Quick Review – Graph Values, Slopes, Areas
Graph Review, Gravity, Constant Acceleration
6
Position (ft)
5
1.
Current position:
2.
Change in position:
3.
Position sign:
4.
Current velocity:
5.
Average velocity:
6.
Change in velocity:
7.
Velocity Sign:
8.
Current acceleration:
9.
Acceleration sign:
4
3
2
1
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
15
Velocity(ft/sec)
10
5
0
-5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
-10
-15
-20
Time (sec)
20
10
2
Acceleration(ft/s )
Coordinate System
a) Zero: ______ Positive: ____
Initial
a) S,V,A ________________
b) S slope at 0 ______
c) V slope at 0 ______
T=0 to 0.6 sec
a) V-T Area ______
b) V-T slope ______
c) Speeding up or down?
T=0.60 to 0.61 (bounce)
a) V before _______
b) V after ______
c) A _______
T=0.61 to 0.96 sec
a) V-T area ______
b) V-T slope ______
c) Speeding up or down?
T=0.96 sec (peak)
a) S,V,A ______________
b) S-T slope ______
c) V-T slope ______
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
-10
-20
-30
-40
Time (sec)
EF 151 Fall, 2016 Lecture 1-6
ef151q@ef.utk.edu
1
Special Unit of Acceleration
EF 151 Fall, 2016 Lecture 1-6
ef151q@ef.utk.edu
2
Area of a Trapezoid
“g” – acceleration caused by gravity (on Earth’s
surface)
 a convex quadrilateral with at least
one pair of parallel sides
 area = average width * height
 area = ½(a+b)h
1 “g” = 9.81 m/s2 = 32.2 ft/s2
The Top Thrill Dragster ride accelerates from 0 to
120 mph in 4.6 seconds.
What is its acceleration in ft/s2, “g”s, and m/s2 ?
What is the area of this trapezoid?
v
v2
v1
vavg
∆s = area
∆t
EF 151 Fall, 2016 Lecture 1-6
ef151q@ef.utk.edu
3
EF 151 Fall, 2016 Lecture 1-6
ef151q@ef.utk.edu
t
4
Deriving Constant Acceleration Equations
Eliminate v2
s2  s1  s
∆
∆
v v 
s2  s1   1 2 t
 2 
v  (v1  at )
s2  s1  1
t
2
s2  s1  v1t  12 at 2
EF 151 Fall, 2016 Lecture 1-6
Solve for ∆
s2  s1
t
v2  v1
aavg 
t
v v 
s2  s1   1 2 t
 2 
s2  s1  v1t  12 at 2
(area under velocity curve)
∆
Eliminate ∆t
v v
s  1 2 t
2
(v1  v2 ) (v2  v1 )
s 
2
a
v22  v12
s 
2a
Constant Acceleration Equations - Summary
∆
: ∆
vavg 
(equation of velocity curve)
s2  s1 
(equation of velocity curve)
a-t curve: ______________
v-t curve: _______________
s-t curve: _______________
v22  v12
2a
You don’t have to memorize these equations, but if you do you will be
able to work problems faster. Working quickly enough is one of the major
issues students have on exams.
ef151q@ef.utk.edu
5
Constant Acceleration Examples
EF 151 Fall, 2016 Lecture 1-6
ef151q@ef.utk.edu
6
One Last Example:
A car is moving at 4.0 m/s. It accelerates for 8 seconds at a constant rate of
2.0 m/s2 and then, over the next 30 seconds, its velocity drops at a constant
rate to 11 m/s.
How far has it gone?
A car accelerates at 6 mph/sec (8.8 ft/s2). If it starts from rest, how
long does it take to go 300 ft?
What is the velocity of the car after it has traveled the 300 ft?
How long does it take to go 300 ft if the car starts at 30 mph?
How long of an acceleration lane is needed for a car to increase its
speed from 30 mph to 60 mph if the acceleration is 8.8 ft/s2?
EF 151 Fall, 2016 Lecture 1-6
ef151q@ef.utk.edu
7
EF 151 Fall, 2016 Lecture 1-6
ef151q@ef.utk.edu
8