Download Acceleration

Chapter 2: 1-D Kinematics
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
Editing by Mr. Gehman
©
2007
The Cheetah: A cat that is built for speed. Its strength and
agility allow it to sustain a top speed of over 100 km/h. Such
speeds can only be maintained for about ten seconds.
Photo © Vol. 44 Photo Disk/Getty
Objectives: After completing this
module, you should be able to:
• Define and apply concepts of average and
instantaneous velocity and acceleration.
• Solve problems involving initial and final
velocity, acceleration, displacement, and time.
• Demonstrate your understanding of directions
and signs for velocity, displacement, and
acceleration.
• Solve problems involving a free-falling body in
a gravitational field.
Uniform Acceleration
in One Dimension:
• Motion is along a straight line (horizontal,
vertical or slanted).
• Changes in motion result from a CONSTANT
force producing uniform acceleration.
• The cause of motion will be discussed later.
Here we only treat the changes.
• The moving object is treated as though it
were a point particle.
Reference Frames and Displacement
Any measurement of position, distance, or
speed must be made with respect to a
reference frame.
For example, if you are sitting on a train and someone
walks down the aisle, their speed with respect to the
train is a few miles per hour, at most. Their speed with
respect to the ground is much higher.
Reference Frames and Displacement
We make a distinction between distance and
displacement.
Displacement (blue line) is how far the object is
from its starting point, regardless of how it got
there.
Distance traveled (dashed line) is measured along
the actual path.
Distance and Displacement
Distance is the length of the actual path
taken by an object. Consider travel from
point A to point B in diagram below:
Distance is a scalar
quantity (no direction):
Contains magnitude only
and consists of a
number and a unit.
(70 m, 40 mi/h, 10 gal)
Distance and Displacement
Displacement is the straight-line
separation of two points in a specified
direction.
A vector quantity:
Contains magnitude
AND direction, a
number, unit & angle.
(40 m east; 8 km/h, N)
Reference Frames and Displacement
d
d2
d1
The displacement is written:
Left:
Right:
Displacement is positive.
Displacement is negative.
Distance and Displacement
• For motion along x or y axis, the displacement is
determined by the x or y coordinate of its final
position. Example: Consider a car that travels 8 m, E
then 12 m, W.
Net displacement D is
from the origin to the
final position:
D = 4 m, W
What is the distance
traveled? 20 m !!
D
8 m,E
x = -4
x
x = +8
12 m,W
The Signs of Displacement
• Displacement is positive (+) or
negative (-) based on LOCATION.
Examples:
The displacement is
the y-coordinate.
Whether motion is
up or down, + or - is
based on LOCATION.
2m
-1 m
-2 m
The direction of motion does not matter!
Definition of Speed
• Speed is the distance traveled per unit of
time (a scalar quantity).
A
s = 20 m
B
v=
s
t
=
20 m
4s
v = 5 m/s
Time t = 4 s
Not direction dependent!
Examples of Speed
Orbit
2 x 104 m/s
Light = 3 x 108 m/s
Jets = 300 m/s
Car = 25 m/s
Speed Examples (Cont.)
Runner = 10 m/s
Glacier = 1 x 10-5 m/s
Snail = 0.001 m/s
Definition of Velocity
• Velocity is the displacement per unit
of time. (A vector quantity.)
A
d = 12 m
B
D 12 m
v 
t
4s
v = 3 m/s, East
Time t = 4 s
Direction required!
Average Speed & Velocity
Speed: how far an object travels in a given time interval
s d
v 
t t
Velocity includes directional information:
D d
v 
t
t
Avg. Speed vs. Avg. Velocity
s 100m
v 
t
4s
v = 25 m/s
Not direction dependent!
s 40m
v 
t
4s
v = 10 m/s, East
Direction required!
Speed Example Exercise
FloJo, ’88 Olympics
d 100.00m
m
m
v 
 9.48766  9.488
t
10.54s
s
s
Converting to km/hr:
m  1km  3600s 
km
v  9.488 

  34.16
s  1000m  1hr 
hr
Example 1. A runner runs 200 m, east, then
changes direction and runs 300 m, west. If
the entire trip takes 60 s, what is the average
speed and what is the average velocity?
Recall that average
s2 = 300 m
speed is a function
only of total distance
start
and total time:
s1 = 200 m
Total distance: s = 200 m + 300 m = 500 m
total path 500 m
Average speed 

time
60 s
Direction does not matter!
Avg. speed
8.33 m/s
Example 1 (Cont.) Now we find the average
velocity, which is the net displacement divided
by time. In this case, the direction matters.
v
x f  x0
t = 60 s
xf = -100 m
x1= +200 m
t
x0 = 0 m; xf = -100 m
100 m  0
v
 1.67 m/s
60 s
xo = 0
Direction of final
displacement is to
the left as shown.
Average velocity: v  1.67 m/s, West
Note: Average velocity is directed to the west.
Example 2. A sky diver jumps and falls for
600 m in 14 s. After chute opens, he falls
another 400 m in 150 s. What is average
speed for entire fall?
Total distance/ total time:
xA  xB 600 m + 400 m
v

t A  tB
14 s + 150 s
1000 m
v
164 s
v  6.10 m/s
Average speed is a function
only of total distance traveled
and the total time required.
14 s
A
600 m
B
400 m
150 s
Average Speed and
Instantaneous Velocity
 The average speed depends ONLY on the
distance traveled and the time required.
A
s = 20 m
C
Time t = 4 s
B
The instantaneous
velocity is how fast
and in what direction
an object is moving
in a particular
instant. (v at point C)
The Signs of Velocity
 Velocity is positive (+) or negative (-)
based on direction of motion.
+
-
+
-
+
First choose + direction;
then v is positive if motion
is with that direction, and
negative if it is against that
direction.
Velocity in Position-Time Graphs
This is a graph of p vs. t for an
object moving with const.
velocity. If we take the slope of
this line (“rise over run”) we get
d
slope 
 velocity
t
Notice that the slope of the p-t
curve can be (+) or (-).
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.
Example of Acceleration
+
Force
t=3s
v0 = +2 m/s
vf = +8 m/s
The wind changes the speed of a boat
from 2 m/s to 8 m/s in 3 s. Each
second the speed changes by 2 m/s.
Wind force is constant, thus acceleration is constant.
Acceleration
 You are driving in a car. What can you do
to cause the car to accelerate?
• Speed it up (step on gas)
• Slow it down (step on brake)
• Change the direction (turn the wheel)
The Signs of Acceleration
• Acceleration is positive (+) or negative
(-) based on the direction of force.
+
F
F
a (+)
a(-)
Choose + direction first.
Then acceleration a will
have the same sign as
that of the force F —
regardless of the
direction of velocity.
Negative. Acceleration vs. Deceleration
There is a difference between negative acceleration
and deceleration:
Negative acceleration is acceleration in the
negative direction as defined by the coordinate
system.
Deceleration occurs when the acceleration is
opposite in direction to the velocity.
Acceleration
• To calculate acceleration, we use the
following formula:
aavg
v v f  v0


t t f  t0
• This can also be written as
v f  v0  at
Example Exercises from Notes
• Accelerating Plane
v
km  1  1000m  1hr 
m
a
 235



  4.35 2
t
hr  15.0s  1km  3600s 
s
• An acceleration of +4.35 m/s2 can also be
written as +4.35 m/s/s or +4.35 m s-2. This
means that every second, the velocity changes
by +4.35 m/s
Acceleration
 Which one of these accelerations is the
largest?
• 60 mi/hr-s
• 60 mi/hr-min
• 60 mi/hr-hr
Ex. Exercises from Notes, cont.
• Car Slowing Down
v 
m
m 1
m
a
  33.2  85.5 
 41.8 2
t 
s
s  1.25s
s
• Notice that in computing Δv, you always subtract
final from initial: v-v0
• Question: An object moving at -25 m/s with an
acceleration of +5 m/s/s. What is its velocity
after 7 s?
Example 3 (No change in direction): A constant
force changes the speed of a car from 8 m/s to
20 m/s in 4s. What is the 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
v v2  v1


t 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, v = vf - v0
v = (-5 m/s) - (+20 m/s) = -25 m/s
Example 4: (Continued)
+
vf = -5 m/s
aavg =
v
t
E
vo = +20 m/s
Force
v = (-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 = -25 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 moving from?
A to B
B to C
C to D
+
D
vf = -25 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 = -25 m/s
C
A
vo = +20 m/s
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.
Distance and Acceleration
• A car starting from rest accelerates at 2 m/s2.
Which of the following is true?
A. The car covers 2 m of distance every second.
B. The car covers 2 m more distance than in the
previous second (2m, then 4m, then 6m, etc)
C. The car covers more and more distance each
second (2m, then 4m, then 8m, etc.)
Equations for Constant Acceleration
To find the distance an object covers while
accelerating, use this equation:
1 2
d  d0  vot  at
2
(2)
Notice that, if a =0 and you start from the origin,
this equation becomes more recognizable:
0
1 20
d  d 0  vot  at  vot
2
d
v
t
Eqns for Motion at Constant Acceleration
We can combine equations (1) & (2) so as to
eliminate t:
v  v  2a ( d  d 0 )
2
2
0
(3)
We now have all the equations we need to solve
constant-acceleration problems.
v  vo  at
1 2
d  d 0  v0t  at
2
2
2
v  v0  2ad
(1)
(2)
(3)
Example 5: A ball 5 m from the bottom of an
incline is traveling initially at 8 m/s. Four seconds
later, it is traveling down the incline at 2 m/s. How
far is it from the bottom at that instant?
+
d
vo
5m
8 m/s
d = do +
vo + vf
2
F
vf
-2 m/s
Careful
t=4s
t =5m+
8 m/s + (-2 m/s)
2
(4 s)
(Continued)
+
F
d
vo
vf
-2 m/s
5m
t=4s
8 m/s
d=5m+
d=5m+
8 m/s + (-2 m/s)
2
8 m/s - 2 m/s
2
(4 s)
(4 s)
x = 17 m
Use of Initial Position d0 in Problems
0
d  d0 
v0  v f
2
0
t
d  d 0  v0t  at
1
2
0
2
2a(d  d0 )  v  v
v f  v0  at
2
f
2
0
If you choose the
origin of your x,y
axes at the point of
the initial position,
you can set d0 = 0,
simplifying these
equations.
The do term is very
useful for studying
problems involving
motion of two bodies.
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)
Review sign convention for each symbol
Problem Solving Strategy:
 Draw and label sketch of problem.
 Set up coordinate system (which way is + and
where is the origin)
 List givens and state what is to be found.
Given: ____, _____, _____ (d,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
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
+
F
vo
X0 = 0
Step 1. Draw and label sketch.
Step 2. Indicate + direction and F direction.
Example: (Cont.)
v=0
+400 ft/s
300 ft
+
Step 3. List given; find
information with signs.
List t = ?, even though
time was not asked for.
F
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
+
vo
F
X0 = 0
Step 4. Select equation
that contains a and not t.
a=
-vo2
2d
=
-(400 ft/s)2
2(300 ft)
0
v2 =vo2
0
+ 2a(d
-do)
Initial position and
final velocity are zero.
a = - 267 ft/s2
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 = 9.80 m/s2 or 32 ft/s2
Directed downward (usually negative).
Experimental Determination
of Gravitational Acceleration.
The apparatus consists of a
device which measures the time
required for a ball to fall a given
distance.
Suppose the height is 1.20 m
and the drop time is recorded
as 0.650 s. What is the
acceleration due to gravity?
t
y
Experimental Determination of
Gravity (y0 = 0; y = -1.20 m)
t
y = -1.20 m; t = 0.495 s
y  v0t  at ; v0  0
1
2
2
2 y 2(1.20 m)
a 2 
2
t
(0.495 s)
Acceleration
2
of Gravity: a  9.79 m/s
Acceleration a is negative
because force W is negative.
y
+
W
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
(Earth pulling on ball).
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 Velocity:
Step 4. 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
Finding Displacement:
Step 5. 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
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
= 3.06 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 = 3.06 s
Omitting units, we obtain:
+
a=g
y  (30)(3.06)  (9.8)(3.06)
1
2
y = 91.8 m - 45.9 m
vo =+30 m/s
ymax = 45.9 m
2
Summary of Formulas
x  x0 
v0  v f
2
t
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
Summary: Procedure
 Draw and label sketch of problem.
 Indicate + direction and force direction.
 List givens and state what is to be found.
Given: ____, _____, ______
Find: ____, _____
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.
CONCLUSION OF
Chapter 6 - Acceleration