Today’s Topics
• Review s-v-a graphs
• g’s
• Constant acceleration
Bouncing Tennis Ball
2.
6.
Current position:
7.
Change in position:
Change in velocity:
Velocity Sign:
3.
Position sign:
8.
Current acceleration:
4.
Current velocity:
9.
Acceleration sign:
5.
Average velocity:
Position (ft)
4
3
2
1
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1
1.2
1.4
1
1.2
Time (sec)
15
10
Velocity(ft/sec)
1.
5
5
0
-5
0
0.2
0.4
0.6
0.8
-10
-15
-20
Time (sec)
20
10
2
Quick Review – Graph Values, Slopes, Areas
6
Acceleration(ft/s )
Setup
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
1.4
-10
-20
-30
-40
Time (sec)
EF 151 Fall, 2015 Lecture 1-6
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Special Unit of Acceleration
EF 151 Fall, 2015 Lecture 1-6
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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
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EF 151 Fall, 2015 Lecture 1-6
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Examples:
Deriving Other Constant Acceleration Equations
During a panic stop, a car goes from 60 mph to a stop in 4.0 s.
At what rate was the car accelerating? How far did it travel?
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
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. What is the acceleration over the second time interval? How
far has it gone?
Eliminate ∆t
v1 + v2
∆t
2
(v + v ) (v2 − v1 )
∆s = 1 2
2
a
2
2
v −v
∆s = 2 1
2a
∆s =
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EF 151 Fall, 2015 Lecture 1-6
∆ =
=(
∆
+ ∆ )
(area under velocity curve)
(equation of velocity curve)
=(
+ ∆ ) (equation of velocity curve)
Solve for ∆ : ∆ =
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Constant Acceleration Equations - Summary
Constant Acceleration Examples
s2 − s1
∆t
v2 − v1
aavg =
∆t
v +v 
s2 = s1 +  1 2 ∆t
 2 
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?
vavg =
a-t curve: ______________
v-t curve: _______________
s-t curve: _______________
What is the velocity of the car after it has traveled the 300 ft?
s2 = s1 + v1∆t + 12 a∆t 2
How long does it take to go 300 ft if the car starts at 30 mph?
v2 − v2
s2 = s1 + 2 1
2a
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?
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.
EF 151 Fall, 2015 Lecture 1-6
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EF 151 Fall, 2015 Lecture 1-6
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