1. Straight Line Motion (continued)
Chapter 2
Straight Line Motion (continued)

2. Review Displacement: change in position
Average velocity: rate of change in position over time
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3. Average Velocity Black line is the actual path
Green line is the “average” path
Not completely descriptive t
x(t) Dx Dt t1 t2 x1 x2
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4. Ball Thrown in the Air “On average” it’s velocity is zero (Dx=0)
We can do better by drawing two lines
As we draw more lines, the approximation gets better
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5. Smaller and Smaller Dt
The smaller Dt is, the better our “straight line” approximation looks
Dt = 5 s
Dt = 2.5 s
Dt = 0.5 s
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6. Instantaneous Velocity
What if we want to look at really small Dt?
Look familiar?
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7. Position vs. Time Graph
Velocity is equal to the slope of the tangent line to x(t)
Positive (negative) velocity means the cart is moving in the positive (negative) direction
v = 0 at a direction change
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8. Bungee Jumper Consider the velocity of a bungee jumper
When is the velocity negative?
When is the velocity positive
When is the velocity zero?
What does this tell us?
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9. x(t) = 5 cm + (36 cm/s) t – (6 cm/s2) t2.
Example
You observe that a flea running on your ruler has a position given by
x(t) = 5 cm + (36 cm/s) t – (6 cm/s2) t2.
What is the instantaneous velocity of the flea at t=2 s? At what time does it change direction?
We know that velocity is the time derivative of position. The derivative is linear, so
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10. x(t) = 5 cm + (36 cm/s) t – (6 cm/s2) t2
Solution
Taking the derivative of
x(t) = 5 cm + (36 cm/s) t – (6 cm/s2) t2
Adding the terms together:
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11. Solution
Finally, we plug in t = 2 s and solve:
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12. Solution At what time does the flea change direction? When v(t) = 0
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13. Displacement as an Integral of Velocity
Consider a particle with a constant velocity
Rearrange to give x(t2)
Knowing v gives x(t) (almost – still need x(t1))
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14. Displacement as an Integral of Velocity
For constant velocity:
Dx=v Dt
The area under the curve is the displacement! t
v(t) v Dt
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15. Displacement as an Integral of Velocity
v(t) t
Imagine breaking the a complex curve into lots of little rectangles
This is just an integral!
Displacement is the area under the curve
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16. Displacement as an Integral of Velocity
We can arrive at the same result using calculus…
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17. Acceleration Time rate of change of velocity Units: m/s2
Average acceleration
Instantaneous acceleration
Units: m/s2
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18. Velocity vs. Time Graph (cont)
Acceleration is equal to the tangent line to the slope of v(t)
Note: An object can have a negative acceleration and still have a positive velocity! t
v(t)
Zero accel.
Positive accel.
Negative accel.
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19. Velocity as an Integral of Acceleration
Same procedure as before…
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20. Constant Velocity Graph
v(t) t
Velocity vs Time
x(t) t
Position vs Time
v(t) = constant
x(t)=v t + x(0)
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21. Kinematics variables
In one dimension Position x(t) meters Velocity v(t) meters/second Acceleration a(t) meters/second2
All depend on time
All are vectors (have direction and magnitude)
Position
Velocity
Acceleration d dt
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22. Physical Interpretation of Zero
x = 0
Object is at the origin
v = 0
Object stopped or changing directions
a = 0
Velocity is constant
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23. Special Case: Constant Acceleration
In many cases, acceleration will be constant. We can use this to make our lives easier.
Be careful: The following equations ONLY work if acceleration is constant.
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24. What Does It Look Like? Acceleration vs Time Velocity vs Timex t
Position vs Time
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25. Constant Acceleration
Given initial position and velocity, we can calculate position:
Define:
x(0) = x0
v(0) = v0
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26. Constant Acceleration
We can use calculus to get this equation, too!
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27. Constant Acceleration
So with constant acceleration, the fundamental equations are
Should be able to solve any constant acceleration problem from these two, but…
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28. Constant Acceleration
We can derive variants of these equations which are also useful
Equation
Missing Quantity
vf = v0 + a t
x – x0
x – x0 = v0 t + ½ a t2 vf
vf 2 = v a (x – x0) t
x – x0 = ½ (v0 + vf) t a
x – x0 = vf t – ½ a t2 v0
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29. Free-Fall Acceleration
Constant acceleration caused by gravity
Only true if there is no air resistance
We’ll ignore the effects of air resistance most of the time
Near Earth’s surface, the acceleration due to gravity is approximately constant,
g = 9.8 m/s2 DOWN
This is a pretty good approximation
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30. If You Believe…
A popular demonstration of free fall was conducted by Commander David Scott on the moon…
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31. Problem Solving Tips Read the problem carefully Set up the problem
What facts are you given?
What is asked for?
Set up the problem
Sketch a diagram
Decide where x = 0, when t = 0
Decide which way is positive x direction
Work from fundamental equations
Don’t try to memorize every formula.
Don’t just search the book for any formula containing the right symbols!
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32. Problem Solving Tips Reason symbolically as long as possible
Calculator-punching errors are harder to find than algebra errors
Use standard units
Check result for correct units
Practice builds skill and confidence
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33. Example:
150 ft
It is rumored that Galileo dropped steel balls off the Tower of Pisa to determine that g was the same for all objects. How long does it take for a ball to reach the ground and how fast is it moving when it gets there? (neglect air resistance)
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34. The Solution 45.7 m Convert to standard units
x=0 +x
Convert to standard units
150 ft=45.7 m
2. Choose a coordinate system
Origin: x=0 on ground Direction: up positive Drop begins at t=0
3. Write down the equation for position
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35. The Solution 45.7 m 4. Note that v0=0 5. Find t when x(t)=0 +x x=0
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36. The Solution x x=0 45.7 m 6. Solve for t with x0=45.7 m
7. Solve for v(t = 3.05 s)
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37. Question:
45.7 m
x=0 x
Let’s say Galileo dropped two steel balls. After dropping the first ball he waited one second and dropped the second ball
How long after the first hits the ground does the second hit?
Easy! 1 s
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38. The student is caught when xp = xs
Example
A stopped police car can accelerate at 30 m/s2. A PSU student passes the stopped police car going at the constant speed of vs = 50 m/s (110 mph). If the policeman takes 2 seconds to start his engine, how long does it take him to catch up to the student?
Let xp be the position of the policeman and xs be the position of the student.
The student is caught when xp = xs
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39. Solution: There is more than one way to “skin a cat”
I think the easiest solution is to start the clock when the policeman begins accelerating, so we have the following:
x0s=?
v0p=0 m/s
a0p=30 m/s2
v0s=50 m/s
a0s=0 m/s2 +x
Need to find x0s first
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40. Solution
The car travels for 2 s at v = 50 m/s before we start the clock, so
Now we can write down everyone’s position as a function of time
This simplifies to:
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41. Solution Setting xp equal to xs: We have a quadratic equation. Recall:
Solving equation 1 for t:
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42. Solution Simplifying and plugging in values
Remember, this is 4.7 s after the engine was started, so it is 6.7 s after the student passed the policeman
Why are there two answers, and why is one of them negative?
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43. Solution
There are two solutions because the lines meet in two places
The negative solution is non-physical x t
t = 0 s
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44. One more Example
How much acceleration must be provided by the brakes to stop a car traveling 30 m/s in 100 m?
X=0 x
Let x=0 m when braking begins.
Then xf=100 m, vi=30 m/s, vf= 0 m/s
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45. Bouncing a Tennis ball
Sketch x(t), v(t), a(t) for a basketball during one ‘dribble’. a -g
bounce
down up v0 v t t
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46. Bouncing a Tennis ball
Sketch x(t), v(t), a(t) for a basketball during one ‘dribble’.
bounce
down up v0 v t x x0 t
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