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Kinematics in 1-Dimension

Slide 3 / 113 How to Use this File

· Each topic is composed of brief direct instruction · There are formative assessment questions after every topic denoted by black text and a number in the upper left. > Students work in groups to solve these problems but use student responders to enter their own answers. > Designed for SMART Response PE student response systems. > Use only as many questions as necessary for a sufficient number of students to learn a topic. · Full information on how to teach with NJCTL courses can be found at njctl.org/courses/teaching methods

Slide 4 / 113 Table of Contents

· What is Kinematics?

Click on the topic to go to that section

· Velocity and Speed · Acceleration · Free Fall · Displacement and Distance · Velocity and Position by Integration · Kinematics Equations · Position-Time Graph Interpretation

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What is Kinematics?

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Slide 6 / 113 Terms

Kinematics is the study of the motion of objects. It doesn't care what is making the objects move - that's the topic of the next chapter - Dynamics. Before we can launch into the study of Kinematics, we need to define position, and the mathematical terms scalar and vector.

Slide 7 / 113 Position

To know how an object moves, you need to know its position at all times. You measure position by choosing a reference frame and noting how far and in what direction the object is from a specific point on the reference frame - typically the origin. Most of the time a Cartesian coordinate system works (x, y, z), but you can also use a polar coordinate system (r, θ).

Slide 8 / 113 Position - Time graph and chart

This chapter will focus on 1- Dimensional motion, so an appropriate reference frame will be the x axis (or y axis if the object is moving up and down). Since we need to know its position at all times, we can either use a table, listing its position, or a graph. It is easier to visualize the actual motion using a graph - a Position - Time graph.

Slide 9 / 113 Scalar and Vector

· Scalar - this gives the magnitude of the motion - it has no

• direction. Scalar variables include time and energy -

quantities that have no direction. Note - scalar values may be positive or negative, but they don't indicate direction. · Vector - this is a combination of the magnitude and the direction in which the motion changes. Most physics problems will involve the use of vectors. One more formalism before we get into describing motion. There are two terms that are used to describe where an object is located and how it moves.

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1 At what time is the object's position equal to 6.0 m? A 2.0 s B 2.5 s C 3.0 s D 3.5 s

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1 At what time is the object's position equal to 6.0 m? A 2.0 s B 2.5 s C 3.0 s D 3.5 s

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Answer B

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2 What is the object's position at t = 3.4 s? A 8.0 m B 10 m C 12 m D 14 m

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2 What is the object's position at t = 3.4 s? A 8.0 m B 10 m C 12 m D 14 m

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Answer C

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3 Compare and contrast the properties of scalar and vector quantities.

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3 Compare and contrast the properties of scalar and vector quantities.

Students type their answers here

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Answer Both quantities have a magnitude (size), but a vector has a direction, and a scalar does not have a

• direction. A scalar may be positive
• r negative, and they add

algebraically, but there is no direction implied.

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Displacement and Distance

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Slide 14 / 113 Displacement and Distance

Distance is a measure of how far an object travels without regard to its initial or final position. It is a scalar quantity, and is frequently labeled "d". Displacement is quite different. It measures the difference between an object's final and initial positions. It is a vector and has direction and magnitude and is labeled . Different variables are used in 2 and 3 dimensional problems. Both quantities are measured in meters (m). Displacement ≡

Slide 15 / 113 Displacement and Distance

(Displacement) Motion is in this direction To show the difference between distance and displacement, consider the below sketch. It shows the motion of an object from to . The displacement is in the positive direction (or the east or to the right or any relevant direction description). In this case, the magnitude of the displacement is equal to the distance traveled by the object. The distance is just a number with appropriate units (m, in this case) - it has no direction.

Slide 16 / 113 Displacement and Distance

5

• 5

10 Consider the following case where the magnitude of the displacement will NOT equal the distance traveled. An object starts at zero, moves 9 units to the right (red), then 13 units to the left (green). What is its displacement, ? What is the distance, d, traveled?

Slide 17 / 113 Displacement and Distance

Distance does not have to equal the magnitude of the displacement!

= (-4 - 0) m = -4 m, or 4 m in the negative direction

• r 4 m to the left or 4 m West - depending on how the problem

is asked - please check with your instructor for his/her preference (represented by the blue line). d = (9 + 13) m = 22 m (represented by the sum of the red and green lines, without regard to sign, and with no direction - the total distance traveled). 5

• 5

10

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4 How far your final position is from your initial position is known as: A distance B displacement C scalar D vector

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4 How far your final position is from your initial position is known as: A distance B displacement C scalar D vector

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Answer B

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5 How far you physically travel is called: A distance B displacement C scalar D vector

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5 How far you physically travel is called: A distance B displacement C scalar D vector

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Answer A

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6 A car travels 5500 m to the north, and then 3200 m to the

• south. What distance was traveled by the car?

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6 A car travels 5500 m to the north, and then 3200 m to the

• south. What distance was traveled by the car?

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Answer 8700 m

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7 A car travels 5500 m to the north, and then 3200 m to the

• south. What was the displacement of the car?

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7 A car travels 5500 m to the north, and then 3200 m to the

• south. What was the displacement of the car?

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Answer 2300 m to the north

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8 You run around a 400 m track. At the end of the race, what is the distance that you traveled?

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8 You run around a 400 m track. At the end of the race, what is the distance that you traveled?

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Answer 400 m

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9 You run around a 400 m track. At the end of the race, what is your displacement?

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9 You run around a 400 m track. At the end of the race, what is your displacement?

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Answer 0 m

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Velocity and Speed

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Slide 25 / 113 Average Velocity and Speed

The next quantities to be defined are average velocity and average speed, and they represent, respectively, the change in displacement and the change in distance over time. Average velocity (x direction): Average speed: Time is a scalar. So what does that tell you about average velocity and average speed? Velocity and speed are used in everyday language, but they have distinct definitions in physics.

Slide 26 / 113 Average Velocity and Speed

Since displacement is a vector, when you divide it by a scalar, the result is a vector; average velocity is a vector. Since distance is a scalar, when you divide it by a scalar, the result is a scalar; average speed is a scalar. Speed can, but does not have to, equal the magnitude of velocity. Velocity is more useful in Physics.

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10 One object has a velocity of 5.0 m/s. Another object has a velocity of -5.0 m/s. Do they have the same speed? Yes No

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10 One object has a velocity of 5.0 m/s. Another object has a velocity of -5.0 m/s. Do they have the same speed? Yes No

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Answer Yes

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11 A particle is at position, x0 = 19 m at t0 = 1.0 s, and is at position, xf = 277 m at t = 4.0 s. Find the average velocity and the average speed.

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11 A particle is at position, x0 = 19 m at t0 = 1.0 s, and is at position, xf = 277 m at t = 4.0 s. Find the average velocity and the average speed.

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Answer in the positive x direction In this case, the speed is equal to the magnitude of the velocity.

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12 A particle is at position, x0 = 19 m at t0 = 1.0 s, and is at position, xf = -277 m at t = 4.0 s. Find the average velocity and the average speed.

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13 A particle is at position, x0 = 0 m at t0 = 1.0 s, and is at position, xf = -277 m at t = 4.0 s. Find the average velocity and the average speed.

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13 A particle is at position, x0 = 0 m at t0 = 1.0 s, and is at position, xf = -277 m at t = 4.0 s. Find the average velocity and the average speed.

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Answer in the negative x direction In this case, the speed is equal to the magnitude of the velocity.

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14 A particle is at position, x0 = 19 m at t0 = 1.0 s, moves to position x1 = 184 m, then to position x2 = -34 m and comes to rest at xf = 19 m at tf = 6.1 s. Find the average velocity and the average speed.

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14 A particle is at position, x0 = 19 m at t0 = 1.0 s, moves to position x1 = 184 m, then to position x2 = -34 m and comes to rest at xf = 19 m at tf = 6.1 s. Find the average velocity and the average speed.

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Answer In this case, the speed is not equal to the magnitude of the velocity.

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15 What is the average velocity of a particle, whose motion is described by the below position time graph, from t = 0 s to t = 4 s?

t (s) x (m)

time position Position-Time graph

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15 What is the average velocity of a particle, whose motion is described by the below position time graph, from t = 0 s to t = 4 s?

t (s) x (m)

time position Position-Time graph [This object is a pull tab]

Answer

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16 What is the average velocity of a particle, whose motion is described by the below position time graph, from t = 2 s to t = 4 s?

t (s) x (m)

time position Position-Time graph

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16 What is the average velocity of a particle, whose motion is described by the below position time graph, from t = 2 s to t = 4 s?

t (s) x (m)

time position Position-Time graph [This object is a pull tab]

Answer

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17 Explain, using the definition of slope, why the average velocity from t = 0 s to t =2 s was less than the average velocity from t = 2 s to t = 4 s.

x (m)

Position-Time graph time position

t (s)

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17 Explain, using the definition of slope, why the average velocity from t = 0 s to t =2 s was less than the average velocity from t = 2 s to t = 4 s.

x (m)

Position-Time graph time position

t (s)

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Answer The slope of the line connecting the particle's position change between 0 and 2 s was less than the slope of the line connecting the particle's position change from 2 to 4 s. The position change was greater in the second interval for the same period of

• time. Hence, the average velocity

was greater for the 2 to 4 s interval.

Slide 39 / 113 Instantaneous Velocity

Average velocity is used to define a particle's motion over a specified interval of time, but that is not always what is most

• interesting. For example, a police officer is not interested in your

average velocity over a two hour period. But, he is very interested in your velocity at a specific given time. This is more fundamental then it seems - the officer, even though he probably doesn't realize it, actually wants your velocity at a time interval of zero seconds! The officer is looking for your Instantaneous Velocity to ensure you're obeying the speed limit.

Slide 40 / 113 Instantaneous Velocity

Let's see how the average velocity equation works for a zero second time interval. It's problematic because algebra doesn't handle fractions with a zero in the denominator. Sir Isaac Newton recognized this problem and came up with the concept of a zero time interval and invented calculus so he could solve problems like this. Calculus involves taking a time interval and shrinking it until it is infinitesimally close to zero. You'll learn more about this in your Calculus class - but you can tell the teacher you first saw it here. Calculus was invented for Physics. Δt = 0 is not good

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Slide 44 / 113 Slide 45 / 113 Instantaneous Velocity

Switching to algebra and a bit of calculus, the average velocity equation was used as Δt kept decreasing until it almost hit zero, at which point the instantaneous velocity of the particle was found. We call that a "limit," as follows: Instantaneous velocity is labeled vx, the limit was discussed above, and the dx/dt term is called the "derivative of x with respect to t." The derivative of a graphed function is also the slope of the tangent line to the function at that point. The derivative of the position is the velocity. Your calculus course will cover this in more depth.

Slide 46 / 113 Instantaneous Velocity Notation Convention

Since we assume that the time interval Δt is positive, vx has the same algebraic sign as Δx. For example, a positive vx means that x is increasing and the motion is toward the right. Whereas a negative v x means x is decreasing and the motion is toward the left. Instantaneous velocity is a vector quantity like average velocity. To show instantaneous velocity along the x-axis, use v x.

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18 Explain how instantaneous velocity is derived by starting with average velocity. Use algebra, graphical analysis and a little bit of calculus in your response.

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18 Explain how instantaneous velocity is derived by starting with average velocity. Use algebra, graphical analysis and a little bit of calculus in your response.

Students type their answers here

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Answer Use the definition for average velocity, but then keep shrinking the time interval; this will also decrease the change in position. When the time interval is close to zero, average velocity is now the instantaneous

• velocity. The slope of the tangent line

at the point of interest is the instantaneous velocity.

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20 What is the instantaneous velocity of the particle whose motion is defined by the position-time graph at point P1 = (1,1)?

tangent line to (1,1) motion curve

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20 What is the instantaneous velocity of the particle whose motion is defined by the position-time graph at point P1 = (1,1)?

tangent line to (1,1) motion curve

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Answer The slope of the tangent line to the motion curve is the instantaneous velocity.

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21 If the position of a particle as a function of time is x(t) = 2t3 + 4t2 + t + 18, what is its velocity at t = 3 s?

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21 If the position of a particle as a function of time is x(t) = 2t3 + 4t2 + t + 18, what is its velocity at t = 3 s?

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Answer

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22 If the position of a particle as a function of time is x(t) = 2t4 + 6t2 + 8t + 7 what is its velocity at t = 4 s?

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22 If the position of a particle as a function of time is x(t) = 2t4 + 6t2 + 8t + 7 what is its velocity at t = 4 s?

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Answer

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23 If the position of a particle as a function of time is x(t) = 4t2, what can you say about its velocity? A The velocity is a non zero constant. B The velocity increases linearly with time. C The velocity increases proportional to the square of the elapsed time. D The velocity is in the negative direction.

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23 If the position of a particle as a function of time is x(t) = 4t2, what can you say about its velocity? A The velocity is a non zero constant. B The velocity increases linearly with time. C The velocity increases proportional to the square of the elapsed time. D The velocity is in the negative direction.

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Answer B

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Acceleration

Slide 54 / 113 Acceleration

Velocity was defined as the change in position of an object over time. Velocity can also change over time. This quantity is called Acceleration. Like velocity, acceleration is a vector quantity.

Slide 55 / 113 Average Acceleration

Suppose the velocity of a particle changes from t1 to t2. The average acceleration can be found from the change in velocity Δvx divided by time interval Δt. Average acceleration is defined as: Looking at the equation, we see that the units of acceleration are: Acceleration is measured in m/s2.

Slide 56 / 113 Average Acceleration

If a particle has an acceleration of 3 m/s2, then for every second it travels, it increases its velocity by 3 m/s. The units on either side of the above equality are mathematically

• equivalent. But the left side more clearly reflects what is

physically happening. The right side is easier to write, so that's what we'll use going forward.

Slide 57 / 113 Instantaneous Acceleration

We can define instantaneous acceleration using the same procedure to find instantaneous velocity. The instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero. In this course, unless otherwise stated, acceleration will be assumed to be constant, so average acceleration will equal instantaneous acceleration. From your previous physics knowledge, what common force produces a constant acceleration?

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24 A horse gallops with a constant acceleration of 3 m/s2. Which of the following statements is true? A The horse's velocity remains constant. B The horse's position changes by 3 m every second. C The horse's velocity increases 3 m every second. D The horse's velocity increases 3 m/s every second.

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24 A horse gallops with a constant acceleration of 3 m/s2. Which of the following statements is true? A The horse's velocity remains constant. B The horse's position changes by 3 m every second. C The horse's velocity increases 3 m every second. D The horse's velocity increases 3 m/s every second.

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Answer D

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25 You are in a racing car and your velocity changes from 60 m/s to the right to 100 m/s to the right in 20 s. What is the average acceleration?

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25 You are in a racing car and your velocity changes from 60 m/s to the right to 100 m/s to the right in 20 s. What is the average acceleration?

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Answer

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26 You are in a racing car and your velocity changes from 60 m/s to the right to 20 m/s to the right in 20 s. What is the average acceleration?

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26 You are in a racing car and your velocity changes from 60 m/s to the right to 20 m/s to the right in 20 s. What is the average acceleration?

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Answer

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27 You are in a racing car and your velocity changes from 50 m/s to the left to 10 m/s to the right in 15 s. What is the average acceleration?

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27 You are in a racing car and your velocity changes from 50 m/s to the left to 10 m/s to the right in 15 s. What is the average acceleration?

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Answer

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28 You are in a racing car and your velocity changes from 90 m/s to the right to 20 m/s to the right in 5 s. What is the average acceleration?

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28 You are in a racing car and your velocity changes from 90 m/s to the right to 20 m/s to the right in 5 s. What is the average acceleration?

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Answer

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29 If the velocity of a particle as a function of time is v(t) = t3 + 2t2 + 6t + 8, what is its acceleration at t = 5 s?

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29 If the velocity of a particle as a function of time is v(t) = t3 + 2t2 + 6t + 8, what is its acceleration at t = 5 s?

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Answer

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30 If the velocity of a particle as a function of time is v(t) = t3 - 2t2 - 6t + 9, what is its acceleration at t = 2 s?

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30 If the velocity of a particle as a function of time is v(t) = t3 - 2t2 - 6t + 9, what is its acceleration at t = 2 s?

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Answer

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32 Using the below velocity-time graph, find the instantaneous acceleration of the object at t = 2 s. Is this equal to the average acceleration of the object between 0 and 2 s?

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32 Using the below velocity-time graph, find the instantaneous acceleration of the object at t = 2 s. Is this equal to the average acceleration of the object between 0 and 2 s?

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Answer

Find the slope of the tangent line at t = 2s Yes, they are equal. The slope of the line connecting the points at t = 0 s and t = 2 s is the slope of the tangent line at t = 2 s. This is a case of constant acceleration.

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33 Using the below velocity-time graph, find the instantaneous acceleration of the object at t = 2 s.

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33 Using the below velocity-time graph, find the instantaneous acceleration of the object at t = 2 s.

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Answer The acceleration is 0 m/s2. The tangent to the velocity- position curve is horizontal, with zero slope. This is true for any time, hence this is a case of zero acceleration.

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Free Fall

Slide 72 / 113 Freely Falling Bodies

Freely falling bodies are those that are subject only to the gravitational force, and have a constant acceleration, regardless

• f mass. (Fg = mg = ma, thus a = g).

Aristotle believed that heavier objects fell faster than light

• bjects depending on their weight.

Galileo later disagreed with Aristotle's belief, and experimented to find the truth. His results found that the motion was independent of weight. The strength of gravity is constant as long as the fall height is considerably less than the radius of the planet.

Slide 73 / 113 Freely Falling Bodies

The picture to the left, shows the affect of gravity on a falling body. The basketball travels much faster per time interval as the ball accelerates due to gravity. The constant acceleration of a freely falling body is called the acceleration due to gravity. Its magnitude is represented by the letter g. The approximate value of g is 9.8 m/s2. Since the value of g is dependent on the planet, the moon and the sun have different values of g. Click on this paragraph to see a great illustration

• f the effect of g and friction on falling objects.

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34 Galileo proved that objects of different masses fall with the same accleration. You want to demonstrate this in class, and drop a sheet of paper at the same time as a

• ball. You observe that the ball hit the ground first. What

happened? Is Galileo wrong? How would you modify the experiment to show Galileo was really right?

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34 Galileo proved that objects of different masses fall with the same accleration. You want to demonstrate this in class, and drop a sheet of paper at the same time as a

• ball. You observe that the ball hit the ground first. What

happened? Is Galileo wrong? How would you modify the experiment to show Galileo was really right?

Students type their answers here

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Answer The problem is that in addition to gravity, there is another force - air

• friction. The larger surface area of

the sheet of paper exposes it to a greater frictional force, so its acceleration is less than the ball. Reduce the air friction on the paper by crumpling it up, and repeat the experiment.

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35 You throw a ball up in the air. Which of the following describe the motion of the ball? A The acceleration on the way up is less than the acceleration

• n the way down.

B The acceleration on the way up is greater than the acceleration on the way down. C The acceleration on the way up is the same as the acceleration on the way down. D The magnitude of the velocity increases on the way up.

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35 You throw a ball up in the air. Which of the following describe the motion of the ball? A The acceleration on the way up is less than the acceleration

• n the way down.

B The acceleration on the way up is greater than the acceleration on the way down. C The acceleration on the way up is the same as the acceleration on the way down. D The magnitude of the velocity increases on the way up.

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Answer C

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Kinematics Equations

Slide 77 / 113 Kinematics Equations Derivation

There are three main equations that will enable us to solve kinematics problems, assuming a constant acceleration. We'll start with three equations: Definition of acceleration: Average velocity (click here to see where this was shown):

Slide 78 / 113 First Kinematics Equation Derivation

Then make t1 = 0 and t2 be any later time t.

• r

First Kinematics Equation

Slide 79 / 113 Second Kinematics Equation Derivation

Substitute the vx equation into the vx,avg equation: After some algebra, we can arrive at position: Second Kinematics Equation Set this equal to the other vx,avg equation

Slide 80 / 113 Slide 81 / 113 Kinematics Equations Summary

Second Kinematics Equation Third Kinematics Equation First Kinematics Equation

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36 A car and a truck start from rest and at the same position. They then both accelerate at the same rate. The car accelerates for twice the time that the truck accelerates. What is the final velocity of the car compared to the truck? A Half as much B The same C Twice as much D Four times as much

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36 A car and a truck start from rest and at the same position. They then both accelerate at the same rate. The car accelerates for twice the time that the truck accelerates. What is the final velocity of the car compared to the truck? A Half as much B The same C Twice as much D Four times as much

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Answer C

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37 A car and a truck start from rest and at the same position. They then both accelerate at the same rate.The car accelerates for twice the time that the truck accelerates. What is the final position of the car compared to the truck? A Half as far B The same C Twice as far D Four times as far

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37 A car and a truck start from rest and at the same position. They then both accelerate at the same rate.The car accelerates for twice the time that the truck accelerates. What is the final position of the car compared to the truck? A Half as far B The same C Twice as far D Four times as far

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Answer D

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38 Two cars start from rest and accelerate at the same rate. The second car accelerates for four times the distance that the first car accelerates. What is the velocity of the second car compared to the first car after that acceleration? A The same B Twice as fast C Four times as fast D Eight times as fast

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38 Two cars start from rest and accelerate at the same rate. The second car accelerates for four times the distance that the first car accelerates. What is the velocity of the second car compared to the first car after that acceleration? A The same B Twice as fast C Four times as fast D Eight times as fast

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Answer B

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Velocity and Position by Integration

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Slide 88 (Answer) / 113 Slide 89 / 113 Slide 89 (Answer) / 113 Slide 90 / 113 Velocity and Position by Integration

Without constant acceleration, many of the previously derived equations cannot be used (the three Kinematics Equations were all based on constant acceleration). Calculus needs to be used to find velocity and position. Previously, calculus was used to go from x(t) to v(t) to a(t) by taking derivatives: Used when position x is a known function of time Used when velocity v

x is a known function of time

Slide 91 / 113 Velocity and Position by Integration

What if we had acceleration as a function of time and wanted to find velocity and position? We want to go from a(t) to v(t) to x(t). We're looking for an anti-derivative! Really, that's the name. Another name for that is the integral. And it involves summing the area underneath a curve (as was done a few slides ago). When acceleration is constant or a linear function, it's relatively easy - you have a geometrical shape such as a triangle or rectangle. But what if acceleration is not constant or linear?

Slide 92 / 113 Velocity and Position by Integration

Let's begin with a look at the graph of a particle with non-constant and non-linear acceleration. When acceleration was constant, we found the change in velocity by finding the area of a rectangle - that was the shape under the curve describing a(t). If you needed to find the area under the curve to the left from 0 to 3 s, is there a simple equation to accomplish that?

Slide 93 / 113 Velocity and Position by Integration

• No. You can't even build

that shape from adding a triangle to a rectangle. It's a curve. But, we know the area of rectangles - just base times

• height. That's going to be

the trick we use. How? See if you can come up with a proposal before going to the next page.

Slide 94 / 113 Slide 95 / 113 Slide 96 / 113 Slide 97 / 113 Velocity and Position by Integration

As the Δt intervals decrease in size and increase in number, the average acceleration approaches the instantaneous acceleration at a specific time. When this limit is reached, the area under the curve is the integral of ax(t) over time, or the change in velocity. an(t) is the value of ax(t) for each little rectangle when it intersects the acceleration curve. The formal equation for this area is represented as shown - and this equals the change in velocity for the time interval t1x to t2x. Fun observation - the integral represents an infinite SUM, and the integral sign looks like a stretched out S.

Slide 98 / 113 Velocity and Position by Integration

We can use the same procedure to find total displacement if we know the velocity as a function of time. Here's the result:

Slide 99 / 113

This equation can be used if acceleration ax as a function of t is known. This equation can be used if velocity vx as a function of t is known.

Velocity and Position by Integration

These equations can be rewritten for an initial position of x0, and an initial velocity of v0x, and t1 = 0, which unless other specified, may be assumed.

Slide 100 / 113

Position Velocity Acceleration

Acceleration, Velocity, and Position by Integration and Differentiation Slide 101 / 113

41 An object starts at an initial position of x0 = 0 m. Its velocity as a function of time is v(t) = 3t2 + 2t + 6. What is its position at t = 4 s?

Slide 101 (Answer) / 113

41 An object starts at an initial position of x0 = 0 m. Its velocity as a function of time is v(t) = 3t2 + 2t + 6. What is its position at t = 4 s?

[This object is a pull tab]

Answer

Slide 102 / 113

42 A particle starts at rest at x0 = 0 m. If its acceleration as a function of time is a(t) = 12t + 6, what is the velocity and position of the particle at t = 3 s?

Slide 102 (Answer) / 113

42 A particle starts at rest at x0 = 0 m. If its acceleration as a function of time is a(t) = 12t + 6, what is the velocity and position of the particle at t = 3 s?

[This object is a pull tab]

Answer

Slide 103 / 113

Return to Table

• f Contents

Position-Time Graph Interpretation

Slide 104 / 113 Slide 105 / 113 Second Derivative of Position

Since and Acceleration equals the second derivative of x with respect to t: The second derivative is relative to the concavity (curvature)

• f a graph. Your calculus course will cover this in more detail, but it

will be summarized on the next slide.

Slide 106 / 113 Second Derivative of Position

The slope of the curve on a position time graph gives us the velocity (positive slope = positive velocity, negative slope = negative velocity). The concavity gives us information about the acceleration: Concave up - acceleration greater than zero Concave down- acceleration less than zero Zero curvature - acceleration equals zero

Slide 107 / 113 Slide 108 / 113

43 At which points is the acceleration of the object 0 m/s2? A A and B B A and D C C and E D C and D

A B C D E

t x

Slide 108 (Answer) / 113

43 At which points is the acceleration of the object 0 m/s2? A A and B B A and D C C and E D C and D

A B C D E

t x

[This object is a pull tab]

Answer B

Slide 109 / 113

44 At which points is the velocity of the object 0 m/s2? A A and B B A and D C C and E D C and D

A B C D E

t x

Slide 109 (Answer) / 113

44 At which points is the velocity of the object 0 m/s2? A A and B B A and D C C and E D C and D

A B C D E

t x

[This object is a pull tab]

Answer C

Slide 110 / 113

45 Describe the motion of the object at point B completely - include its position, velocity and acceleration.

Students type their answers here

A B C D E

t x

Slide 110 (Answer) / 113

45 Describe the motion of the object at point B completely - include its position, velocity and acceleration.

Students type their answers here

A B C D E

t x

[This object is a pull tab]

Answer The object is moving in the negative direction, but it is still at a positive displacement. The velociy is negative and decreasing. The acceleration is positive.

Slide 111 / 113

46 Describe the motion of the object at point C completely - include its position, velocity and acceleration.

Students type their answers here

A B C D E

t x

Slide 111 (Answer) / 113

46 Describe the motion of the object at point C completely - include its position, velocity and acceleration.

Students type their answers here

A B C D E

t x

[This object is a pull tab]

Answer The object is momentarily at rest, stopping its motion in the negative direction and will start moving in the positive direction. The position is still negative with reference to its position at t = 0 s. Its velocity is 0 m/s, but is changing from a negative to a positive velocity. Its acceleration is positive.

Slide 112 / 113

47 Describe the motion of the object at point D completely - include its position, velocity and acceleration.

Students type their answers here

A B C D E

t x

Slide 112 (Answer) / 113

47 Describe the motion of the object at point D completely - include its position, velocity and acceleration.

Students type their answers here

A B C D E

t x

[This object is a pull tab]

Answer The object is moving in the negative direction, but it is still at a positive displacement. The velociy is negative and constant. The acceleration is 0 m/s2.

Slide 113 / 113

48 A student in a physics lab uses a battery powered lab car that has a pen attached that marks its position on a roll of tape as it moves. The pen makes a mark every one

• second. The results are tabulated below. Is the car

moving with a constant acceleration? Explain what technique you used to answer the question.

Students type their answers here

t(s) x(m) 1 0.5 2 2.0 3 4.5 4 8.0 5 12.5

Answer