Unit B01 – Motion in One Dimension [UNAUTHORIZED COPYING OR USE OF ANY PART OF ANY ONE OF THESE SLIDES IS ILLEGAL.]

Section 2-1: Reference Frames and Displacement Explain what “frame of reference” is in terms a junior-high student could understand. A frame of reference is like a point of view. Specifically, it is the point of view of an observer making measurements and describing what is going on.

Section 2-1: Reference Frames and Displacement Explain what “frame of reference” is in terms a junior-high student could understand. Example: Adam (red) says the lightning bolt struck 2 mi north and 2 mi west of him. Betty (blue) says the lightning struck 4 mi south and 3 mi west. Both give a different position for the lightning bolt. Who is right? Both observed the same thing, but made different measurements based on their different frames of reference.

Section 2-1: Reference Frames and Displacement Explain the difference between “distance” and “displacement”. Distance – The length of the path an object travels. Displacement – Distance straight from start to end, with direction specified.

Section 2-1: Reference Frames and Displacement Explain the difference between “distance” and “displacement”. What is Adam’s (red) distance from the lightning strike? What is Adam’s displacement from the lightning strike? What is Betty’s (blue) distance from the lightning strike? What is Betty’s displacement from the lightning strike? What is the lightning’s displacement from Adam? What is Adam’s distance from Betty?

Section 2-1: Reference Frames and Displacement Explain what a vector is. A vector is any measured quantity that has both a magnitude (amount) and a direction. Explain what a scalar is. A scalar is any measured quantity that has only a magnitude (amount) but no meaningful direction.

Section 2-1: Reference Frames and Displacement What letter is the symbol for displacement used in mathematical equations? x Distance and displacement are measured in what units? Meters

Section 2-1: Reference Frames and Displacement Example: Consider the coordinate axes below. Each box represents 1 meter. y x Fred’s position is Point A, which is at (x = 3, y = –3). Mark this point. A

Section 2-1: Reference Frames and Displacement Example: Consider the coordinate axes below. Each box represents 1 meter. Fred walks 5 meters to the left, 5 meters up, and 5 meters to the right, arriving at point B. Mark point B and Fred’s path on the diagram. y x A B A.(2, 2) B.(–2, 2) C.(–2, –3) D.(–3, –2) E.(2, 3) F.(3, 2)

Section 2-1: Reference Frames and Displacement Example: Consider the coordinate axes below. Each box represents 1 meter. What distance did Fred travel? y x A B 15 m What is the distance between points A and B? 5 m What is Fred’s displacement from his original position? 5 m North A.5 meters B.10 meters C.15 meters D.5 meters North E.10 meters West F.15 meters North

Section 2-2: Average Velocity Give a word-equation for average speed. Give a word-equation for average velocity.

Section 2-2: Average Velocity What is the difference between speed and velocity? Speed is a scalar—it only has a magnitude (such as 60 mph). Velocity is a vector—it has magnitude and direction (such as 60 mph north).

Section 2-2: Average Velocity Which is a true statement? (A) Average speed can be greater than average velocity. (B) Average velocity can be greater than average speed. A person travels 10 m east, then 5 m west in 10 seconds. His displacement is 5 m (east), but the distance he traveled is 15 m. The first one. This makes his average velocity (5 m east)/(10 sec) = 0.5 m/s east. His average speed is (15 m)/(10 sec) = 1.5 m/s.

Section 2-2: Average Velocity The symbol for time is t. Time is measured in seconds. The symbol for velocity is v. Velocity is measured in meters/second. “Average” is indicated by putting a straight bar over the symbol v. “Change” is represented by the letter “delta” (  ) “Change” is determined by subtracting:  (value) = (final value) – (initial value)

Section 2-2: Average Velocity Example: Consider the coordinate axes below. Each box represents 1 meter. y x Frank’s position is Point C, which is at (x = 4, y = 4). Mark this point. C

Section 2-2: Average Velocity Example: Consider the coordinate axes below. Each box represents 1 meter. y x Frank walks down at a speed of 2 m/s for 4 seconds. He then immediately walks left 3 m/s for 3 seconds. He waits for 5 seconds, and then walks 1 m/s upward for 8 seconds and stops at point D. Plot point C, point D, and Frank’s path. CD A.(4, –4) B.(–4, 4) C.(–5, –4) D.(–5, 4) E.(5, –4) F.(–4, –4)

Section 2-2: Average Velocity Example: Consider the coordinate axes below. Each box represents 1 meter. y x During this time, what distance did Frank travel? CD 25 m What is the distance between points C and D? 9 m What is Frank’s displacement from his original position? 9 m West A.8 meters B.9 meters C.25 meters D.8 meters North E.9 meters West F.25 meters West

Section 2-2: Average Velocity Example: Consider the coordinate axes below. Each box represents 1 meter. y x What is Frank’s average speed during this time? CD (Avg. Speed) = (Distance)/(Time) (Avg. Speed) = (25)/(20) (Avg. Speed) = 1.25 m/s A.0.45 m/s B.1.25 m/s C.1.67 m/s D.0.45 m/s West E.1.25 m/s West F.1.67 m/s West

Section 2-2: Average Velocity Example: Consider the coordinate axes below. Each box represents 1 meter. y x What is Frank’s average velocity during this time? CD (Avg. Vel.) = (Displacement)/(Time) (Avg. Velocity) = (9)/(20) (Avg. Velocity) = 0.45 m/s West A.0.45 m/s B.1.25 m/s C.1.67 m/s D.0.45 m/s West E.1.25 m/s West F.1.67 m/s West

Section 2-2: Average Velocity Originally, Frank waited for 5 seconds (below point D). If he wanted to make his average speed half as much, how long should he have waited? Explain your reasoning. A.5 seconds B.10 seconds C.15 seconds D.20 seconds E.25 seconds F.30 seconds

Section 2-2: Average Velocity Originally, Frank waited for 5 seconds (below point D). If he wanted to make his average speed half as much, how long should he have waited? Explain your reasoning. …double the denominator.In order to cut this in half… The total time from before was 20 seconds. To cut the original avg. speed in half, the total time needs to be 40 sec. This adds 20 seconds to the original wait time of 5 seconds. New wait time: 25 seconds.

Section 2-2: Average Velocity Would this amount of wait-time have also halved his average velocity? This was doubled……so this is also cut in half. Answer: YES.

Section 2-3: Instantaneous Velocity Explain what “instantaneous velocity” is in the simplest terms possible. Make sure you point out the difference between this and “average velocity”. “Instantaneous velocity” is the rate at which an object moves at a particular point in time. “Average velocity” is the rate at which an object moves over a particular interval of time.

Section 2-3: Instantaneous Velocity What does a speedometer in a car measure? Instantaneous Speed – It tells you your rate of motion right now, but does not give you any information about direction. A.Average Speed B.Average Velocity C.Instantaneous Speed D.Instantaneous Velocity

Section 2-3: Instantaneous Velocity What two instruments in a car can be used to measure the car’s average speed during a long car trip? Explain how to use these instruments to calculate average speed. Odometer (mileage gauge) – Measures the distance of the car trip. Clock– Measures the time of the car trip. Take the total distance traveled and divide by the total time of the trip to get the average speed over a long car trip.

Section 2-4: Acceleration Write a word-equation for acceleration. The symbol for acceleration is a. The units are m/s 2.

Section 2-4: Acceleration Explain where the “square” in the units for acceleration comes from. These units are m/s These units are s Plug in the units of m/s and s to get:

Section 2-4: Acceleration Explain the difference between velocity and acceleration using the words “rate of change of”. Velocity – Rate of change of an object’s position. (Or: the rate at which an object’s position changes.) 6 m/s means that the object gains 6 meters of position every second. Acceleration – Rate of change of an object’s velocity. (Or: the rate at which an object’s velocity changes.) 6 m/s 2 means that the object gains 6 m/s of speed every second.

Section 2-4: Acceleration Acceleration is a vector. Explain how to determine the direction of acceleration. If an object is speeding up, then acceleration points in the same direction as the object’s velocity. If an object is slowing down, then acceleration points in the direction opposite the object’s velocity.

Section 2-4: Acceleration Example: Each diagram shows a car’s position at one second intervals as it travels along a line in the same direction. The solid [white] car is the final position at time = 5 seconds. For each diagram, make a graph of position vs. time, and draw an arrow indicating the direction of velocity and an arrow for acceleration: 0510152025 x t 0 5 10 15 20 25 12345 VelocityAcceleration (A)  (B)  (C)  (D)  (E)  0 (F)  0

Section 2-4: Acceleration Example: Each diagram shows a car’s position at one second intervals as it travels along a line in the same direction. The solid [white] car is the final position at time = 5 seconds. For each diagram, make a graph of position vs. time, and draw an arrow indicating the direction of velocity and an arrow for acceleration: x t 0 5 10 15 20 25 12345 0510152025 VelocityAcceleration (A)  (B)  (C)  (D)  (E)  0 (F)  0

Section 2-5: Motion at Constant Acceleration Write the equation for velocity as a function of time for constant acceleration. Write the equation for average velocity under constant acceleration.

Section 2-5: Motion at Constant Acceleration Write the equation for position as a function of time for constant acceleration. Write the equation that relates velocity, acceleration, and position (the “no time” equation).

Section 2-5: Motion at Constant Acceleration Identify every symbol that you used in the four equations above. x = Position x 0 = Initial Position v = Velocity v 0 = Initial Velocity a = Acceleration t = Time

Section 2-5: Motion at Constant Acceleration

Section 2-6: Solving Problems Example: A runner finishes a 100-meter dash in 8 seconds. Determine his average speed. x = 100 m, t = 8 sec v = ? (100) = v(8) G: U: E: SS: v = 12.5 m/s

Section 2-6: Solving Problems Example: A car can go from zero to 30 m/s in 2 seconds. Determine the car’s average acceleration. v 0 = 0, v = 30 m/s, t = 2 sec a = ? (30) = a(2) G: U: E: SS: a = 15 m/s 2

Section 2-6: Solving Problems Example: Another car can go from zero to 30 m/s in 3 s. How far does the car go while it accelerates? v 0 = 0, v = 30 m/s, t = 3 sec x = ? x = (15)(3) G: U: E: SS: x = 45 m Avg. speed = average of start & end speeds

Section 2-6: Solving Problems Example: The radius of Earth’s orbit is 1.5  10 11 m. How fast does Earth move as it orbits the Sun? r = 1.5  10 11 m v = ? 2  (1.5  10 11 ) = v(31536000) G: U: E: SS: v = 29886 m/s x = circumference = 2  r t = 1 year = 365 days= 31536000 seconds

Section 2-7: Falling Objects What type of motion is free-fall? Constant Acceleration A.Constant Position B.Constant Velocity C.Constant Acceleration D.Changing Acceleration

Section 2-7: Falling Objects What does the  symbol mean in the top paragraph on page 33 that states “d  t 2 ”? How is that different than the = symbol? The  symbols says “is proportional to”. It means that two symbols have the same ratio all the time. “d  t 2 ” means that d/t 2 = (the same number all the time). When accelerating, d = ½at 2, so d/t 2 = ½a (the same number all the time). If d = t 2, then d/t 2 = 1 all the time. For a circle, A =  r 2. This means that A/r 2 =  (the same number all the time), so A  r 2

Section 2-7: Falling Objects Explain Figure 2-17 on page 33. Why does the paper fall differently than the ball in one case, but not in the other? The paper’s surface area contributes to the force of air resistance on it. When the paper is balled up, air resistance is greatly reduced and it falls at the same acceleration as the ball and everything else in free-fall.

Section 2-7: Falling Objects What is the value of the acceleration of gravity on Earth’s surface? What symbol do we give this number? The acceleration of gravity on earth is 9.8 m/s 2. We give this value the special symbol g. When a calculator is not available to us (or sometimes when it is just convenient), we use g = 10 m/s 2.

Section 2-7: Falling Objects Example: A person throws a ball upward near the edge of a building as shown. Ignore air resistance. The first white dot is the position of the ball when it leaves the thrower’s hand. Each white dot after that represents the ball’s position every second after the ball leaves the thrower’s hand. In the table below, the time represents how many seconds after the ball leaves the thrower’s hand. Time (seconds) Position x (meters) Speed (m/s) Velocity (m/s) Accel. (m/s 2 ) 0 1 2 3 4 5 00 –10 –20 –30 10 20 10 20 30 0 15 20 15 0 –25 –10

Section 2-7: Falling Objects Fill in the boxes representing the ball’s velocity at each time. When the ball is at ____________________, we know for sure its velocity is ________________. Each second, the ball’s velocity ______________________________________________________. The speed is almost the same as velocity, except __________________________________________________. Fill in the ball’s acceleration at each time. How does the ball’s acceleration vary with time? ____________. the highest height zero decreases by 10 m/s (has –10 m/s added to it every sec) speed has no direction and is always positive It doesn’t

Section 2-8: Graphical Analysis of Linear Motion How can a curve be considered to have a slope? We look at a point on the curve, and create a line tangent to the curve at that point. The slope of the tangent line is the same as the slope of the curve at that point. y x

Section 2-8: Graphical Analysis of Linear Motion How can velocity at a certain time be found by looking at a position vs. time graph? Velocity is the slope of a position vs. time graph. x t Going slow in the positive direction Stop for an instant Going fast in the positive direction Going fast in the negative direction

Section 2-8: Graphical Analysis of Linear Motion How can the change in position between two times be found from a velocity vs. time graph? On a velocity vs. time graph, the area between the graph and the horizontal axis gives the change in position of an object. v (m/s) t (m/s) 0 1 2 3 4 5 –1–1 –2–2 –3–3 –4–4 –5–5 12345678910 This area (of 12) means the object went 12 m in these 5 sec. This area of –6 means the object went 6 m to the left in these 3 seconds.

Section 2-8: Graphical Analysis of Linear Motion How can acceleration at a certain time be found by looking at a velocity vs. time graph? Acceleration is the slope of a velocity vs. time graph.

Section 2-8: Graphical Analysis of Linear Motion How can the change in velocity between two times be found from an acceleration vs. time graph? On an acceleration vs. time graph, the area between the graph and the horizontal axis gives the change in velocity of an object.

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following x vs. t graph. x (m) t (sec) 5 0 –5 1020 Estimate the velocity at: t = 4 ________ t = 10 ________ t = 12 ________ –2 m/s0 m/s1 m/s

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following x vs. t graph. x (m) t (sec) 5 0 –5 1020 During what time intervals is the velocity positive? ______________________________ negative? ___________________________ zero? ___________________________ 10 to 21 sec 0 to 8 sec 8 to 10 sec

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following x vs. t graph. x (m) t (sec) 5 0 –5 1020 During what time intervals is the acceleration positive? ____________________________ negative? ___________________________ zero? ___________________________ 4 to 8, 10 to 12, 14 to 16 sec 0 to 4, 12 to 14, 19 to 21 sec 8 to 10, 16 to 19 sec

Section 2-8: Graphical Analysis of Linear Motion x (m) t (sec) 5 0 –5 1020 Velocity NegativePositive Acceleration Negative Positive

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following x vs. t graph. x (m) t (sec) 5 0 –5 1020 What is the object’s average velocity between 0 and 10 seconds? _______________ What is the object’s acceleration between 0 and 4 seconds? _______________ –0.8 m/s –1 m/s

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following v vs. t graph. v (m/s) t (sec) 5 0 –5 1020 What is the acceleration at: t = 11 s? __________ t = 15 s? __________ –2 m/s 2 0 m/s 2

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following v vs. t graph. v (m/s) t (sec) 5 0 –5 1020 How far, and in which direction, did the object travel: from t = 4 to t = 11 s? __________ from t = 13 to t = 18 s? __________ 24 m right 16 m left Area = 24 Area = –16

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following v vs. t graph. v (m/s) t (sec) 5 0 –5 1020 During what time intervals is the object speeding up? ______________________________ slowing down? ______________________________ 0 to 4, 11 to 13 seconds 9 to 11, 16 to 18 seconds

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following v vs. t graph. v (m/s) t (sec) 5 0 –5 1020 During what time intervals is the acceleration positive? ______________________________ negative? ____________________________ zero? ____________________________ 0 to 4, 16 to 18 seconds 9 to 13 seconds 5 to 9, 13 to 16 seconds

Section 2-8: Graphical Analysis of Linear Motion Example: Consider the following v vs. t graph. v (m/s) t (sec) 5 0 –5 1020 If the object is at x = –5 m at t = 0 s, where is the object at t = 20 s? __________ Area = 8Area = 20Area = 4 Area = –4 Area = –12 Area = –4 Net Area = 12 +7 m

v (m/s) t (sec) 5 0 –5 1020 a (m/s 2 ) t (sec) 1 0 –1 1020 2 –2 Slope = 1 Value = 1

v (m/s) t (sec) 5 0 –5 1020 a (m/s 2 ) t (sec) 1 0 –1 1020 2 –2 Slope = –2 Value = –2

v (m/s) t (sec) 5 0 –5 1020 x (m) t (sec) 5 0 –5 1020 10 15 20 25 Area = 8  Value = 8 Now x =3

v (m/s) t (sec) 5 0 –5 1020 x (m) t (sec) 5 0 –5 1020 10 15 20 25 Area = 20  Value = 20 Now x = 23

v (m/s) t (sec) 5 0 –5 1020 x (m) t (sec) 5 0 –5 1020 10 15 20 25 Area = 4  Value = 4 Now x = 27

v (m/s) t (sec) 5 0 –5 1020 x (m) t (sec) 5 0 –5 1020 10 15 20 25 Area = –4  Value = –4 Now x = 23

v (m/s) t (sec) 5 0 –5 1020 x (m) t (sec) 5 0 –5 1020 10 15 20 25 Area = 0  Value = 0 Still x = 7

Example: For each pair of graphs, use the given graph to make a sketch of the other graph. Dashed lines are for reference. x t v t SLOPE VALUE SLOPE starts positive and decreases to zero SLOPE goes from zero into the negatives. VALUE starts positive and decreases to zero VALUE goes from zero into the negatives.

Example: For each pair of graphs, use the given graph to make a sketch of the other graph. Dashed lines are for reference. v t a t SLOPE VALUE SLOPE starts negative and goes to zero SLOPE goes from zero into the positives. SLOPE is zero. VALUE starts negative and goes to zero VALUE is zero. VALUE goes from zero into the positives.

Example: For each pair of graphs, use the given graph to make a sketch of the other graph. Dashed lines are for reference. a t v t VALUE SLOPE VALUE starts at zero and goes positive VALUE is a constant positive VALUE goes from positive to zero SLOPE starts at zero and goes positive SLOPE is a constant positive SLOPE goes from positive to zero

Example: For each pair of graphs, use the given graph to make a sketch of the other graph. Dashed lines are for reference. v t x t VALUE SLOPE VALUE starts negative and goes to zero VALUE goes from zero to a maximum positive VALUE goes from zero to negative VALUE goes from maximum positive to zero SLOPE starts negative and goes to zero SLOPE goes from zero to a maximum positive SLOPE goes from maximum positive to zero SLOPE goes from zero to negative

Unit B01 – Motion in One Dimension [UNAUTHORIZED COPYING OR USE OF ANY PART OF ANY ONE OF THESE SLIDES IS ILLEGAL.]

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Unit B01 – Motion in One Dimension [UNAUTHORIZED COPYING OR USE OF ANY PART OF ANY ONE OF THESE SLIDES IS ILLEGAL.]

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