Motion diagrams and position-time graphs: an example

key idea: for a motion in one direction, the slope of a position-time graph tells you the velocity of the motion

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Thus, for a motion with constant velocity, the position-time graph is a straight line.

Note that the actual motion takes place on the road, which is represented in a fairly immediate way by the motion diagram. The position-time graph is not a picture of the path of the object, but is rather a more abstract (but very useful) representation of the motion.

Examples of position-time graphs:

Remember that in the following position-time graphs, the actual motion occurs on the road (labelled by the x-axis), so the position-time graph is not a picture of the path of the motion.

For each example, work towards understanding what story the position-time graph tells you about the motion.

In the previous chapter we began to study the key concepts of kinematics (position, displacement, velocity, speed, acceleration, and time), mainly using motion diagrams as a concrete representation of motion. In this chapter, we'll learn how to connect motion diagrams with position-time graphs, a very useful (if slightly more abstract) means for representing motion.

Chapter 2: One-Dimensional MotionSaturday, September 07, 20139:41 PM

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In the position-time graph above, A and B both travel at constant speeds; we know this because each of the graphs is a straight line. We know that the speed of B is greater than the speed of A, because the slope of B's graph is greater than the slope of A's graph. Both objects begin at the zero position, and they both move in the + direction on the road.

In the following example, D gives C a head-start.

In the position-time graph above, C and D both travel at constant speeds, and D's speed is greater than C's speed. They both travel in the + direction. Note that at t = 2 s, D is still at position zero, and C is at position 2 m. Thus, C has a "head start." At t = 4 s, both C and D are at position 4 m, so D has caught up to C. After t = 4 s, D is further ahead.

In the following example, the two motions are in the opposite direction along the straight road.

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In the position-time graph above, E and F travel in opposite directions; the direction of motion is indicated by the sign of the slope of the position-time graph. Thus, E travels in the + direction and F travels in the - direction. To compare the speeds of E and F, you'll have to get out your ruler and use rise-over-run to determine the speeds from the graph.

Note that at t = 2 s, E and F are at the same position; this is the time at which they pass each other. After t = 5 s, the position of F is negative; this just means that F moves onto the part of the road labelled with negative positions.

Also note that the speed of F is NOT zero at t = 5 s; it's the position of F that is zero at t = 5 s, not its speed. Remember that F moves at a constant speed, because the slope of its position-time graph is constant.

The next example shows the position-time graph for an object that is not moving:

The position-time graph above is a horizontal straight line, so its slope is zero. Because the slope of the position-time graph is the velocity, this means that G has zero velocity. Another way to say this is that the position of G is constant. Another way to say this is that the displacement of G is zero.

However you say it, G is not moving.

The following two position-time graphs are not physically realistic:

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Kinematics equation for motion with constant speed (the following forms are all equivalent):

Neither of the position-time graph above makes sense. The graph for H fails the vertical line-test; that is, there are some times for which the object H has two different positions, which is not physically realistic. The graph for J suffers from the same defect, but is even worse, because it indicates that the object blinks into existence at a certain time, exists in all positions on the road at that one instant, and then blinks out of existence again.

While we're on the topic of what is physically realistic, note that many of the position-time graphs shown in the textbook have sharp corners, which are not physically realistic. Consider the following position-time graph. The sharp corner to the left indicates that the object moves in the positive direction at a constant speed and then INSTANTLY stops. This extreme "stopping on a dime" does not happen; there is always a slowing-down time period, which should be indicated on the position-time graph as a curve. That is, the sharp corner should actually be curved.

Why does the textbook display these physically unrealistic graphs? The authors are trying to help us by starting us off with graphs that do not contain curves, which are harder to analyze (not for calculus lovers, though) than straight lines. OK, we'll accept with thanks, but keep in mind that sharp corners are not realistic.

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velocity-time diagrams

if the velocity is constant, then the velocity-time graph is a horizontal straight line:•

The matching position-time graph is a straight line; we know this because the slope of the position-time graph is the velocity, and if the velocity has a constant value of 0.8 m/s, then the slope of the position-time graph has a constant value of 0.8 m/s.

However, we can't know the initial position of the position-time graph from the information contained in the velocity-time graph. Knowing that the speed of the moving object is 0.8 m/s, we can calculate the displacement for any time interval; for example, we can conclude that the displacement during a time interval of 3 s will be (0.8 m/s )(3 s) = 2.4 m, but we don't know where the object is. This means that if we start from the given velocity-time graph, we can know the slope of the position-time graph, but the actual position-time could be any one of an infinite number of parallel lines, all of which have a slope of 0.8 m/s:

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Calculus lovers will be able to understand the previous paragraph in mathematical terms. Going from the velocity-time graph to the position-time graph is equivalent to antidifferentiating (integrating) the velocity function. But we know that antidifferentiating results in a constant of integration, which can be interpreted as the initial position. The constant of integration could have any value; this corresponds to the possibility of having an infinite number of position-time graphs, all parallel lines with different initial positions, that match the given velocity-time graph.

Physically, the presence of the constant of integration makes sense, because it allows us to model motions with a given constant velocity and having any initial position, as could be possible in reality.

velocity-time diagram if velocity is not constant•

Here's an example of a velocity-time graph with velocity that is not constant. What is the corresponding position-time graph?

In class we argued that the corresponding position-time graph should be curved. The velocity increases as time passes; this means that the slope of the position-time graph increases as time passes. The slope starts out positive and gradually increases; note how the slope of the position-time graph corresponds to the value of the velocity-time

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acceleration•

The velocity tells us how fast the position is changing; that is, velocity is the rate of change of position. If the velocity is zero, then the position is constant, which means that the object does not move.

If the velocity is constant, and has a value of 2 m/s, this means that the position gradually increases; more specifically, every second the position increases by 2 m.

The acceleration is the rate of change of velocity. In other words, the acceleration tells us how fast the velocity is changing.

If the acceleration is constant, and has a value of 3 m/s2, this means that the velocity gradually increases; more specifically, every second the velocity increases by 3 m/s.

(Back when I was a high-school student it was popular to write the acceleration in the previous paragraph as 3 m/s/s, to emphasize the meaning described in the previous paragraph. However, this way of writing the unit is ambiguous (does it mean (m/s)/s or m/(s/s)?), and so writing it as 3 m/s2 is better. Nevertheless, reflecting on the unit as written in this old-fashioned way may be useful in helping you understand acceleration.)

a formula for acceleration, valid when acceleration is constant:•

graph at the same times:

Remember that the position-time graph sketched above is only one possibility; we don't know anything about the initial position, so the graph could be translated upwards or downwards and still have the same corresponding velocity-time graph.

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this is rise-over-run = slope, again, but this time on the velocity-time graph:

Examples of velocity-time diagrams and their interpretations:

speed is increasing•acceleration is constant (a < 0)•object moves in - direction•

speed is increasing•acceleration is constant (a > 0)•object moves in + direction•

speed is decreasing•acceleration is constant (a < 0)•object moves in + direction•

speed is decreasing•acceleration is constant (a > 0)•object moves in - direction•

Here are position-time graphs that match the four velocity-time graphs above. Each graph represents a motion with constant, non-zero acceleration, and the four graphs cover the four possible combinations of signs of velocity and acceleration:

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determining displacement from a velocity-time diagram •

Suppose that the velocity is constant for a motion. Then the velocity-time graph is a horizontal line, as in the following diagram. If we rearrange the basic kinematics equation for constant velocity,

which is v = ( x)/( t), then we obtain x = v t. The displacement x can be thought of as the area of the rectangle in the figure below,

which has height v and width t.

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If the velocity is NOT constant, the same area idea still works; that is it's still true that the displacement can be calculated as the area under the velocity-time graph.

Furthermore, we can use this idea to develop a kinematics equation valid when acceleration is constant. Consider the following figure:

A third useful kinematics equation (valid for constant acceleration) can be obtained by eliminating time from the previous two equations. We

can derive this third equation by solving equation (1) for t and substituting the resulting expression into equation (2):

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going from position-time graphs to velocity-time graphs and vice-versa•

Examples: given the position-time graphs, draw the matching velocity-time graphs, and describe the motion in words.

Examples: given the velocity-time graphs, draw the matching position-time graphs, and describe the motion in words

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free fall•

If we ignore air resistance, and no other forces act except for gravity (this is what "free" means in "free fall"), then the acceleration is the same for any object, no matter what its mass is, and is approximately constant, with value 9.80 m/s2. How do we know this? An enormous number of very careful experiments verify this fact.

Examples of solving one-dimensional motion problems

You might find the following strategy helpful in solving kinematics problems:

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Example: You drop a baseball from the top of a three-storey building (i.e., from a height above the ground of 10 m). (a) Determine the time needed for the ball to reach the ground.(b) Determine the speed of the ball just before it hits the ground.

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Example: You are standing at the top of a three-storey building (at a height above the ground of 10 m). You lean over the edge of the building and throw a baseball straight upwards with an initial speed of 24 m/s. The ball goes up, reaches a maximum height, and then falls back down all the way to the ground.

(a) Determine the time when the ball reaches its maximum height.(b) Determine the maximum height above the ground.(c) Determine the time when the ball reaches the ground.(d) Determine the speed of the ball just before it hits the ground.

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The positive solution is relevant for this problem; thus, it takes for 5.3 s for the baseball to reach the ground.

The negative solution is usually discarded as "extraneous." However, it is meaningful, even if it is not relevant for this problem. If a second baseball had been thrown upwards from the ground, instead of from the top of the building like the first baseball, in such a way that it reached the top of the building at time t = 0 s at a speed of 24 m/s, then it would have copied the motion of the first baseball exactly. The negative solution tells us when the second baseball would have had to have been thrown to accomplish this copying of the first baseball's motion.

This example should help give the lie to the incorrect phrase "time cannot be negative."

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In this chapter we've emphasized qualitative analysis of kinematics in one dimension. Qualitative analysis of graphs is a skill that is well worth developing, as it is useful in many fields of study, and in understanding information that we receive from news media as well. A good example is inflation; we can think of the average cost of the necessities of life (which economists call the consumer price index) as the "cost position." The rate at which the consumer price index increases is known as the inflation rate; this is analogous to the "cost velocity." But we often hear in the news media that this month inflation has increased or decreased; this rate of increase or decrease is analogous to the "cost acceleration."

If you can understand the graphical analysis we do in physics class, it will help you in reading the financial pages of your newspaper. And it will also help you interpret graphs that you encounter in biology, chemistry, mathematics, etc.

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