Chapter 6A. Acceleration. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University. © 2007.
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A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 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
s = 20 m
A
Distance and DisplacementDistance is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below:
Distances is a scalar quantity (no direction):
Contains magnitude only and consists of a number and a unit.
(20 m, 40 mi/h, 10 gal)
D = 12 m, 20o
A
q
Distance and DisplacementDisplacement is the straightline separation of two points in a specified direction.
A vector quantity:
Contains magnitude AND direction, a number,unit & angle.
(12 m, 300; 8 km/h, N)
8 m,E
x = 4
x = +8
12 m,W
Distance and DisplacementNet displacement D is from the origin to the final position:
x
D = 4 m, W
What is the distance traveled?
20 m !!
The displacement is the ycoordinate. Whether motion is up or down, + or  is based on LOCATION.
The Signs of Displacement2 m
1 m
2 m
The direction of motion does not matter!
s
t
20 m
4 s
v = =
A
Definition of Speeds = 20 m
v = 5 m/s
Not direction dependent!
Time t = 4 s
B
D=12 m
A
20o
Time t = 4 s
Definition of Velocityv = 3 m/s at 200 N of E
Direction required!
s1 = 200 m
Avg. speed 8.33 m/s
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 speed is a function only of total distance and total time:
start
Total distance: s = 200 m + 300 m = 500 m
Direction does not matter!
xf= 100 m
x1= +200 m
xo = 0
Average velocity:
Example 1 (Cont.)Now we find the average velocity, which is the net displacement divided by time. In this case, the direction matters.
x0 = 0 m; xf = 100 m
Direction of final displacement is to the left as shown.
Note: Average velocity is directed to the west.
A
625 m
B
356 m
142 s
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:
Average speed is a function only of total distance traveled and the total time required.
B
C
A
Time t = 4 s
Average Speed and Instantaneous VelocityThe instantaneousvelocity is the magnitude and direction of the speed at a particular instant. (v at point C)
+
+

+
The Signs of VelocityFirst choose + direction; then vis positive if motion is with that direction, and negative if it is against that direction.
x2
Dx
Dx
Displacement, x
x1
Dt
Dt
t1
t2
Time
Average and Instantaneous vAverage Velocity:
Instantaneous Velocity:
The direction of accel eration is same as direction of force.
A formal treatment of force and acceleration will be given later. For now, you should know that:
a
2F
2a
Acceleration and ForcePulling the wagon with twice the force produces twice the acceleration and acceleration is in direction of force.
t = 3 s
+
v0= +2 m/s
vf= +8 m/s
Example of AccelerationThe 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.
F
F
The Signs of AccelerationChoose + direction first. Then accelerationawill have the same sign as that of the force F —regardless of the direction of velocity.
a ()
a(+)
+
v1= +8 m/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 4 s. What is average acceleration?
t = 4 s
v2= +20 m/s
Step 1. Draw a rough sketch.
Step 2. Choose a positive direction (right).
Step 3. Label given info with + and  signs.
Step 4. Indicate direction of force F.
+
t = 4 s
v2 = +20 m/s
v1= +8 m/s
Step 5. Recall definition of average acceleration.
Example 3 (Continued):What is average acceleration of car?Force
E
Example 4:A wagon moving east at 20 m/s encounters a very strong headwind, 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
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 headwind, 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
tf  to
Dv
Dt
aavg= =
vo= +20 m/s
vf= 5 m/s
+
Force
Dv = (5 m/s)  (+20 m/s) = 25 m/s
E
25 m/s
5 s
a =
Example 4: (Continued)
Acceleration is directed to left, west (same as F).
a =  5 m/s2
D
B
A
+
a =  5 m/s2
vf= 5 m/s
vo= +20 m/s
Force
E
Signs for Displacement
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
C
D
B
A
+
a =  5 m/s2
vf= 5 m/s
vo= +20 m/s
Force
E
Signs for Velocity
What are the signs (+ or ) of velocity at points B, C, and D?
D
B
A
+
a =  5 m/s2
vf= 5 m/s
vo= +20 m/s
Force
E
Signs for Acceleration
What are the signs (+ or ) of acceleration at points B, C, and D?
F
x
vf
vo
2 m/s
5 m
Careful
t = 4 s
8 m/s
vo + vf
2
x = xo + t
8 m/s + (2 m/s)
2
= 5 m + (4 s)
Example 5:A ball 5 m from the bottom of an incline is traveling initially at 8 m/s. Fourseconds later, it is traveling down the incline at 2 m/s. How far is it from the bottom at that instant?
F
+
x
vf
vo
2 m/s
5 m
t = 4 s
8 m/s
8 m/s + (2 m/s)
2
x = 5 m + (4 s)
8 m/s  2 m/s
2
x = 5 m + (4 s)
x = 17 m
Acceleration:
Setting to = 0 and solving for v, we have:
Final velocity = initial velocity + change in velocity
F
x
v
vo
2 m/s
5 m
t = 4 s
8 m/s
Acceleration in our ExampleWhat is the meaning of negative sign for a?
The force changing speed is down plane!
a = 2.50 m/s2
0
0
0
Use of Initial Position x0in Problems.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.
The xoterm is very useful for studying problems involving motion of two bodies.
Review sign convention for each symbol
The displacement is the ycoordinate. Whether motion is up or down, + or  is based on LOCATION.
1 m
2 m
The Signs of Displacement+
+
First choose + direction; then velocity vis positive if motion is with that + direction, and negative if it is against that positive direction.

+
The Signs of VelocityA push or pull (force) is necessary to change velocity, thus the sign ofais same as sign of F.
F
a()
F
a(+)
Acceleration Produced by ForceMore will be said later on the relationship between Fand a.
Given: ____, _____, _____ (x,v,vo,a,t)
Find: ____, _____
v = 0
300 ft
vo
+
F
X0 = 0
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?Step 1. Draw and label sketch.
Step 2. Indicate + direction and F direction.
v = 0
300 ft
vo
+
F
X0 = 0
Example: (Cont.)Step 3. List given; find information with signs.
Given:vo = +400 ft/s
v = 0
x = +300 ft
List t = ?, even though time was not asked for.
Find:a = ?; t = ?
+400 ft/s
v = 0
300 ft
vo
+
0
0
F
X0 = 0
vo2
2x
(400 ft/s)2
2(300 ft)
a = =
Continued . . .
Step 4. Select equation that contains aand not t.
2a(x xo) = v2  vo2
Initial position and final velocity are zero.
a =  267 ft/s2
Why is the acceleration negative?
Because Force is in a negative direction!
W
Earth
Acceleration Due to Gravitya = g = 9.80 m/s2 or 32 ft/s2
Directed downward (usually negative).
y
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?
y
+
Acceleration of Gravity:
W
Experimental Determination of Gravity (y0 = 0; y = 1.20 m)y = 1.20 m; t = 0.495 s
Acceleration ais negative because force W is negative.
a = 
v = 0
y = +
a = 
v = +
y = +
y = +
a = 
v = 
v = 
y = 0
a = 
y = 0
Release Point
y = Negative
v= Negative
a = 
Tippens
Given: ____, _____, a =  9.8 m/s2
Find: ____, _____
a = g
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.
Step 3. Given/find info.
a = 9.8 ft/s2 t = 2, 4, 7 s
vo = + 30m/sy = ?v = ?
vo= +30 m/s
Step 4. Select equation that contains y and not v.
+
a = g
0
vo= 30 m/s
Finding Displacement:y = (30 m/s)t + ½(9.8 m/s2)t2
Substitution of t = 2, 4, and 7 s will give the following values:
y = 40.4 m; y = 41.6 m; y = 30.1 m
a = g
vo= 30 m/s
Finding Velocity:Step 5. Find v from equation that contains vand not x:
Substitute t = 2, 4, and 7 s:
v = +10.4 m/s; v = 9.20 m/s; v = 38.6 m/s
a = g
vo = +96 ft/s
Example 7: (Cont.) Now find the maximum height attained:Displacement is a maximum when the velocity vf is zero.
To find ymaxwe substitute t = 3.06 s into the general equation for displacement.
y = (30 m/s)t + ½(9.8 m/s2)t2
a = g
vo =+30 m/s
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:
y = 91.8 m  45.9 m
ymax = 45.9 m
Draw and label sketch of problem.
Given: ____, _____, ______
Find: ____, _____