left, so acceleration does not change.
Constant Acceleration
Acceleration:
a avg
Dv v f  v0


Dt t f  t0
Setting to = 0 and solving for v, we have:
v f  v0  at
Final velocity = initial velocity + change in velocity
Formulas based on definitions:
v f  v0  at
Derived formulas:
x  x0  v0t  12 at 2
2a( x  x0 )  v  v
2
f
2
0
For constant acceleration only
Use of Initial Position x0 in Problems.
0
x  x0  v0t  at
1
2
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
v f  v0  at
0
2a( x  x0 )  v  v
2
f
2
0
The xo term is very
useful for studying
problems involving
motion of two bodies.
Review of Symbols and Units
• Displacement (x, xo); meters (m)
• Velocity (vf, vo); meters per second (m/s)
• Acceleration (a); meters per s2 (m/s2)
• Time (t); seconds (s)
Review sign convention for each symbol
The Signs of Displacement
• Displacement is positive (+) or
negative (-) based on LOCATION.
2m
-1 m
-2 m
The displacement is
the y-coordinate.
Whether motion is
up or down, + or - is
based on LOCATION.
The Signs of Velocity
• Velocity is positive (+) or negative (-)
based on direction of motion.
+
-
+
-
+
First choose + direction;
then velocity v is positive
if motion is with that +
direction, and negative if
it is against that positive
direction.
Acceleration Produced by Force
• Acceleration is (+) or (-) based on
direction of force (NOT based on v).
F
F
a(-)
a(+)
A push or pull (force) is
necessary to change
velocity, thus the sign of
a is same as sign of F.
More will be said later
on the relationship
between F and a.
Problem Solving Strategy:
 Draw and label sketch of problem.
 Indicate + direction and force direction.
 List givens and state what is to be found.
Given: ____, _____, _____ (x,v,vo,a,t)
Find: ____, _____
 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
m/s lands on a carrier deck and stops in a
distance of 300 m. What is the acceleration?
+400 m/s
v=0
300 m
+
F
vo
X0 = 0
Step 1. Draw and label sketch.
Step 2. Indicate + direction and F direction.
Example: (Cont.)
v=0
+400 m/s
300 m
+
Step 3. List given; find
information with signs.
List t = ?, even though
time was not asked for.
F
vo
X0 = 0
Given: vo = +400 m/s
v=0
x = +300 m
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 m/s)2
2(300 m)
0
0
2a(x -xo) = v2 - vo2
Initial position and
final velocity are zero.
2
a = - 267 m/s
Why isForce
the acceleration
negative?
Because
is in a negative
direction!
CONCLUSION OF
Chapter 6 - Acceleration