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Updated 11/11/04
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Ch 4a Test Practice
Ch 4b Acceleration
OBJ: Calculate the average acceleration.
1
a]
Acceleration:
the rate of change in velocity.
It is more than a change in
speed!
Word Eq:
Acceleration = (Vfinal - Vinitial)/elapsed time
Symbol Eq:
a = (Vf - Vi)/t
Unit Eq:
m/s2
= m/s/s
b] Translate
3 ft/s and 4ft/s2
into
two sentences.
Time (s) | 0 | 1 | 2 | 3 | . . .
3 ft/s says an
object moves 3 ft every second
Dist (ft) | 0 | 3 | 6 | 9 | . . .
4 ft/s2
says an object increases its speed by 4 ft/s every second.
Vel (ft/s)| 0 | 4 | 8 | 12 | . . .
2
| DISTANCE | SPEED | ACCELERATION
METRIC | meter |
m/s |
m/s/s
ENGLISH |
foot | ft/s
| ft/s/s
b]
ASSUMPTION IN S = D/t, Constant
Speed, NOT constant direction or constant accel
3
DRAW & LABEL THE 3 ACCELEROMETER DIAGRAMS.
4
FIND THE ACCELERATION OF THE FOLLOWING OBJECTS.
a] THE SPACE
SHUTTLE TAKES FIVE MINUTES TO REACH 18,000 mi/h.
b] A '91
LAMBORGHINI AT A STOP LIGHT ACCELERATES TO 60 mi/hr IN 4.40 s
c] A CAR
MOVING AT 60.0 mph SLOWS TO 25.0 mi/hr IN FIVE SECONDS.
5
a] State and give examples of two types of uniform acceleration.
Linear Accel:
Changing speed & constant direction;
A car moving 0 to 60 on a straight away.
Circular Accel:
Constant speed & changing direction
A car making a U-turn
b] Non-uniform acceleration:
A rocket taking off.
c] The slope of a velocity-time graph = ΔV/t
= acceleration
d] Graphs for acceleration that is:
increasing curves up
decreasing curves down
constant straight line slopes up
zero:
flat horizontal line
6 a]
You determine the acceleration graphically by finding the slope of a
velocity-time graph.
b] Find the average acceleration for each section
in the graph.
c] You find the displacement graphically by
finding the area under the curve of a velocity-time graph.
d] Find the displacement for each section in the
graph.
e] Describe the motion (position, d, v,
a w/ t) in each section.
50|----|----+----+----|----|----|----|----|----|----|----|----|----|
V
| | |
| | |
| |
| |
| | |
|
E
40|----|----|----|----|----|----|----|----|----|----|----|----|----+
L
| A | B
| C | | |
| |
| | |
|
O
30|----|----|----|----|----+----|----|----|----|----|----|----|----|
C
| | |
| |
| | |
| |
| |
F
| |
I
20|----|----|----|----|----|----|----|----|----|----|----|----|----|
T
| | |
| |
| |
| |
| | |
| |
Y
10|----|----|----|----|----|----|----|----|----|----|----|----|----|
m/s | |
| |
| | |
| | E
| | |
| |
+----|----|----|----|----|----|----|----+----+----|----|----|----|
T-->0 2
4 6
8 10 12 14
16 18
20 22 24
26s
A:
a = (Vf - Vi)/t = (50 - 00)/( 4 -
0) = +12.5 m/s/s
d = 1/2 base x height =
(4 s x 50m/s)/2 = 100 m
B:
(50 - 50)/( 8 - 6) = +0.00 m/s/s
d = base x height = 2 s x
50 m/s = 100m
C:
(30 - 50)/(10 - 6) = -5.00 m/s/s
d = bh + bh/2 = 4 s x 30
m/s + (4 s x (50-20)m/s)/2 = 120 + 60 = 180 m
D: a decreases, Avg a =
(00 - 30)/(16 - 10) = -5.00 m/s/s
d = Area of Square - Area
of 1/4 Circle
d = (30 - 0)x(16-10) -
pR2/4
d = 180 m -
p32/4
Squares
d = 180 m - 7.065
Squares: 1 Square = 2 s x 10 m/s = 20 m
d = 180 m - 7.065 Squares
x 20 m/Square
d = 180 m - 141.3 m
d = 38.7 m
E:
a decreases, Avg a =
(40 -
00)/(26 - 18) = +5.00 m/s/s
d = Area of 1/4 Circle
d =
pR2/4
Squares
d =
p42/4
Squares
d = 12.56 Squares x 20 m/Square
d = 251.2 m
Note: down hill = negative slope = slowing down
up hill = positive slope = speeding up
7
a] Positive acceleration is called acceleration and means the
object is speeding up
b] Negative
acceleration is deceleration and means the object is slowing down
c] Negative
accel.: Word Eq.
a = (Final Velocity - Initial Velocity)/Elapsed Time
Symbol Eq. a = (Vf - Vi)/t
Note: The definition of acceleration is independent of the type of
change.
Positive & Negative Acceleration have the same equations!
d] An object
with zero acceleration means the object is moving in a straight line
at constant speed.
8
Construct a velocity-time graph (include:
V, a, t, d) for:
a] A plane
that flies at constant velocity of 300 mi/h for 4.00 h then slows at a
uniform rate
and stops in 1/2 h.
Section #1:
Constant velocity; Horizontal line
Vf = Vi =
300 mi/h, a = 0, t = 4 h, d = 1200 mi
Section #2:
Slows: Straight Line sloping down from 300 to 0 mi/h:
Vi = 300
mi/h, Vf = 0;
(300 - 0)mi/h
a =
------------- = 10 mi/h/min;
30 min
t = 0.5 h
D = 1/2 (0.5
h)(300 mi/h) = 75 mi
b] A car
starts from 6.0 ft/s, accelerates at 2.0 ft/s2 for 12.0 sec
maintains
its speed for 6.0 s,
then slows to a 12.0 ft/s at 6.0
ft/s2.
Sec #1: Increasing Speed:
Vi = 6 ft/s, Vf
= Vi + at = 6 ft/s + (2.0 ft/s2)(12.0 s) = 30 ft/s
a = 2.0 ft/s2
t = 12 sec
D = Rectangle
+ Triangle
D
= (6.0 ft/h)(12 s)
+ 1/2(12s)(30 ft/s - 6 ft/s)
D
= 72 ft
+ 144 ft
D
= 216 ft
Section
#2: Constant Speed:
Vi = Vf = 30
ft/s
a = 0
t = 6 s
D =
Rectangle = 30 ft/s x 6 s =
180 ft
Section #3:
Decreasing speed
Vi = 30
ft/s, Vf = 12 ft/s
a = -6 ft/s2
t = (Vf - Vi)/a
t = [(12
- 30)ft/s]/[-6 ft/s2]
t =
[-18 ft/s]/[-6 ft/s2]
t = +3 s
D = Rectangle
+ Triangle
D =
3 s x 12 ft/s
+
1/2(3 s) (30 - 12)ft/s
D = 36
ft + 27 ft
D = 63 ft
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1 Define, write the
word, symbol, & unit equations for acceleration. 4
2 STATE THE METRIC &
ENGLISH UNITS FOR
a] DISPLACEMENT, VELOCITY, AND ACCELERATION.
3
b] STATE THE ASSUMPTION IN S =
D/t 1
3 a] DRAW & LABEL THE 3
ACCELEROMETER DIAGRAMS. 3
b] State the
implications. 2
4 A car moving at 65.0
mi/h slows to 20.0 mi/hr in five seconds.
a] Graph the motion. b] Find the acceleration.
4
5 a] You find the
displacement graphically by _?_. 1
b] Explain the motion of a
rocket. 2
c] You find the acceleration graphically by
_?_ 1
d] State and give examples of two types of uniform
acceleration. 3
e] You find the velocity graphically by
_?_. 1
f] Draw & label 7 types of velocity-time
graphs. 4
6 For each section in the graph: Find the
average acceleration & displacement,
then describe the motion (position, d, v, a w/ t). 12
25+----|----|----|----|----|----|----|----|----+----|----|----|----|
V | | | | | | | | | | | | | |
E 20|----|----|----|----|----|----|----|----|----|----|----|----|----|
L | | | A | | | | | C | | | | | |
O 15|----|----|----|----|----|----|----|----|----|----|----|----|----|
C | | | | | | | | | | | D | | |
I 10|----|----|----|----+----|----+----|----|----|----|----|----|----|
T | | | | | B | | | | | | | |
Y 5|----|----|----|----|----|----|----|----|----|----|----|----|----|
m/s | | | | | | | | | | | | | |
|----|----|----|----|----|----|----|----|----|----|----|----|----+
T-->0 3 6 9 12 15 18 21 24 27 30 33 36 39s
7 An object with zero
acceleration means the object is _?_ 1
8 a] A plane that flies
at constant velocity of 450 mi/h for 3.00 h then slows at a uniform rate and
stops in 1/4 h.
b] A car starts at 10 ft/s, accelerates at 12 ft/s2 for 15 s
maintains its speed for 10 s, slows to a rest at 5 ft/s2.
i] Construct a velocity-time graph (include: V, a, d, t) for:
(11 lines: Δt,
clock time)
ii] Calc the missing V & Δt
iii] Calc the distance for each section (ft)
9 P 20|----|----|----|----|----|----|----+
a] Find the average velocity
O | | | | | | | | between 2 and 11 s?
S 15|----|----|----|----|----|----|----| b] Find the instantaneous
I | | | | | | | | velocity at 11 s.
T 10|----|----+----|----+----|----|----|
c] Find the avg acceleration
I | | | | | | | | between 2 and 11 s.
O 5|----|----|----|----|----|----|----| d] Find the instantaneous
N | | | | | | | | acceleration at 11 s.
(m) +----|----|----|----|----|----|----|
0 2 4 6 8 10 12 14 t(s)
Ans: 4] -9 mi/h/s
6A] Constant accel at -1.25m/s2,
d = 210 m, V decreases 25 to 10 m/s
B Zero Accel, d = 60 m, Constant velocity at 10 m/s
C Decreasing accel, avg +1.67 m/s2,
d= 196 m, V increases 10 - 25 m/s
D Constant accel at -2.08 m/s2,
d = 150 m, V decreases 25 to 0 m/s
8a] a = -1,800 mi/h/h; d1 = 1,350 mi, d2 = 56.25 mi t2 = 3.25 h
b] V2= 190 ft/s,
Δt
= 36 s, d1 = 1500 ft, d2 = 1900 ft, d3 = 3420 ft
9a] V avg = (15 - 5)m/(11 - 2)s = 1.11 m/s
b] Inst V = V = (20-10)m/(14-8) = 1.67 m/s
c] Avg accel. = (1.67 - 2.5)/(11-2) = 0.0944 m/s2
d] Velocity = constant ---> acceleration = 0
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Ch 4 b Acceleration
Updated 12/18/03
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Equations Motion with Uniform
Acceleration
OBJ: Relate final velocity to
initial velocity, time, displacement, and acceleration.
1
A] Given a = (Vf - Vi)/t find [a] Vf = ?
[b] Vi = ? [c] t = ?
a]
a = (Vf - Vi)/t
Multiply both sides by t
at = Vf - Vi
Add Vi to both sides
Vi + at = Vf
Ans: Vf = Vi + at
b] Given Vf = Vi + at find Vi
Vf = Vi + at
Subtract at from both sides
Vf - at = Vi
Ans: Vi = Vf - at
c] a = (Vf - Vi)/t
Multiply both sides by t
at = Vf - Vi
Divide both sides by a
t = (Vf - Vi)/a
B] Translate the final velocity equation into a
sentence.
Vf = Vi + at
Final speed = start
speed + accel x time
2*
a]* Use a graphical analysis to derive the equation for displacement.
Eq #1:
d = (Vf + Vi)t/2 = Avg V x t
Note:
Similar to D = Speed x Time
b] Assumption:
acceleration = constant
c] Sub Vf = Vi + at into d = (Vf + Vi)t/2 -->
d = Vi(t) + at2/2
Eq #2
Sub Vi = Vf - at into d = (Vf + Vi)t/2 -----> d = Vf(t) - at2/2
Eq #3
Sub t = (Vf - Vi)/a into d = (Vf + Vi)t/2 -->
d = (Vf2
- Vi2)/2a
Eq #4
3
WHAT IS THE DISPLACEMENT OF A TRAIN AS IT IS ACCELERATED UNIFORMLY FROM
13 m/s TO
25 m/s, NORTH IN A 8 s INTERVAL?
a = ?-->Can’t use Eq 2,3, or 4
d = ?
Vi = 13 m/s
d = (Vf + Vi)t/2 = (13 + 25)m/s(8 s)/2 = 152 m
Vf = 25 m/s
t = 8
s Ans:
d = 152 m
4
AIRPLANES CAN ACCELERATE AT 2 m/s2,
AND MUST GAIN A GROUND SPEED OF
70 m/s TO TAKE OFF.
a]
HOW LONG WILL IT TAKE THEM TO ACHIEVE TAKE OFF SPEED?
t = ?
t = (Vf - Vi)/a
a = 2 m/s2
t = (70 - 0)m/s/2 m/s/s = 35 s
Vf = 70 m/s
Ans: t = 35 s
Vi = 0 m/s (all planes start from rest)
b]
WHAT MUST BE THE MINIMUM LENGTH OF THE RUNWAY?
t = ? (Don’t trust ans in a.)-->can’t
use Eq 1,2, or 3
d = ?
d = (Vf2
- Vi2)/2a
a = 2 m/s2
d = (702 - 02)m2/s2/2(2
m/s2)
Vf = 70 m/s
d = 4900 m/4 = 1225 m
Ans:
d = 1225 m
5
ON PLANET X A STONE DROPPED FROM A HEIGHT OF 245 m HITS THE GROUND IN 5.0
s.
WHAT IS THE ACCELERATION OF GRAVITY?
d = 245 m
d = Vit + at2/2
Eq #2
t = 5 s
245 m = 0 + a(5s)2/2
a = g = ?
245 m = a 12.5 s2
Vi = 0
245 m/12.5 s2
= a
Ans:
a = 19.6 m/s2
6
A CAR STARTING FROM 20 m/s ACCELERATES AT 7.0 m/s2.
WHAT IS THE CAR'S DISPLACEMENT DURING THE FIRST
5.0 s?
Vi = 20 m/s
d = Vit + at2/2
Eq #2
a = 7.0 m/s2
d = 20 m/s(5.0 s) + 7.0 m/s2(5.0
s)2/2
d = ?
d = 100 m + 87.5 m =
187.5 m
Ans:
d = 187.5 m
t = 5.0 s
7
AN AIRPLANE MUST ACHIEVE A VELOCITY OF 50.0 m/s FOR TAKEOFF.
IF THE RUNWAY IS 1,100 m LONG, WHAT MUST THE
ACCELERATION BE?
Vf = 50.0 m/s
d = (Vf2
- Vi2)/2a
d = 1,100 m
1,100 m = (502
- 0)m2/s2/2a
Vi = 0 m/s
1,100 m = 1,250 m2/s2a
Ans:
a = 1.14 m/s2
a = ?
a = 1,250 m2/s2/1,100
m = 1.14 m/s2
8 A BOX
FALLS OFF A TRUCK AND SLIDES ALONG THE STREET FOR 144 m.
FRICTION
DECELERATES THE BOX AT 16.0 m/s2.
a] DRAW & LABEL THE DIAGRAM;
b] AT WHAT
SPEED WAS THE TRUCK MOVING?
d = 144 m
d
= (Vf2 - Vi2)/2a
a = -16.0
m/s2
144
m = (02
- Vi2)/2(-16.0
m/s2)
Note: -Vi2/-16.0
= +Vi2/16.0
Vf = 0 144
m = Vi2/32
m/s2
.
Vi = ? 144
m x 32 m/s2
= Vi2
--> Vi = √144
x 32 m2/s2
Ans:
Vi =
67.9 m/s
9
FOR A BALL ROLLING DOWN AN INCLINE AT 4.0 m/s2
FOR 5.0 SECONDS:
a] Construct a table (Note a/2 = 4/2 =2):
|Vf=at|D=at2/2|
t |Vf=4t| D=2t2
| ΔD | Δ(ΔD)
|
0 |
0
|
0
|
|
|
|
|
|
2 |
|
1 |
4
|
2
|
|
4
|
|
|
|
6 |
|
2 |
8
|
8
|
|
4
|
|
|
| 10 |
|
3 | 12
|
18
|
|
4
|
|
|
| 14 |
|
4 | 16
|
32
|
|
4
|
|
|
| 18 |
|
5 | 20
|
50
|
|
|
b] Draw and label the diagram.
Draw a 25 cm diagonal line & mark off 0, 1,
4, 9, 16, & 25 cm (= 12,
22,
32,
42,
& 52)
Give it 5 labels:
D 0 2 m 8 m 18
m 32
m 50
m
V 0 4 m/s 8 m/s 12
m/s 16
m/s 20 m/s
t 0 1 s 2 s 3
s
4
s 5
s
ΔD
|-|<-6->|<--10
m->|<----14 m--->|<----18
m------->|
|-|-----|---------|-------------|-----------------|
cm 0 1 cm 4 cm 9
cm 16
cm 25 cm
c] Scale:
1 cm = 2 m (see distance after 1 sec)
Check #1 ΔD:
Graph ?=? Calculation
For t = 3 to 4 sec
ΔD
= (16 cm - 9cm)x 2m/cm ?=? Δd in the table
7 cm x
2m/cm ?=? 14 m
14 m = 14 m (it checks)
Check #2 Total D: Graph ?=? Calculation
For t = 0 to 3 Sec
Total D
= 9 cm x 2m/cm ?=? Total D
in Table
18 m ?=? 18 m (it checks)
d] The assumption?:
acceleration is constant
e] How far does the ball roll during the 5th second.
At t = 4, D = 32 m
At t = 5, D = 50 m
During the 5th second the ball rolls 50 - 32 ft = 18 ft
Changing
Acceleration
Obj: a] Use a velocity-time graph
to find the average acceleration.
b] Determine
acceleration at any point on a velocity-time graph by finding the
slope
of the tangent line.
10
a] The average acceleration for an interval is a cord from time = zero
to the end
time on a velocity-time graph.
[diagram].
b] The instantaneous acceleration at a point in
time is a line tangent to the curve
at that time on a velocity-time
graph [diagram].
11
a] DRAW HALF CIRCLE VELOCITY vs TIME GRAPH| LABLE AXES.
(PROTRACTOR|
curve down| bottom 2 lines up| 1 line= 5 m/s| 1 cm= 2 s).
b] DESCRIBE
THE MOTION: Since the slope is
positive, levels off and then
negative the object speeds up then
slows down.
c] LABEL
TRIANGLE & FIND THE AVERAGE ACCELERATION AT 26 s = the slope o the line
from t = 0 to t =26 s
d] LABEL
TRIANGLE & FIND THE INSTANTANEOUS ACCELERATION AT 10 s.
PRAC: Curve up: Since
the slope is negative, levels off and then positive
the
object slows down then speeds up.
Tailgating: http://www.physics.umd.edu/rgroups/ripe/perg/abp/aha/tail.htm
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