Motion in One Dimension Study Guide (page 3)
We start the article by defining the notions of displacement, velocity, speed, and acceleration, Then we look into graphical representation of motion as a way of describing the motion of an object or particle, We consider a special case of unidimensional motion where the acceleration is constant and show an application to the free fall of objects in the gravity field of Earth,
Displacement, Velocity, and Acceleration
Motion involves an object changing its position during some finite time interval with respect to a given reference frame. For most of the time, the object's position may be associated with just one set of coordinates at a time—that would be just one coordinate in the case of a one-dimensional motion. Also, motion is a relative notion: You may be sitting still in a car, so you do not move, but the car may be speeding on the highway, and therefore, you are in fast motion with respect to the road.
We say that displacement is the change in position of a particle or object with respect to a given reference frame. Because the position is a vector, the displacement is also a vector: It points from the initial to the final position and has a magnitude equal to the distance between the two positions. We can always associate both a direction and a size with a displacement.
For a particle moving along a straight line in one dimension, the displacement can be expressed in terms of the initial and final coordinates as:
Δx = xf – xi
If xf is greater than xi, then we have a positive displacement; otherwise, the displacement is negative.
For any object in motion for a finite duration of time, we can define an average speed as the ratio of the distance traveled over the time elapsed:
This formula assumes our moving object starts at the time zero at some point P and then moves a distance d during a time interval t.
You can go from Lansing to Detroit, Michigan, in, say, 2 hours, but out of these 2 hours, you spend 1 hour going 60 miles from Lansing to Novi and the next hour going the rest of 40 miles from Novi to Detroit (mostly stuck in traffic). Find the average speed from Lansing to Novi, the average speed from Novi to Detroit, and the overall average speed from Lansing to Detroit.
Since all quantities are given in miles and hours, we will keep these units instead of using SI units.
Your average speed from Lansing to Novi is 60 miles pee hour. From Novi to Detroit, as you are stuck in traffic, your average speed is only 40 miles per hour. The overall average speed from Lansing to Detroit is:
We note that for calculating the average speed, only the total distance traveled and the total time are important. The example also shows that during a long time interval, we may at times move faster and at other times slower; the overall average speed gives us an idea of the rate at which we move on average during the whole trip.
Now let's look for a moment at what happens when the time interval t is made almost zero. In this case, the distance traveled d also becomes very small (for normal-range finite speed values). Therefore, the start point and the end point are very close to each other. Our average speed is calculated over a very small space interval around the point P. In this case, our average velocity is a very close approximation to the instantaneous speed at point P.
Back to our example, a good practical approximation of the car's instantaneous speed at every moment of time during the trip from Lansing to Detroit is the value we read on the car's speedometer.
Let's note that speed is always a positive quantity. The 51 unit for measuring speed follows from its definition: meter per second. Other convenient units may be miles per hour or kilometers per hour.
As far as the direction of the movement is concerned, we need to extend the speed to the concept of velocity. We can reach this by replacing the "distance" in the definition of the average speed with the "displacement" vector. The instantaneous velocity is a vector obtained by taking the ratio of the displacement vector d to the time t elapsed for that displacement to occur, when the time interval is very small:
Therefore, the instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time.
For the case of motion in one dimension along a straight line, the distance and the magnitude of the displacement are the same, and therefore, the speed is the magnitude of the velocity. This statement remains true in the general case of an arbitrary motion in two or three dimensions. The proof is a trivial exercise of infinitesimal calculus. Is it really a surprise that Sir Isaac Newton is, at the same time, one of the inventors of the infinitesimal calculus and the famous father of the laws of mechanics?
Instantaneous velocity indicates how fast the ear moves and the direction of motion at each instant of time.
As the velocity shows the change in an object's position, the acceleration shows how the velocity of the object changes. As the velocity is a vector, it can change either its magnitude or its direction. In either case, the object will feel an acceleration a given by:
Therefore, in the International System, or SI, unit for acceleration is m/s2.
Suppose a car starts from rest. The car accelerates straight down the road and at t = 30 s attains a speed of v = 30 km/h. Determine the average acceleration of the car.
(30 km/h – 0 km/h)/(30 s – 0 s) = 1.0(km/h)/s = 1,000 m/3,600 S2 0.28 m/s2
Similar to the case of velocity, we start by defining an average acceleration over a finite time interval Δt and then get to the concept of instantaneous acceleration, taking this time interval to zero:
Whenever the acceleration and velocity vectors have opposite directions, the object slows down and is said to be decelerating.
Graphs and Motion
An easy and clear method of quickly analyzing relationships among large quantities of numerical data is to plot the experimental data as a function of time or of some other variable. Straight lines on a graph can be easily recognized and interpreted, and they play an important role in graphically analyzing experimental data.
By far, the most usual kind of graph is the Cartesian one, where both axes have linear scales. The horizontal axis is also called abscissa, and the vertical axis is called ordinate. The graph shows the dependency of the quantity, which is represented as the ordinate on the qualitity, which is represented as the abscissa.
We define the slope of a straight line on the graph by choosing any two points on the line and taking the ratio of the ordinate separation of these two points to their abscissas separation. If the graph is not made up of straight lines, then we can define a slope at any point along the curve by drawing a tangent to the curve and calculating the slope of this tangent, using the slope definition for a straight line.
In the case of position versus time graphs, the slope of the graph at a certain time is immediately associated with the instantaneous velocity at that time, as the ratio of position coordinate difference versus the time interval to go from the initial position to the other is an exact reformulation of the definition of speed. By similar reasoning, the slope of a velocity versus time graph is associated with the instantaneous acceleration of the object at that time.
The two graphs in Figure 3.1 show the position versus time for two swimmers racing across a 35-meter pool. Interpret the information in the graphs.
There is a lot of information we can read from these graphs.
The first athlete starts at time t = 0 and swims the first 25 meters of the race at constant speed v0 = 25 m/15 s = 1.6 m/s. He gets tired and goes the last 10 meters of the race at a much slower pace, at a speed v = 10 m/20 s = 0.5 m/s.
The second athlete starts with a handicap of 5 seconds and initially goes slowly. For example, from point C to point D, his speed is almost constant. His speed at point D, after 35 seconds from the beginning of the race, is given by the slope of the tangent and is vD = 7.5 m/2s s = 0.3 m/s. But at point E, 50 seconds after the beginning of the race, his instantaneous speed increased to vE = 16.5 m/15 s = 1.1 m/s. The last 10 seconds of the race, he rides Flipper the dolphin, and at point F, his final speed is around 2.8 m/s. This reminds us somewhat of the Hare and Tortoise fable but with an aquatic spin!
Motion with Constant Acceleration
Let's examine now the case of constant acceleration motion. This is an important kind of motion encountered in frequent approximations to real situations. An object sliding down an incline or multiple objects falling near the surface of the earth when we neglect air resistance are both situations where we encounter constant acceleration motion.
As the acceleration is constant, the instantaneous and average accelerations are equal, and we can invert the definition to get:
Δv = a Δt
If we consider that at time t = 0, the initial velocity is v0, we can write this as:
v (t) = v0 + a · t
For the case of the one-dimensional motion, this becomes a scalar relation between speeds, with a positive acceleration a when it is directed in the sense of the motion, and a negative acceleration a when its direction is against the direction of the motion.
If we consider that at the initial time t = 0, there is an initial displacement x0, then the position at time t is given by:
Solving for t in the velocity equation and inserting the result in the displacement equation, we obtain another useful relation that is independent of time:
v2 = v20 + 2aΔx
Freely Falling Objects
Any object found in the neighborhood of the surface of the earth suffers a gravitational pull with a downward acceleration g = 9.80 m/s2. The exact value slightly varies depending on various factors such as the latitude and the mineral composition of the earth's crust, which finally affects the mass distribution of the earth. But for the purpose of most calculations, we may assume all objects fall down with a constant acceleration g = 9.80 m/s2. The direction is always downward, actually toward the center (or mass—for pedantic eyes) of the earth.
The main experimental result for the freely falling motion was obtained by Galileo hundreds of years ago and states that when we can neglect air resistance, all bodies fall at the same rate, no matter how heavy or light they are. He reached this conclusion by dropping at the same time different objects from the top of the Tower of Pisa and observing that they always hit the ground almost at the same time. If the experiment could be done so that there is no air drag, such as in a vacuum tube, the times of the falls would certainly be identical, and they would hit the ground exactly at the same time. And that is exactly what Robert Boyle (1627–1691) did when he perfected his vacuum pumps in the mid-1600s and had a golden opportunity to test Galileo's work for the first time.
This may sound counterintuitive at some point: If we drop a quarter and a sheet of paper from the same height at the same time, the quarter quickly reaches the floor, while the paper still slowly flutters through air for a period of time. But if we crush the paper into a wrinkled ball and therefore cut its air drag without changing its weight, it will hit the ground almost at the same time as the quarter.
In general, for a body moving along the vertical plane near the surface of Earth, we can apply the rules of constant acceleration motion. Considering the direction of motion is positive from the surface of Earth upward, the acceleration will be a = g = –9.80 m/s2 for an ascending body, and a = +9.80 m/s2 for a falling body. NOTE: The direction of the acceleration due to gravity is always directed toward the center of Earth.
A bullet blasts from the barrel of a gun upward in the vertical direction with an initial speed of 700 m/s. Find the maximum altitude reached by this bullet and the time needed to reach it.
At the highest point of its path, the bullet reverses direction from moving upward to falling downward. Therefore, at this point, its instant velocity (and speed) should be zero. Using the formula linking square speeds to acceleration and distance, we find:
– 2 · 9.8m/s2 · h
Solving for h, we find:
h = 25,000 m = 25 km
To find the time needed to reach that altitude, we simply use:
v = v0 + a · t
or, plugging in the numbers:
0 = 700 m/s – 9.8 m/s2 · t
Solving for the time t, we find:
t = 71.43 s
Practice problems of this concept can be found at: Motion in One Ditnension Practice Questions
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