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Ch 02: Motion Along a Straight Line

Chapter 2, Problem 2

A cat walks in a straight line, which we shall call the x-axis, with the positive direction to the right. As an observant physicist, you make measurements of this cat's motion and construct a graph of the feline's velocity as a function of time (Fig. E2.30). (d) Assuming that the cat started at the origin, sketch clear graphs of the cat's acceleration and position as functions of time.

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Hey everyone today, we're being told that a physicist has measured a dog's path of motion in a straight line and uses a coordinate system to where motion is parallel to the X axis and the right is the positive X direction. Now, with these measurements, the physicist has plotted the dogs motion as a graph of velocity as a function of time are being asked to sketch a graph that represents both the acceleration and position as functions of time of the dog, assuming that the dog starts his motion at the origin first of all, let's start with the acceleration first. And we can recall that acceleration on a velocity versus time graph is nothing but the slope because the slope is defined as the change in y over the change in X, which in this case is the change in velocity over the change in time, which is the definition of acceleration. So our graph of velocity versus time is a straight line, which means it has the same slope throughout, which therefore means that a philosophy or that the acceleration must be constant since the velocity itself remains constant as well or not constant. But the velocity versus time graph follows a straight line. So writing that out and taking the triangle formed here to calculate price overrun the change in why? Or the change in velocity goes from negative 5 to 0. So it will be zero minus Now you have five and change in the X direction and time will be 3. 0 0. And this is the change in time or a change of velocity over change in time, which will give us 5/3 0.5 which equals 1.43 m per second. So this is the value of our acceleration and we remember we stated that acceleration is constant, which means if we are to draw this out as a graph, this as zero with that as the oops, that is the origin. Let us say this is the .43 and this is this is acceleration in meters per second squared and this is time in seconds. Then our graph will look like this Because our acceleration is constant at 1.4, m/s. No, taking a look at the uh position. Well again recalling some of our rules. Excuse me, We know that the area under a graph to the actual area. The area under a graph is the distance that is traveled by whatever is being or whatever is being observed. So we know a few things. The velocity starts velocity starts At a very negative velocity and that zero At 3.5 seconds, which means if we were to write this out and let me let me scroll down a bit. So we have a little more space for us to try this out. Oops, so you get a little straighter, making it a little nicer. Okay, cool, so this is our uh, this is our change in position which will denote with X in mirrors and this is our change in time seconds. So we have a few key points at 3.5 seconds right here, it's when uh the negative velocities which is to positive velocity and that happens Uh that happens as velocity approaches zero. So what does that mean? Well, negative velocity in this case means moving away from the origin. Right. And we've determined that moving away from the origin is in the negative X direction. Right? And or like negative access in negative uh distance, we're moving away from the origin in the negative value. So considering the first half of this path of travel From time 0-3.5 seconds and we'll write this as 3.5 seconds, the dog will start at the origin and increase in velocity Right? We're going from negative 5-0. So its velocity will increase. But its position when we're moving away from should be moving a way from the origin. And remember on a position versus time graph, the velocity Is represented as the slope of the line at that point. So at 3.5 seconds, the velocity which was negative all the way up until there Which is slowly becoming more and more more or less and less uh large and less in magnitude. If we consider the slope at each point will finally reach a plateau at 3.5 seconds, it'll be zero. We'll have zero velocity after that point, the velocity continues in the positive direction, which means the dog would actually be going in the opposite direction. Now the velocity is in the opposite direction and it continuously increases, which means the slope will also increase as it goes on. So that would be the graph for both acceleration and position as functions of time for the path of motion of the dog. I hope this helps, and I look forward to seeing you all in the next one.
Related Practice
Textbook Question
A cat walks in a straight line, which we shall call the x-axis, with the positive direction to the right. As an observant physicist, you make measurements of this cat's motion and construct a graph of the feline's velocity as a function of time (Fig. E2.30). (a) Find the cat's velocity at t = 4.0 s and at t = 7.0 s.

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Textbook Question
A cat walks in a straight line, which we shall call the x-axis, with the positive direction to the right. As an observant physicist, you make measurements of this cat's motion and construct a graph of the feline's velocity as a function of time (Fig. E2.30). (b) What is the cat's acceleration at t = 3.0 s? At t = 6.0 s? At t = 7.0 s?

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Textbook Question
A cat walks in a straight line, which we shall call the x-axis, with the positive direction to the right. As an observant physicist, you make measurements of this cat's motion and construct a graph of the feline's velocity as a function of time (Fig. E2.30). (c) What distance does the cat move during the first 4.5 s? From t = 0 to t = 7.5 s?

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Textbook Question
The Fastest (and Most Expensive) Car! The table shows test data for the Bugatti Veyron Super Sport, the fastest street car made. The car is moving in a straight line (the x-axis). (a) Sketch a vx–t graph of this car's velocity (in mi/h) as a function of time. Is its acceleration constant? (b) Calculate the car's average acceleration (in m/s2) between (i) 0 and 2.1 s; (ii) 2.1 s and 20.0 s; (iii) 20.0 s and 53 s. Are these results consistent with your graph in part (a)? (Before you decide to buy this car, it might be helpful to know that only 300 will be built, it runs out of gas in 12 minutes at top speed, and it costs more than $1.5 million!)

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Textbook Question
High-speed motion pictures (3500 frames/second) of a jumping, 210–μg flea yielded the data used to plot the graph in Fig. E2.54. (See 'The Flying Leap of the Flea' by M. Rothschild, Y. Schlein, K. Parker, C. Neville, and S. Sternberg in the November 1973 Scientific American.) This flea was about 2 mm long and jumped at a nearly vertical takeoff angle. Use the graph to answer these questions: (a) Is the acceleration of the flea ever zero? If so, when? Justify your answer.

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Textbook Question
High-speed motion pictures (3500 frames/second) of a jumping, 210–μg flea yielded the data used to plot the graph in Fig. E2.54. (See 'The Flying Leap of the Flea' by M. Rothschild, Y. Schlein, K. Parker, C. Neville, and S. Sternberg in the November 1973 Scientific American.) This flea was about 2 mm long and jumped at a nearly vertical takeoff angle. Use the graph to answer these questions: (b) Find the maximum height the flea reached in the first 2.5 ms.

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