1. Status: Unit 1, Chapter 2 Reference Frames and Displacement
Average Velocity
Instantaneous Velocity
Acceleration
Motion of Constant Acceleration
Solving Problems
Free Fall
A real-world application of the equations.
Variable Acceleration
An important generalization of our results.
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2. Just a quick summary…
Remember also, the prescription for solving problems
1) Setup
2) Selection of Equations
3) Check
Often you'll get some wrinkles due to occurrence of relationships between objects and segments, but don't panic these are usually very helpful.
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3. Falling objects or Free Fall
First formally studied by Galileo Galilei ( )
Postulated that all objects would fall with the same constant acceleration in the absence of air or other resistance.
He showed that this assumption predicts that for an object falling from rest the distance traveled would be proportional to the square of the time:
This is completely consistent with our third equation of motion:
Today we are going to do something quite fun, follow in the steps of the “father of science” and study the equations of motion for freely falling bodies.
This is the most familiar and common example of uniformly accelerated motion.
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4. Some preliminaries
In the absence of air resistance and at the surface of the earth all objects fall with the same constant acceleration
Notice the caveats, "no air resistance" and "near the surface of the earth".
Obviously a feather falls more slowly than a rock because of the air.
Also far away from the earth there is no gravity (and also at the center of the earth!).
As long as the air resistance is minimized and distance of the fall is small compared to the radius of the earth the acceleration is essentially constant throughout the fall.
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5. Idealized motion (where air resistance is neglected and the acceleration is constant) is known as freefall.
The magnitude of the acceleration due to gravity at the earth’s surface is denoted:
Since the acceleration is constant we can apply the equations of motion!
Because objects fall in the vertical or y direction it's also natural to use the symbol y for displacement
This does not really change the equations. Remember x meant displacement in the direction of constant acceleration. y can mean the same thing, displacement in the direction of the constant acceleration of gravity. So we change x y and a  g:
g = 9.80 m/s2 or 32.2 ft/s2
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6. Example 1: A dropped ball
Step 1: Suppose a ball is dropped from a tall building. What would be its position and speed after 1.00, 2.00, and 3.00s?
Step 2:
Note that we’ve chosen +y to be down.
For each problem there is complete freedom to choose the direction, however we must be consistent through the problem since position, velocity, and acceleration are all vectors.
In this case then the acceleration would be g = m/s2. +y -y
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7. Step 5: The equations are Step 6: A sample calculations gives:
Step 4: Acceleration and time are known, we need to find a relationship between those two and displacement and velocity.
Step 5: The equations are
Step 6: A sample calculations gives:
So for:
1 sec, x1 = 4.90 m and v1 = 9.80m/s
2 sec, x2 = 19.6 m and v2 = 19.6m/s
3 sec, x3 = 44.1 m and v3 = 28.4m/s
Known
Unknown
y0=0
y = ?
v0=0
v = ?
g=-9.80 m/s2
t = 3.00 s
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8. Take a graphical look… v m/s As we expect: Position proportional to t2
Velocity proportional to t
Velocity the derivative dx/dt
v m/s
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9. Example 2: Ball Toss
Step 1:Consider a ball tossed in the air with initial speed 15.0 m/s. How high does it rise above the release point before beginning to fall back to the release point? How long does the flight take?
Step 2:
Here it’s more convenient to choose +y to be upward. In that case g = m/s2.
Steps 3 and 4:Let’s attack the 1st question
As the ball rises its speed will decrease until it reaches the highest point and momentarily stop. It will then start moving downward with increasing speed. That’s why v=0 in the above table.
Known
Unknown
y0=0
y = ?
v0=+15.0m/s
t = ?
g=-9.80 m/s2
v = 0 +y
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10. Now we take on the second question, regarding the total time of flight
Step 5: Equation three relates the velocity and acceleration to the displacement:
Step 6:
Now we take on the second question, regarding the total time of flight
Steps 3 and 4: All of the initial information is the same but here we set the final displacement to zero:
Known
Unknown
y0=0
v = ?
v0=+15.0m/s
t = ?
g=-9.80 m/s2
y = 0
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11. Step 5: We can find the total time of flight using: Step 6:
This is interesting, there are two roots t = 0 and t =3.06 s.
The two solutions correspond to the initial throw and the return.
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12. Some comments
Notice here that “freely falling” may refer to an object falling or rising. What is important is that the acceleration is constant and downward.
In the ball toss, the ball first had a positive velocity, then a zero velocity, and finally a downward velocity.
In all cases the acceleration was constant and downward.
This is free fall. Even though the ball's velocity changes moment to moment the acceleration vector does not! It is always down and 9.8m/s2.
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13. More on the ball toss
Step 1: A) How much time does it take for the ball to reach the maximum height? B) What is the velocity when the ball returns to the initial point? C) At what time does the ball pass the point 8.00 m above the initial point?
Step 2: See slide 9
Steps 3 and 4: For this case we consider the interval from the toss to the apex. This gives us the same table as with the initial problem statement.
Known
Unknown
y0=0
y = ?
v0=+15.0m/s
t = ?
g=-9.80 m/s2
v = 0
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14. Step 5 and 6: The needed equation is the first one
This is interesting, the flight up takes 1.53 s and the round trip 3.06s. The time of travel up equals the time of travel down.
Steps 7 and 8: The magnitude and units are certainly reasonable
Next up: velocity when the ball returns to the start.
Steps 3 and 4:The table contains the initial information plus a final displacement of y=0.
Known
Unknown
y0=0
v = ?
v0=+15.0m/s
t = ?
g=-9.80 m/s2
y = 0
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15. But since the motion must be down the velocity is -15.0 m/s
Step 5 and 6: Given this set of information we can use the third equation:
But since the motion must be down the velocity is m/s
We could also derive this using the total travel time we calculated earlier:
On the way up the velocity is positive and equal to m/s, on the way down it's negative but of the same magnitude!
This would be true for any distance y. This symmetry occurs because the coin loses 9.8 m/s in speed each second on the way up and gains the same amount each second on the way down
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16. Finally we ask at what time does the ball pass 8.00m?
Steps 3 and 4:The table contains the initial information plus a final displacement of y=8.00m.
Steps 5 and 6: The second equation has time if displacement and acceleration are given.
For our trouble we get a quadratic equation:
Two solutions correspond to the ball going up past 8.00m and returning down past 8.00m.
Steps 7 and 8: The magnitudes and direction for all three questions seem fine.
Known
Unknown
y0=0
v = ?
v0=+15.0m/s
t = ?
g=-9.80 m/s2
y = 8.00m/s
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17. Velocity up and down equal at a given point
Two clear symmetries
Time up and down equal
Velocity up and down equal at a given point
Since v = dx/dt both must be symmetric
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18. Variable Acceleration
The equations we derived are clearly very useful, but are also a limiting case.
We need to use the calculus to generalize the equations of motion.
We simply start with the definition of acceleration a = dv/dt and recast it as dv=(a)dt.
Already this structure is suggestive, an infinitesimal change in velocity depends on acceleration times an infinitesimal time interval. The integration suggests itself…
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19. Now if a were indeed constant, Likewise,
And once again assuming constant acceleration,
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20. Time varying acceleration
The calculus quickly leads to the equations of motion for constant acceleration.
But now we’ve added to our tool kit and can deal with varying acceleration. Let’s consider a problem:
Step 1: A car starts from rest and accelerates as a function of time such that a = (7.00 m/s3)t. What are the velocity and displacment at t=2.00s?
Step 2:
(7.00m/s3)t
v = ?
x = ?
t=2.00s
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21. Step 3:
Step 4: We’ll this is a pretty straightforward problem, it’s just that we won’t be selecting from our trusty foursome, but from the integral equations.
Step 5 and 6:
Known
Unknown
x0=0
x = ?
v0=0
v = ?
a=(7.00 m/s3)t
t = 2.00s
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22. Step 5 and 6:
Step 7 and 8: We did the same problem last lesson with a=2.00m/s2, after 5.5s the displacement and velocity were 30m and m/s, respectively. Given that the average acceleration in this problem was larger at 7 m/s2 a smaller displacement and larger velocity seems reasonable.
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23. Let’s explore a simple model for terminal velocity (Problem 67).
When an object falls it does meet air resistance, eventually the resistance will be such that there is no net acceleration.
For a wide body like a parachute this occurs quickly and at a slow velocity, for a narrow object like a meteor this takes some time and occurs at a high velocity.
Let’s explore a simple model for terminal velocity (Problem 67).
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24. Manipulating infinitesimals
We really just need to modify the form of the acceleration .A reasonable model would hold that the magnitude of the acceleration decreases linearly with velocity.
Manipulating infinitesimals
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25. What is the terminal velocity?
Integrating
What is the terminal velocity?
Well this occurs when the acceleration reaches zero. So solving:
It also occurs at t = infinity
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26. v
g/k t
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27. Where we’ve been and are going…
We’ve developed and explored quite thoroughly the equations of motion for constant acceleration in one dimension.
And even shown how to generalize to varying acceleration.
Next we’ll take a trip into higher dimensions.
But first we’ll need to add vectors to our tool box.
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Physics 253