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Accelerated Motion Chapter 3

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**Chapter Objectives Describe accelerated motion**

Use graphs and equations to solve problems involving moving objects Describe the motion of objects in free fall (separate presentation - not on this test)

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**Section 3.1 Acceleration Define acceleration**

Relate velocity and acceleration to the motion of an object Create velocity-time graphs

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**Uniform Motion Nonuniform Motion**

Moving at a constant velocity If you close your eyes, you feel as though you are not moving at all Moving while changing velocity Can be changing the rate or the direction You feel like you are being pushed or pulled

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Changing Velocity Consider the following motion (particle model) diagram Constant Velocity Not moving Increasing Velocity Decreasing Velocity

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**Changing velocity You can indicate change in velocity by**

the motion diagram spacing the magnitude (length) of the velocity vectors. If the object speeds up, each subsequent velocity vector is longer. If the object slows down, each vector is shorter than the previous one.

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Velocity-Time Graphs Distance being covered is longer, thus the runner is speeding up. Distance being covered is longer, thus the runner is speeding up.

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**Velocity-Time Graphs Time (s) Velocity (m/s) 1 5 2 10 3 15 4 20 25**

Slope??? Time (s) Velocity (m/s) 1 5 2 10 3 15 4 20 25 Area???

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**Change in Position → can’t get position**

Velocity-Time Graphs Analyze the units Slope = rise over run m = ∆y / ∆x Slope = m/s/ s = m/s^2 m/s^2 is the unit for acceleration Area = ½ b h b = s * m/s = m m is the unit for displacement The slope of a velocity-time graph is the ACCELERATION and the area is DISPLACEMENT. Change in Position → can’t get position

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Area = Displacement Slope = Acceleration

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**Velocity – Time Graphs How Fast something is moving at a given time?**

Average Acceleration Use the information on the x & y axis to plug into the equation a = ∆v / t Slope Instantaneous Acceleration Find the slope of the line (straight line) Find the slope of the tangent (curve) Displacement Find the area under the curve You do not know the initial or final position of the runner, just the displacement.

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**Velocity-Time Graphs Describe the motion of each sprinter. A**

Constant velocity Zero Acceleration Positive displacement B Constant Acceleration Starts from Rest Positive displacement Describe the motion of each sprinter.

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**Velocity-Time Graphs D Constant Acceleration Positive Acceleration**

Comes to a Stop Zero displacement E Constant Velocity Zero Acceleration Negative displacement C Constant Acceleration Negative Acceleration Comes to a Stop Positive displacement

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Sample Question What is the average acceleration of the car shown on the graph below? A. 0.5 m/s2 B m/s2 C. 2 m/s2 D. -2 m/s2

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**Acceleration The rate at which an object’s velocity changes**

Variable: a Units: m/s2 It is the change in velocity which measures the change in position. Thus, it is measuring a change of a change, hence why the square time unit. When the velocity of an object changes at a constant rate, it has constant acceleration

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**Motion Diagrams & Acceleration**

In order for a motion diagram to display a full picture of an object’s movement, it should contain information about acceleration by including average acceleration vectors. The vectors are average acceleration vectors because motion diagrams display the object at equal time INTERVALS (intervals always mean average) Average acceleration vectors are found by subtracting two consecutive velocity vectors.

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**Average Acceleration Vectors**

You will have: Δv = vf - vi = vf + (-vi). Then divide by the time interval, Δt. The time interval, Δt, is 1 s. This vector, (vf - vi)/1 s, shown in violet, is the average acceleration during that time interval.

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**Average Acceleration Vectors**

vi = velocity at the beginning of a chosen time interval vf = velocity at the end of a chosen time interval. ∆v = change in velocity * Acceleration is equal to the change in velocity over the time interval ** Since the time interval is 1s, the acceleration is equal to the change in velocity ***Anything divided by 1 is equal to itself…

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**Average vs. Instantaneous Acceleration**

Average Acceleration Change in velocity during some measurable time interval divided by the time interval Found by plugging into the equation Instantaneous Acceleration Change in velocity at an instant of time Found by calculating the slope of a velocity-time graph at that instant 𝐚= ∆𝒗 𝒕

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**Velocity & Acceleration**

How would you describe the sprinter’s velocity and acceleration as shown on the graph?

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**Velocity & Acceleration**

Sprinter’s velocity starts at zero Velocity increases rapidly for the first four seconds until reaching about 10 m/s Velocity remains almost constant

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**Average vs. Instantaneous Acceleration**

What is the acceleration for the first four seconds? Refers to average acceleration because there is a time interval Solve using the equation a = ∆v /t vi = 0 m/s; vf = 11 m/s; t = 4s a = (11m/s – 0 m/s)/ 4s a = 2.75 m/s2

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**Average vs. Instantaneous Acceleration**

What is the acceleration at 5s? Refers to instantaneous acceleration because it is looking for acceleration at an instant Need to find the slope of the line to solve for acceleration Slope is zero; thus instantaneous acceleration is zero at the instant of 5s.

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**Instantaneous Acceleration**

Solve for the acceleration at 1.0 s Draw a tangent to the curve at t = 1s The slope of the tangent is equal to the instantaneous acceleration at 1s. a = rise / run

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**Instantaneous Acceleration**

The slope of the line at 1.0 s is equal to the acceleration at that instant .

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**Positive & Negative Acceleration**

These four motion diagrams represent the four different possible ways to move along a straight line with constant acceleration.

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**Object is moving in the positive direction**

Displacement is positive Thus, velocity is positive Object is getting faster Acceleration is positive

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**Object is moving in the positive direction**

Displacement is positive Thus, velocity is positive Object is getting slower Acceleration is negative

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**Object is moving in the negative direction**

Displacement is negative Thus, velocity is negative Object is getting faster Acceleration is negative

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**Object is moving in the negative direction**

Displacement is negative Thus, velocity is negative Object is getting slower Acceleration is positive

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**Positive & Negative Acceleration**

When the velocity vector and acceleration vector point in the SAME direction, the object is INCREASING SPEED When the velocity vector and acceleration vector point in the OPPOSITE direction, the object is DECREASING SPEED

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**+ UP - Down Displacement Velocity Acceleration Speeding UP Or**

Displacement & Velocity always have the same sign Displacement Velocity Acceleration Speeding UP Or Slowing Down + UP - Down Up = same Down = Different

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Sample Question How can the instantaneous acceleration of an object with varying acceleration be calculated? A. by calculating the slope of the tangent on a distance versus time graph B. by calculating the area under the graph on a distance versus time graph C. by calculating the area under the graph on a velocity versus time graph D. by calculating the slope of the tangent on a velocity versus time graph

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Practice v-t graph A B C D E

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Segment t (s) vi (m/s) vf (m/s) ∆v avg. a (m/s2) ins. A Xi (m) Xf ∆X A 0 B C D E **Can not assume position on graph. Velocity time graphs can only be used to figure out displacement. You must be given an initial position.

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**3.2 Motion with Constant Acceleration**

Interpret position-time graphs for motion with constant acceleration Determine mathematical relationships among position, velocity, acceleration, and time Apply graphical and mathematical relationships to solve problems related to constant acceleration.

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**Constant acceleration: x-t Graphs**

Velocity is constantly increasing, which means more displacement. Slope must be getting steeper. Results in a curve that is parabolic.

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**Constant acceleration: x-t Graphs**

x (m) t (s) Concave UP = +a t (s) Concave UP = + a x (m) x (m) t (s) Concave Down = -a Concave Down = -a x (m) t (s)

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Kinematics Equations Three equations that relate position, velocity, acceleration, and time. First two are derived from a v-t graph and the third is a substitution. Total of five different variables. Δx (displacement), vi (initial velocity), vf (final velocity), a (acceleration), and t (time). Must know any three in order to solve for the other two.

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**First Kinematics Equation**

Remember that the slope of a v-t graph is the average acceleration. Rearranging the equation, gives us the first kinematics equation. Replace tf – ti with t vf = vi + at

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**Second Kinematics Equation**

We remember that area of a v-t graph equals displacement Break into two known shapes (rectangle & triangle). Total Area = Area of rectangle + area of triangle Δx = vit + ½ (vf –vi)t vf – vi = at (substitute) Δx = vit + ½ at2

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**Third Kinematics Equation**

First equation substituted into the second to cancel out the time variable. vf = vi + at t = (vf – vi) / a Δx = vit + ½ at2 Δx = vi((vf – vi)/a) + ½ a ((vf – vi)/a) 2 Δx = vivf – vi2 + ½ a (vf2 – 2 vivf + vi2 )/a2 2a Δx = 2 vivf - 2 vi2 + vf2 – 2 vivf + vi2 2a Δx = - vi2 + vf2 (rearrange) Simplify Multiply by 2a to get rid of fraction Combine like terms vf2 = vi2 + 2a Δx

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**No displacement (position)**

Kinematics Equation Use when: NO ∆x No displacement (position) No vf No final velocity No t No time vf2 = vi2 + 2a Δx Δx = vit + ½ at2 vf = vi + at Must have 3 of the 5 variables to start. (∆x, vi, vf, a, t) Must always know acceleration. If acceleration is not given, then must solve for it first.

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**Sample Problem Given Unknown Formula Substitution Answer**

An automobile starts at rest and speeds up at 3.5 m/s2 after the traffic light turns green. How far will it have gone when it is traveling at 25 m/s? vi = 0 m/s vf = 25 m/s a = 3.5 m/s2 ∆x = ? m vf2 = vi2 + 2a ∆x (25m/s)2 = (0 m/s)2 + 2 (3.5 m/s2)(∆x ) ∆x = 89.3 m Given Unknown Formula Substitution Answer

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Sample Problem You are driving a car, traveling at constant velocity of 25 m/s, when you see a child suddenly run onto the road. It takes you 0.45 s to react and apply the brakes. As a result, the car slows with a steady acceleration of 8.5 m/s2 and comes to a stop. What is the total distance that the car moves before it stops? Reaction Stopping t =0.45 s v = 25 m/s vi = 25 m/s vf = 0 m/s a = -8.5 m/s2 ∆xreaction = ? m ∆xstopping = ? m 𝑣= ∆𝑥 𝑡 25 𝑚 𝑠 = ∆𝑥 0.45𝑠 (0 m/s)2 = (25 m/s)2 + 2(-8.5m/s2)(∆x) ∆x = m ∆xstopping = m vf2 = vi2 + 2a Δx xtotal = m

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