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**AP C UNIT 1 1-Dimension A quick review of basic kinematic variables**

Difference in math application towards physics Differentiation (Calculus)

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**Definitions Distance (x) Displacement ( ) Speed (v) Velocity ( )**

Avg = 2 pts in time, Instantaneous = 1pt in time

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**Definitions Acceleration ( ) If v = constant, then a = ?**

Velocity is maximum when acceleration is Position is maximum when velocity is Be careful of signs zero zero

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Calculus vs Algebra In HP, all problems were solved using a constant variable in the equations. Graphical analysis used lines of constant slope. In APC, it requires utilizing the tool of differential & integral calculus to further probe physics principles where variables might be changing. The “down & dirty” calculus skills presented here are not meant to replace those you learned/about to learn in your calculus class. Note a mixed ability group of Calc AB and BC.

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**Calculus – Differentiation & Slope**

Last year we formulated linear regression lines of best fit, chose 2 points on the line, and determined the slope. No matter what 2 points were chosen, obviously, the slope was always the same since it was constant.

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**But what happens when we have a non-linear function**

But what happens when we have a non-linear function? In the graph below, the slope varies from point to point. Trying to calculate Δy/Δx from A to B we would yield a negative slope, from B to C it would yield a positive slope, and from A to C, a zero slope. To find the slope at ‘A’, we can examine a point just above and just below ‘A’. The closer those points are to ‘A’ (and to each other), the more accurately they would describe the slope at that point.

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This idea is to compute the rate of change as the limiting value of the ratio of the differences Δy/Δx as Δx becomes infinitely small. It follows that this limit is the exact slope of the tangent. Such a tiny or infinitesimal change in x and y is denoted by dx and dy, where we replace Δy/Δx with dy/dx.

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Differentiating or taking the derivative, f’(x), of a function, f (x), equates to the slope of the tangent line at a point when the run is reduced to such a small value…an instantaneous point To find instantaneous velocity at a point in time, we use the following notation (recall Δ was for 2pts in time) lim Δt dx is a differential distance and dt is a differential time which is a fancy way of saying very, very small. In technical terms, dx is what happens to Δx in the limit when Δx approaches zero.

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**1st derivative of velocity or 2nd derivative of displacement**

To find instantaneous acceleration at a point in time, we use the following notation 1st derivative of velocity or 2nd derivative of displacement

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**Calculus Rules - Differentiation (down & dirty method…your calc teacher will fill in the details)**

Rule #1: Derivative of a Constant *Derivative of a constant is zero If f(x) = 5, then f’(x) = 0. Rule #2: Power Rule If f(x) = xn , then f’(x) = nxn-1. If f(x) = 5x2 + 3, then f’(x) = 2(5x1) + 0 = 10x…using derivative of constant & power rule. NOTE that we are just finding slopes of a tangent using these rules

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**Maximum and Minimum points (ie; when does max velocity occur?)**

When a function reaches its maximum or minimum value, it must turn around where its slope = 0 Relative maxima y relative minimum x x x x Thus, the derivative when evaluated at x1 , x2 ,x3 will equal zero and allows us to determine where max or min occurs.

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Example: f(x) = x3 – 6x2 + 9x + 1 To find max or min values, set f’(x) = 0 and solve. f’ (x) = 3x2 – 12x + 9 = 0 = 3(x2 – 4x + 3) = 0 x = 1 and x = 3 (extreme points) This is where the extreme points are….are they a max or min? 2nd Derivative Test: To find if it’s max or min (concavity), take 2nd derivative of function f’’(x) = 6x-12 Plug in extrema value. If f’’(x) < 0, then it’s a Max If f’’(x) > 0, then it’s a Min x = 1 is max since you get -6 x = 3 is min since you get +6 You could also plug in values above and below the extrema values (1 & 3) into the 1st derivative to determine if slope is incr or decreasing to determine concavity. Essentially, the 2nd derivative is looking at how the 1st derivative is changing.

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**Example The position of a particle is given by:**

x(t) = t – 2.1t3 a) Find the velocity of particle at t = 3.0s. b) Find the maximum position of particle. Practice worksheet

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**Calculus - Integration**

Integration is the reverse of differentiation where an integral is called the anti-derivative. Much like finding the slope of the tangent line at a point for differentiation, integration is finding the area under the curve or function. Like finding area for a v-t graph which equals displacement. Last year we only looked area for linear functions such as:

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**However, if the function is not linear, we must use calculus**

However, if the function is not linear, we must use calculus. We could estimate the area by breaking the curve up into many rectangles and summing them up which can approximate the area. The more rectangles we use (smaller width), the better the estimation.

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Consider the function at the left where we wish to find area from A to C. The problem is that the height keeps changing as we move from A to C along the function. In order to minimize this problem (height changing), we focus on a very small region of the graph, where the height is relatively stable. Start by picking a point x along x-axis and another point extremely close but beyond it, (x+dx). Drawing vertical lines at these two points where they meet the function yields a shaded region.

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If we treat this region as a rectangle, its area is equal to the height f(x) times the tiny width dx. Obviously the shaded region isn't a rectangle, but as dx becomes smaller—as we bring the right side toward the left—the height change becomes less significant, and the region more closely resembles a rectangle. As dx approaches zero, this approximation becomes perfect: the height will equal f(x) on both left and right ends of rectangle…the area of the shaded region = f(x)*dx. We now just use an infinite amount of tiny rectangles and sum them up.

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**‘S’ looking symbol means to sum up**

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**Calculus Rules - Integration**

Rule #1: Power Rule where C is a constant and n ≠ -1 If f(x) = 6x2, then the integral of that function is:

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**(these integrals are evaluated between 2 values)**

DEFINITE INTEGRALS (these integrals are evaluated between 2 values) Consider f(x) = 4x2 + 5x Evaluate the integral between the values of 1 and 2. For each of the following expressions determine the indefinite integral with respect to time: (a) v(t) = 6t3 − 5t (b) a(t) = 3t2 − 4t + 7 (c) a(t) = 10 + t−2 (d) v(t) = 2t5 − t4/3

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**Using the Constant, C. Acceleration of a bus is given by: a(t) = 1.2t**

a) If vbus = 5.0m/s at t = 1.0s, what is vbus at t = 2.0s? b) If position of bus is 6.0m at t = 1.0s, find position at t = 2.0s?

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**Constant Acceleration Eqns**

From above Use substitution to get other two equations

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**Example: A car sits at a light**

Example: A car sits at a light. When light turns green, it accelerates at a constant rate of 2.5m/s2. At the moment the acceleration begins, a truck moves past the car with a uniform speed of 15m/s. At what position beyond the starting point does the car overtake the truck? 23

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QUESTION A pair of identical balls are simultaneously released on a pair of equal length tracks, A and B, as shown. Friction is minimal. Both balls reach the ends of their tracks at the same time speed Both of these None of these 24

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**Graphical Analysis Position vs Time (x vs t):**

Describes position of particle with respect to time Slope (Δx/Δt) indicates the velocity of the particle (mag & dirn) Average Velocity (displacement/time) is slope of the secant line Instantaneous Velocity is slope of the tangent line 25

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**Velocity vs Time (v vs t):**

Describes velocity of particle with respect to time Slope (Δv/Δt) indicates the acceleration of the particle (mag & dirn) Instantaneous Acceleration is slope of the tangent line Average Acceleration (velocity/time) is slope of the secant line 27

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**Note that particle can have negative velocity and positive acceleration**

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**Area indicates the displacement of particle**

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**Acceleration vs Time (a vs t):**

i) Describes acceleration of particle with respect to time ii) Area equates to change in velocity 30

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Sketch the x vs t, v vs t, and a vs t graphs for the diagram assuming the ball stays in contact with the surface of ramp at all times. a v x *Graph worksheet 31

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Freefall Assume no air friction where acceleration due to gravity on Earth is given by a = ‘g’ = -9.8m/s2. Objects in freefall are always accelerating downwards towards center of Earth. Can change based on location. In 1D, what is the value of velocity at apex of flight? What is the value of acceleration at apex?

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A balloonist, riding in the basket of a hot air balloon that is rising at 10.0m/s, releases a sandbag when the balloon is 40.8m above the ground. Determine the velocity of the bag when it hits the ground. Sketch a graph of x, v, and a vs t for motion of sandbag

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A model rocket is fired from rest vertically and ascends with a constant vertical acceleration of 4.0m/s2 for 6.0s. Its fuel is then exhausted and it continues as a free-fall particle. What is the total time elapsed from takeoff to striking the Earth?

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**VECTORS A. TERMS DEFINED *SCALAR *VECTOR**

B. ADDITION / SUBTRACTION METHODS *2 OR MORE VECTORS COMBINED YIELDS RESULTANT *GRAPHICAL HEAD 2 TAIL, PARALLELOGRAM *ANALYTICAL 35

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*SUBTRACTION C. COMPONENTS aX= acosθ ay= asinθ 36

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**Vector with magnitude = 1 w/ direction Lacks dimension & unit **

E. UNIT VECTOR Vector with magnitude = 1 w/ direction Lacks dimension & unit Purpose is to ‘point’ LABELED ˆ = hat, replaces these are vector components of and ax and ay are scalar components of 37

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2-D Motion The position of a particle that moves in both the x & y plane at the same time can be described by: < position vector > Find displacement of particle as it moves from to .

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**Similarly, it’s velocity is described as:**

has a direction that is always tangent to path of particle It’s acceleration is described as: Example: find and for arbitrary times.

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Projectiles

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**Neglecting air resistance,**

Horizontal velocity = constant Vertical velocity changes Horizontal acceleration = zero Vertical acceleration = ‘g’

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a) Rank the paths (use only the symbols > or = , for example, a>b=c) according to time of flight, greatest first. b) Rank the paths (use only the symbols > or = , for example, a>b=c) according to initial speed, greatest first.

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**Derive an expression to solve for minimum initial v to just clear gap (ignoring bike width)**

y x θ

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**Relative Velocity vAC = vAB + vBC**

Used whenever you see “velocity relative to” or “velocity with respect to”, addition and subtraction of velocities is done so by utilizing a subscript method. vAC = vAB + vBC Label each object with its F.O.R. You want the inner subscripts to match up Switching the order of subscripts is like multiplying by negative 1 where vAB = -vBA vAC refers to velocity of A relative to C 45

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Example: A car is moving north at 88km/h when a truck approaches it from the other direction moving at 104km/h. a) What is the truck’s velocity relative to the car? b) What is the car’s velocity relative to truck? c) How do relative velocities change after the pass each other? 46

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2D example: A supersonic aircraft is moving in still air with a constant velocity of Suddenly, at t = 0 a wind gusts with a velocity of Assuming the pilot makes no attempt to correct for wind, find plane’s displacement after 1hr relative to ground.

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**A canoe has a velocity of 0. 40m/s SE relative to the Earth**

A canoe has a velocity of 0.40m/s SE relative to the Earth. The canoe is on a river flowing 0.50m/s East relative to the Earth. Find velocity of canoe relative to the river.

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Kinematics in One Dimension

Kinematics in One Dimension

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