Physics 2

Finding the Slope of a Tangent Line

• In theory:
  • Take the slope of the line passing through two points on a curve.
  • This line is called a "secant line".
  • The two points approach each other until they almost touch.

Notation

• Use "d" instead of "$\Delta$".
• This is called the "derivative".
• $\frac{\Delta y}{\Delta x} \rightarrow \frac{dy}{dx}$
  • $\frac{\Delta y}{\Delta x}$: Slope of a secant line.
  • $\frac{dy}{dx}$: Slope of a tangent line.

Finding the Slope of a Tangent Line In Practice

• On a Graph:
  • Draw the tangent line (blue line) from the curve (red line).
  • The line should touch the curve at one point.
  • Find the change in y and x of the endpoints.
• From a Function:
  • This requires Calculus 1.
  • Not covered in this class.

Clicker Question 1

• What type of line is the red line?
  • A) A secant line
  • B) A tangent line
  • C) A parabolic line
  • D) None of the above

Clicker Question 2

• What is the sign of the slope of the blue curve at “x”?
  • A) Positive
  • B) Negative
  • C) Zero

Plotting the Slope Demonstration

• Simulation available at: https://phet.colorado.edu/sims/html/calculus-grapher/latest/calculus-grapher_all.html
• Observations:
  • The slope on one graph equals the value on the other.
  • They are aligned vertically.
  • Shifting the blue graph vertically leaves the red graph unchanged.
  • Shifting the red graph vertically tilts the blue graph.

Clicker Question 3

• Which graph of the slope matches this graph of the blue function?
  • A)
  • B)
  • C)

Going Backwards

• The function (blue) represents the accumulated amount of slope (red).
  • Positive slope = increasing function.
  • Negative slope = decreasing function.
• Large change in function (blue) occurs when:
  • Slope (red) is far from zero, or
  • Occurs for a large range of x (wide).

Area Under the Curve

• Simulation at: https://phet.colorado.edu/sims/html/calculus-grapher/latest/calculus-grapher_all.html
• Change in value = area under the curve.
  • Positive area is above the horizontal axis.
  • Negative area is under the horizontal axis.
• Demonstration:
  • Simulation at: https://phet.colorado.edu/sims/html/calculus-grapher/latest/calculus-grapher_all.html
  • Value increases by 5 when the area under the curve = 5.

Finer Points

• The slope graph does not indicate the y-intercept.
  • Shifting the function vertically doesn’t change the slope graph.
• Units:
  • Slope units = $\frac{\text{Vertical Units}}{\text{Horizontal Units}}$
  • Horizontal units don’t change. Example: $\frac{m}{s}$, $\\frac{s}{m/s}$

Summary: Notation

• $\frac{\Delta y}{\Delta x}$ vs $\frac{dy}{dx}$
  • $\frac{\Delta y}{\Delta x}$ means the slope of the secant line (between 2 points on the curve).
  • $\frac{dy}{dx}$ means the slope of the tangent line (at one point on the curve).

Summary: Skills You Should Be Able To Do

• Find $\frac{\Delta y}{\Delta x}$ or $\frac{dy}{dx}$ from a y vs x graph.
  • Draw a straight line and find the slope.
• Find $\Delta y$ from a $\frac{dy}{dx}$ vs x graph.
  • Find the area under the curve.
• Roughly sketch a $\frac{dy}{dx}$ graph from a y vs x graph or vice-versa.

Summary: Skills You Are NOT Expected To Do

• Write the function for $\frac{dy}{dx}$ from a $y(x)$ function.
• Write the function for $y(x)$ from a $\frac{dy}{dx}$ function.
• Use calculus terminology and symbols.
  • Derivative, integral, definite vs indefinite integral, $\int y \, dx$, etc.

Position & Distances

Describing Location

• How do you specify a location?
  • Distance relative to an object.
  • Direction matters.
• Physics specifies location using a “reference frame” or “frame of reference”.
• Basic idea
  • Coordinates in space
  • Relative to a physical object
  • Like infinite ruler used to measure position
• Choice of frame is arbitrary
  • Examples: location of zero, speed
  • Necessary to communicate

Position, $x$

• Basic idea:
  • Location measured in frame.
  • Positive or negative based on arbitrary choice.
• SI unit:
  • Meters (m)
• Note:
  • Position is a function of time, $x(t)$
• Example:
  • $x = 120 \, cm$ Position

Moving Frames

• Reference frames can move relative to each other.
  • Relative to car vs road.
  • Equally valid descriptions.
• Motion depends on choice of frame.
  • Truck moves relative to road but stationary relative to car.
• Rest frame:
  • Any frame where object isn’t moving.

Dealing with Time

• INSTANTANEOUS
  • Motion at one instant.
  • Defined at a given time, “t”.
  • Durations are infinitesimal.
  • Use slopes of tangent lines.
• OVER AN INTERVAL
  • Motion over a time interval.
  • Defined over a range, $t<em>i$ to $t</em>f$
  • Durations are finite, $\Delta t = t<em>f - t</em>i$
  • Use slopes of secant lines.

Displacement, $\Delta x$

• Definition
  • Final position minus initial position
  • Often shorter than distance traveled
• $\Delta x \equiv x<em>f - x</em>i$
  • $\equiv$ means “defined as”
  • $\Delta x$: Displacement
  • $x_f$: Final position
  • $x_i$: Initial position
• Example:
  • If $x<em>i = 30 \, cm$ and $x</em>f = 90 \, cm$, then $\Delta x = 60 \, cm$

Clicker Question 4

• You throw a ball in the air and catch it at the same spot. If it reaches a maximum height of 3 m above your hand, what is the displacement over the entire trip? (Positive direction = upward)
  • A) -6 m
  • B) -3 m
  • C) 0 m
  • D) 3 m
  • E) 6 m

Path Length, $s$

• Definition
  • How far the object travelled
• In this example, the path length is 120 cm.
• $s = 120 \, cm$

Graphing Position: Stationary

• If position is constant over time, the object is stationary.
• $x(t) = 60 \, cm$ for all t.
• Stationary = Horizontal line on a position vs time graph.
• Example:
  • At t = 1 s, x = 60 cm
  • At t = 2 s, x = 60 cm
  • At t = 3 s, x = 60 cm

Graphing Position: Moving Right at Constant Speed

• Moving right, constant speed = Straight line with positive slope on a position vs time graph.
• Example:
  • At t = 1 s, x = 30 cm
  • At t = 2 s, x = 60 cm
  • At t = 3 s, x = 90 cm

Graphing Position: Moving Left at Constant Speed

• Moving left, constant speed = Straight line with negative slope on a position vs time graph.
• Example:
  • At t = 1 s, x = 90 cm
  • At t = 2 s, x = 60 cm
  • At t = 3 s, x = 30 cm

Clicker Question 5

• Which direction is the person walking?
  • A) Rightward always
  • B) Leftward always
  • C) Rightward then leftward
  • D) Leftward then rightward
• Use standard convention that +x is to the right

Displacement on x vs t

• Basic idea
  • Displacement is the change in the vertical variable on a position vs time graph.

Speed on x vs t graph

• Basic idea
  • Faster movement is steeper on x vs t graph
• Reason
  • Faster means more distance covered in same amount of time
  • Larger $\Delta x$ for same $\Delta t$
• The steeper the slope, the faster the movement.

Clicker Question 6

• At what time is the person walking the fastest?

Velocity & Speed

(Instantaneous) Velocity, $v$

• Basic idea
  • How quickly & in what direction is it moving at a specific time
• Mathematical definition
  • Slope of tangent line on $x$ vs $t$ graph
• Notes
  • Often just called velocity.
  • SI Unit: meters per second (m/s)
  • Velocity can be positive or negative.
• Sign indicates direction of motion
• $v \equiv \frac{dx}{dt}$
  • $v$: Velocity
  • $\equiv$ means “defined as”
  • $\frac{dx}{dt}$: Slope of tangent line on $x$ vs $t$ graph
• Negative slope & moving in –x direction
• Positive slope & moving in +x direction

Average Velocity, $\bar{v}$

• Basic idea:
  • On average, how quickly & in what direction is it moving over a specific interval of time
• Mathematical definition
  • Slope of secant line on $x$ vs $t$ graph
• Notes
  • Depends on time interval chosen
  • If constant velocity, $v = \bar{v}$
• $\bar{v} \equiv \frac{\Delta x}{\Delta t}$
  • $\bar{v}$: Average velocity
  • $\Delta x$: Displacement
  • $\Delta t$: Duration of time interval

Speed, $v$

• Basic idea
  • Just how fast an object is moving.
  • Does not include direction.
• Mathematical definition
  • Absolute value of velocity (always positive).
• Warning: notation change in next chapter
  • Velocity will change to $\vec{v}$
  • Speed will change to $v$ or $\mid \vec{v} \mid$
  • Speed is $v$

Average Speed, $\bar{v}$

• Basic idea
  • Average value of speed in a specific time interval
• Mathematical definition
  • Path length divided by duration (always positive)
• Warning
  • Not the same as absolute value of average velocity
• Notes
  • If constant speed, $v = \bar{v}$
  • $\bar{v} \neq \mid \bar{v} \mid$
• $\bar{v} = \frac{s}{\Delta t}$
  • $\bar{v}$: Average speed
  • $s$: Path length
  • $\Delta t$: Duration of time interval

Clicker Question 7

• You pace back and forth. First you walk 2 m to the right then 2 m to the left. The round trip takes 4 seconds. What is the average velocity during the round trip?
  • A) -2 m/s
  • B) -1 m/s
  • C) 0 m/s
  • D) 1 m/s
  • E) 2 m/s
  • (Positive direction = rightward)

Clicker Question 8

• You pace back and forth. First you walk 2 m to the right then 2 m to the left. The round trip takes 4 seconds. What is the average speed during the round trip?
  • A) -2 m/s
  • B) -1 m/s
  • C) 0 m/s
  • D) 1 m/s
  • E) 2 m/s
  • (Positive direction = rightward)

Class Problem: Jogging

• A person jogs to the end of a two-mile trail in 40 minutes, rests for 20 minutes, and jogs back to the start of the trail in 60 minutes. Find:
  • A) Their average speed on their way to the end of the trail.
  • B) Their average speed on their way back.
  • C) Their average speed over the entire trip.
  • D) Their average velocity over the entire trip.

Solution

• A) average speed on their way to the end of the trail
  $v_{avg} = \frac{Path \,length}{Duration \,of \,time \,interval} = \frac{2 \,miles}{40 \,min} \cdot (\frac{60 \,min}{1 \,hr})= 3 mph$
• B) average speed on their way back
  $v_{avg} = \frac{Path \,length}{Duration \,of \,time \,interval} = \frac{2 \,miles}{60 \,min} \cdot (\frac{60 \,min}{1 \,hr})= 2 mph$
• C) average speed over the entire trip
  $v_{avg} = \frac{Path \,length}{Duration \,of \,time \,interval}= \frac{4 \,miles}{(40+20+60) \,min} \cdot (\frac{60 \,min}{1 \,hr})= 2 mph$
• D) Their average velocity over the entire trip.
  $Average velocity = 0 mph$

Graphing Velocity: Moving Right at Constant Speed

• Moving right, constant speed = Horizontal line above axis on a velocity vs time graph.
• $v = 30 \, cm/s$ is constant.

Graphing Velocity: Moving Left at Constant Speed

• Moving left, constant speed = Horizontal line below axis on a velocity vs time graph.
• $v = -30 \, cm/s$ is constant.

Graphing Velocity: Moving Right at Increasing Speed

• Moving right, increasing speed = Straight line tilted up away from axis on a velocity vs time graph.

Graphing Velocity: Moving Left at Increasing Speed

• Moving left, increasing speed = Straight line tilted down away from axis on a velocity vs time graph.

$v$ vs $t$ Graphs & $x$ vs $t$ Graphs

• Velocity is the slope of tangent line on $x$ vs $t$ a graph
  • Can get $v$ vs $t$ graph from $x$ vs $t$
• $v \equiv \frac{dx}{dt}$
  • $v$: Velocity
  • $\equiv$ means “defined as”
  • $\frac{dx}{dt}$: Slope of tangent line on $x$ vs $t$ graph

Displacement on $v$ vs $t$ Graph

• Displacement is area under curve on $v$ vs $t$ graph
  • Time interval must match
• Negative velocity produces negative displacement
  • Area counts as negative below axis
    ## Lab 1: Velocity and position
    ## Acceleration

Introducing Acceleration

• Everyday meaning
  • Speeding up
• Physics meaning
  • Speeding up
  • Slowing down
  • Turning (in 2D or 3D)

Clicker Question 9

• Which of the following can accelerate a car when used, according to the physics definition of acceleration?
  • A) The gas pedal
  • B) The brake pedal
  • C) The steering wheel
  • D) All of the above

(Instantaneous) Acceleration, a

• Basic idea
  • Rate of speeding up or slowing down
• Mathematical definition
  • Slope of tangent line on v vs t graph
• Notes
  • Often just called acceleration
  • SI Unit: meters per second per second ($\frac{m}{s^2}$)
  • Acceleration can be positive or negative
• $a \equiv \frac{dv}{dt}$
  • a: Acceleration
  • $\equiv$ means “defined as”
  • $\frac{dv}{dt}$: Slope of tangent line on v vs t graph
• Negative acceleration
• Positive acceleration

Average Acceleration, $\bar{a}$

• Basic idea
  • On average, rate of speeding up or slowing down over a specific interval of time
• Mathematical definition
  • Slope of secant line on v vs t graph
• Notes
  • Depends on time interval chosen
  • If constant acceleration, $a = \bar{a}$
• $\bar{a} \equiv \frac{\Delta v}{\Delta t}$
  • $\bar{a}$: Average acceleration
  • $\Delta v$: Change in velocity
  • $\Delta t$: Duration of time interval

Speeding Up & Slowing Down

• Speeding up
  • Moving away from $v = 0$
• Slowing down
  • Moving toward $v = 0$
• Sign of a $\neq$ speeding up/slowing down
  • Common misconception

Interpreting Sign of Acceleration

• Speeding up if $v$ & $a$ have the same sign/direction
• Slowing down if $v$ & $a$ have the opposite sign/direction

Clicker Question 10

• Which of the following are examples of nonzero acceleration?
  • A) Walking into a wall
  • B) Riding up the elevator at a constant speed
  • C) Sitting in a chair
  • D) Jogging at 10 mph in a straight line
  • E) None of the above

Graphing Acceleration: Moving at Constant Speed

• Constant speed = Horizontal line at zero on an acceleration vs time graph.
• $a = 0 \, cm/s^2$

Graphing Acceleration: Moving Right at Increasing Speed

• Moving right increasing speed (constant rate) = Horizontal line above axis on an acceleration vs time graph.
• $a = 20 \, cm/s^2$

Graphing Acceleration: Moving Right at Decreasing Speed

• Moving right decreasing speed (constant rate) = Horizontal line below axis on an acceleration vs time graph.
• $a = -20 \, cm/s^2$

Graphing Acceleration: Moving Left at Increasing Speed

• Moving left increasing speed (constant rate) = Horizontal line below axis on an acceleration vs time graph.
• $a = -20 \, cm/s^2$

Graphing Acceleration: Moving Left at Decreasing Speed

• Moving left decreasing speed (constant rate) = Horizontal line above axis on an acceleration vs time graph.
• $a = +20 \, cm/s^2$

Acceleration on Other Graphs

• Acceleration on v vs t graphs
  • Slope of tangent line
  • Away from zero = speeding up (same direction)
  • Toward zero = slowing down (opposite direction)
• Acceleration on x vs t graphs
  • Curvature
  • “Smile” = positive & “frown” = negative
  • Flattening = slowing down & getting steeper = speeding up

$\Delta v$ on a vs t Graph

• $\Delta v$ is area under curve on a vs t graph
  • Time interval matches

Going Beyond Acceleration

• Rate of change of acceleration over time called “jerk” or “jolt”
  • Examples: Hitting the brakes or being tackled
• Typically, we rarely discuss anything beyond acceleration
  • Physical cause of motion related to acceleration

Example Problem: Graphing Motion

• A ball is thrown up in the air and its velocity is plotted to the right. Plot the y vs. t and a vs. t graphs if the ball starts at zero height.

Lab 2: Acceleration & Velocity

Motion Diagrams

Motion Diagrams

• Basic idea
  • Several images overlaid taken at constant time intervals
  • Same as a strobe light image

Clicker Question 11

• If the motion diagram shown describes an object moving rightward or leftward, which best describes the acceleration of the object?
  • A) The acceleration is to the right
  • B) The acceleration is zero (no acceleration)
  • C) The acceleration is to the left
  • D) Not enough information
• Note: dots are sequential, but may be either left to right or right to left

Motion Diagrams

• What it shows
  • Position (at those times)
  • Displacement (between those times)
  • Average velocity (between those times)
  • Roughly, average acceleration (as velocity changes)
• $\bar{v} = \frac{\Delta x}{\Delta t}$
• $\bar{a} = \frac{\Delta v}{\Delta t} \sim \frac{\Delta \bar{v}}{\Delta t}$

Kinematic Equations: FOR CONSTANT ACCELERATION

Kinematic Equations

• Purpose
  • Relate the motion variables x, v, a, and t
• Requirements
  • Constant acceleration during time interval
• Notes
  • t is duration of time interval
  • i and f are initial and final values at start & end of time interval
• $\bar{v} = \frac{v<em>f + v</em>i}{2}$
• $v<em>f = at + v</em>i$
• $x<em>f = \frac{1}{2}at^2 + v</em>it + x_i$
• $v<em>f^2 = v</em>i^2 + 2a\Delta x$

Problem Solving Steps

1. Diagram situation
   • Simple picture with path of object
2. Label moments of interest
   • Info known or want to know
3. Identify times with constant acceleration
   • Typically, stated or implied
4. List known information at moments of interest
   • Position, velocity, time

Problem Solving Steps

5. Review kinematic equations & decide on which to use
   • Decide on time interval
   • Identify known variables
   • Identify unknown but desired variables
   • Identify unknown and not desired variables
6. Rewrite with specific values
   • Match notation you chose
7. Solve for desired variable

Example Problem: Hitting the Brakes

• A car going 80 mph slams on its brakes, skidding to a stop. It constantly accelerates to a stop at 6 $\frac{m}{s^2}$.
  • How far does it travel in 1 second?
  • How much time does it take to come to a full stop?
  • How far does it go while stopping?

Solution

Use the formula

• $x<em>f = x</em>i + v_it + \frac{1}{2} a t^2$

• $v<em>f^2 = v</em>i^2 + 2 a (x<em>f - x</em>i)$

• $v<em>f = v</em>i +at$
  Then,

• convert from mph $\rightarrow$ m/s

• $80 \, mph = 35.8 \, m/s$

• final = 1, Vi= 35.8 m/s, $a= -6 \frac{m}{s^2}$

• Travel in 1 second

• $x_f = 35.8 \cdot 1 + \frac{1}{2} \cdot -6 \cdot 1^2 = 32.7 \, m$

• Time to full stop $t= \frac{v<em>f-v</em>i}{a} \therefore t= \frac{0-35.8}{-6} \, s = 5.96 \, s$

• How far it will go wile stopping $x<em>f= 0, x</em>i= ?, v<em>f= 0, v</em>i=35.8, a=-6 \; Then=$ $0=35.8^2 + 2 \cdot -6 \cdot x<em>f \therefore, x</em>f= 107 m$

Example Problem: Hitting the Brakes

• A car going 80 mph slams on its brakes, skidding to a stop. It constantly accelerates to a stop at 6 $\frac{m}{s^2}$.
  • How far does it travel in 1 second? 32.7 m
  • How much time does it take to come to a full stop? 5.96 s
  • How far does it go while stopping? 107 m

Class Problem: Playing with Blocks

• A block rests on an inclined plane. Then it is briefly struck, so that it has some initial velocity ($v_{start}$) toward the top of the ramp. It slides all the way up to the very tip of the ramp before sliding down to the bottom. During this time, it has a constant 4 $\frac{m}{s^2}$ acceleration toward the bottom of the ramp.
  • What is the initial velocity, $v_{start}$?
  • How much time does it take to reach the bottom of the ramp?
    We have the following data:
• $V=0, a= -4, x= (10-3)$ Then by using the following formula
• $v<em>f^2 = v</em>i^2 + 2 a (x<em>f - x</em>i)$
  $0 = v<em>i^2 + 2 (-4) (0.07-0)$$v</em>i = 0.748 m/s$
  $t=? \triangle x = V<em>i t + 0.5 a t^2 , \, x= (10 +3)$ $\rightarrow 0.13= v</em>i t + 0.5 a t^2$
• $x<em>f = x</em>i + v<em>it + \frac{1}{2} a t^2$$\therefore 0.13 = 0.894 t -2 t^2$ $\rightarrow t=0.1 s \, and \, t=0.347 , \triangle V= at + V</em>i = \frac{\triangle x}{t}$

Class Problem: Playing with Blocks

• A block rests on an inclined plane. Then it is briefly struck, so that it has some initial velocity ($v_{start}$) toward the top of the ramp. It slides all the way up to the very tip of the ramp before sliding down to the bottom. During this time, it has a constant 4 $\frac{m}{s^2}$ acceleration toward the bottom of the ramp.
  • What is the initial velocity, $v_{start}$? 0.894 m/s
  • How much time does it take to reach the bottom of the ramp? 0.479 s

Free Fall

Free Fall

• Basic idea
  • Motion only influenced by gravity
  • Effectively, nothing is touching it
  • Not just “falling down”
  • Also rising and orbits
• Caveats
  • Negligible air resistance
  • Small distance vs size of Earth
  • Not orbit of Moon, for example
• BBC Two Demonstration with Brian Cox (at 2:50)

Clicker Question 12

• What is common about all objects in free fall, including objects thrown up, thrown down, let go, etc.?
  • A) All objects have the same position
  • B) All objects take the same time to hit the ground
  • C) All objects have the same velocity
  • D) All objects have the same acceleration

Constant Acceleration in Free Fall

• Basic idea
  • All objects have the same downward acceleration in free fall. On the way up
    • Upwards velocity
    • Slowing down
    • Velocity opposite acceleration so downwards acceleration On the way down
    • Downwards velocity
    • Speeding up
    • Velocity same direction as acceleration so downwards acceleration
      Same acceleration at any point in their motion!

Acceleration Due to Gravity, g

• Alternative name

  • Gravitational field
  • Not “gravity”

• Definition

  • The magnitude of acceleration for all objects in free fall.
  • Acceleration direction is downward
    Varies over long distances

• On Earth vs Moon vs Jupiter vs …

• $g = 9.8 \, \frac{m}{s^2}$

  (Near the surface of Earth) $a = -g \,(in \,free \,fall)$
  (with up as positive x direction)

Clicker Question 13

• Which best describes the acceleration of an object thrown upward in the air?
• A) Upwards as it rises, zero as it stops, downward as it falls
• B) Downward as it rises, zero as it stops, downward as it falls.
• C) Always downward
• D) Always upward

Words of Caution

Word of Caution$v<em>f^2 = v</em>i^2 + 2a\Delta x$

• Calculator failure
  • Example
  • If right hand side negative
  • Indicates either:
    • User error (usually missing sign of a or $\Delta x$)
    • Physically impossible (asking what's the speed where it never was)

Example Problem: Movie Stunt Throw

• In an action movie, a hero or heroine throws something in the air, does some dramatic task, and catches it again after. If the object was in the air for 5 seconds, how fast was it moving as it left their hand? How high did it go?
  *To solve this problem we need to identify variables; Initial = leaves start from the start point xi=? final highest time we reach 5s.
• We know $g=9.8 m/s^2$
• We need to use the following formula:
  • $x<em>f = x</em>i + v_it + \frac{1}{2} a t^2$
  • $V<em>f = v</em>i + at$
• $at^2 + v<em>it + x</em>i$
  or $\triangledown V+= \frac{\triangledown x}{t}$
  $final=\frac{gh^2}{2} \therefore x<em>0= 19, v</em>1 =24.5$
  $V_2= \sqrt{2</em>(-9)(24.5)}$

Example Problem: Movie Stunt Throw

• In an action movie, a hero or heroine throws something in the air, does some dramatic task, and catches it again after. If the object was in the air for