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Which position vs. time graph shown below represents the motion of an object that is moving in the negative direction and slowing down? Warmup Question

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Narrow it down! “Moving in the negative direction” – x vs t graph must have a negative slope

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“moving in the negative direction and slowing down” “slowing down” – Velocity and acceleration must be opposite directions – Since velocity is negative, acceleration is positive – Graph must have a positive concavity

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Also, try to use the graph to visualize what the object is doing along the x-axis as time ticks by As time progresses, the object travels less and less distance each second. Slope approaches zero (object steadily comes to a stop)

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Meet the velocity vs time graph t (s) v (m/s) cheerio

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The velocity vs time graph is tricky This is because is makes things even more abstract! It is actually graphing the object’s velocity as time ticks by!

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Very useful tip: Actually say what the graph represents in words “The object starts out with a velocity of zero, and its velocity steadily becomes more and more positive”

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Try it! “The object starts off with a negative velocity, and its velocity steadily becomes closer to zero until finally it has reached zero velocity”

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Seriously, putting it into words really helps to understand what the graph is saying “The object starts off with a positive velocity. Its velocity steadily decreases until it has reached zero. The object’s velocity continues to steadily become more and more negative, and it ends up with a negative final velocity.”

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Question from 2006 AP Exam 8) What is happening to the car at t = 1 second? (A) It is speeding up. (B) It is slowing down. (C) It is slowing down before t = 1 s and speeding up after t = 1 s. (D) It is speeding up before t = 1 s and slowing down after t = 1 s. (E) It is turning around.

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Question from 2006 AP Exam 8) What is happening to the car at t = 1 second? (A) It is speeding up. (B) It is slowing down. (C) It is slowing down before t = 1 s and speeding up after t = 1 s. (D) It is speeding up before t = 1 s and slowing down after t = 1 s. (E) It is turning around.

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The slope of the v vs t graph represents the object’s acceleration A car was caught in traffic. After 20 s of moving at constant speed, traffic cleared a bit, allowing the car to speed up. The car’s motion is represented by the velocity-time graph below. What was the car’s acceleration while it was speeding up? (A) 0.5 m/s 2 (B) 1.0 m/s 2 (C) 1.5 m/s 2 (D) 2.0 m/s 2 (E) 3.0 m/s 2 a = (v f – v i ) / (t f – t i )

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The slope of the v vs t graph represents the object’s acceleration A car was caught in traffic. After 20 s of moving at constant speed, traffic cleared a bit, allowing the car to speed up. The car’s motion is represented by the velocity-time graph below. What was the car’s acceleration while it was speeding up? (A) 0.5 m/s 2 (B) 1.0 m/s 2 (C) 1.5 m/s 2 (D) 2.0 m/s 2 (E) 3.0 m/s 2 a = (v f – v i ) / (t f – t i )

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The total area of the v vs t graph represents the object’s displacement. In this case, Δx = ½ b*h Δx = ½ t f *v f

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Remember, the area of a graph can be positive or negative Positive area (above axis) Negative area (below axis) In this case, displacement = total area = positive area + negative area (looks like ≈ 0 m)

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The graph above shows an object’s velocity as a function of time. At what time after t = 0 s can the object be found back at its initial position? (A) Between 0 and 1 s (B) 1 s (C) Between 1 and 2 s (D) 2 s (E) Between 2 and 3 s

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The graph above shows an object’s velocity as a function of time. At what time after t = 0 s can the object be found back at its initial position? (A) Between 0 and 1 s (B) 1 s (C) Between 1 and 2 s (D) 2 s (E) Between 2 and 3 s “back at its initial position” Δx = 0 When does the total area = 0?

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Meet the acceleration vs time graph

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The things that you need to know how to do with the a vs t graph Use it to describe the object’s acceleration. Take the area of it to determine the change in the object’s velocity. Use it to determine the slope of a v vs t graph Use it to determine when there is a net force exerted on an object/how much force

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SPH3U Exam Review. 1. The slope of a position-time (i.e. displacement-time) graph is equal to the: A. acceleration B. distance travelled C. time interval.

SPH3U Exam Review. 1. The slope of a position-time (i.e. displacement-time) graph is equal to the: A. acceleration B. distance travelled C. time interval.

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