Technically no. Negatives are usually associated with displacement, where one direction is considered positive and the opposite is considered negative. Distance is just how far you have travelled. Ordinary distance is the magnitude of the vector that connects two points. It is always positive or zero if the two points coincide. Of course in a given coordinate system, you can always move in the negative or negative or negative direction. If a car moves with a constant speed, can you say that it also moves with a constant velocity?
Give an example to support your answer. No, you could be going around a corner at constant speed. Your velocity is then not constant. Distance is defined to be the magnitude or size of displacement between two positions.
Distance traveled is the total length of the path traveled between two positions. Distance traveled is not a vector. It has no direction and, thus, no negative sign. The negative sign associated with PV work done indicates that the system loses energy. Heat released is negative and heat absorbed is positive. This all goes back to the concept of a system and the surroundings.
We think of change in energy from the perspective of the system. Begin typing your search term above and press enter to search. Press ESC to cancel. The verbal representation of the situation has already been translated into a motion diagram. A careful reading of the motion diagram allows the construction of the motion graphs. We already know, from the motion diagram, that the car starts at position zero, then has positive, increasing positions throughout the remainder of its motion.
This information can be transferred onto a position vs. Notice that the position is zero when the time is equal to zero, the position is always positive, and the position increases as time increases.
Also note that in each subsequent second, the car changes its position by a smaller amount. This leads to the graph of position vs. Once the car stops, the position of the car should not change. From the motion diagram, we know that the velocity of the car is always positive, starts large in magnitude, and decreases until it is zero. This information can be transferred onto a velocity vs.
How do we know that the slope of the line is constant? The slope of the line represents the rate at which the velocity is changing, and the rate at which the velocity is changing is termed the acceleration. Since in this model of mechanics we will only consider particles undergoing constant acceleration, the slope of a line on a velocity vs.
From the motion diagram, the acceleration of the car can be determined to be negative at every point. Again, in this pass through mechanics we will only be investigating scenarios in which the acceleration is constant. Thus, a correct acceleration vs. After constructing the two qualitative representations of the motion the motion diagram and the motion graphs , we are ready to tackle the quantitative aspects of the motion.
A glance at the situation description should indicate that information is presented about the car at two distinct events. Information is available about the car at the instant the driver applies the brakes the velocity is given , and the instant the driver stops the position is given. Other information can also be determined by referencing the motion diagram.
To tabulate this information, you should construct a motion table. In addition to the information explicitly given, the velocity at the first event and the position at the second event, other information can be extracted from the problem statement and the motion diagram.
Since you are working under the assumption in this model that the acceleration is constant, the acceleration between the two instants in time is some unknown, constant value. To remind you that this assumption is in place, the acceleration is not labeled at the first instant, a1, or the second instant, a2, but rather as the acceleration between the two instants in time, a You now have a complete tabulation of all the information presented, both explicitly and implicitly, in the situation description.
Moreover, you now can easily see that the only kinematic information not known about the situation is the assumed constant acceleration of the auto and the time at which it finally stops.
Thus, to complete a kinematic description of the situation these two quantities must be determined. What you may not know is that you have already been presented with the information needed to determine these two unknowns. In the concepts and principles portion of this unit, you were presented with two formal, mathematical relationships, the definitions of velocity and acceleration. In the example that you are working on, there are two unknown kinematic quantities.
You should remember from algebra that two equations are sufficient to calculate two unknowns. Thus, by applying the two definitions you should be able to determine the acceleration of the car and the time at which it comes to rest.
Although you can simply apply the two definitions directly, normally the two definitions are rewritten, after some algebraic re-arranging, into two different relationships. This rearrangement is simply to make the algebra involved in solving for the unknowns easier. It is by no means necessary to solve the problem. In fact, the two definitions can be written in a large number of different ways, although this does not mean that there are a large number of different formulas you must memorize in order to analyze kinematic situations.
There are only two independent kinematic relationships. The two kinematic relationships [1] we will use when the acceleration is constant are:. Thus, the car must have accelerated at 3. The kinematic description of the situation is complete. When she is 10 m from the light, and traveling at 8. She instantly steps on the gas and is back at her original speed as she passes under the light.
Notice that between the instant she hits the brakes and the instant she steps on the gas the acceleration is negative, while between the instant she steps on the gas and the instant she passes the light the acceleration is positive. Thus, in tabulating the motion information and applying the kinematic relations we will have to be careful not to confuse kinematic variables between these two intervals.
Below is a tabulation of motion information using the coordinate system established in the motion diagram. Recall that by using your two kinematic relations you should be able to determine these values. Second, notice that during the second time interval again two variables are unknown.
Once again, the two kinematic relations will allow you to determine these values. Thus, before I actually begin to do the algebra I know the unknown variables can be determined! Now, using these results, examine the kinematics between stepping on the gas and passing the light.
We now have a complete kinematic description of the motion. The driver of an automobile suddenly sees an obstacle blocking her lane. Determine the total distance the auto travels between seeing the obstacle and stopping d as a function of the initial velocity of the car vi and the magnitude of its acceleration while stopping as. Rather than calculate the stopping distance for particular values of initial velocity and acceleration, the goal of this activity is to determine, in general, how the stopping distance depends on these two parameters.
If we can construct this function we can then use the result to calculate the stopping distance for any car if we know its initial velocity and stopping acceleration. Since our goal is to determine d as a function of vi and as, we must eliminate t2. To do this, solve for t2 in the left equation and substitute this expression into the right equation. Thus, the stopping distance appears to be proportional to the square of the initial velocity and inversely proportional to the stopping acceleration.
Does this make sense? To determine if a symbolic expression is sensible it is often useful to check limiting cases. A limiting case is when one of the variables in the expression takes on an extreme value, typically zero or infinity.
For example, if the initial velocity of the car was zero the stopping distance would have to also be zero, since the car was never moving! Another limiting case would be setting the acceleration of the car equal to zero. With no acceleration, the car should never stop. In our expression, setting the acceleration equal to zero results in an infinite stopping distance, which again agrees with commonsense. Confusion about the meaning of algebraic signs is common among beginning physics students.
The best way to clarify this confusion is to remember that algebraic signs are simply a mathematical way to describe direction. The key to the translation is the coordinate system. A coordinate system is very similar to the English-French dictionary you might take with you on your first trip to France.
A negative acceleration, for example, does NOT imply that the object is slowing down. It implies an acceleration that points in the negative direction. It is impossible to determine whether an object is speeding up or slowing down by looking at the sign of the acceleration! I can decelerate in the positive direction as easily as I can decelerate in the negative direction. In a later chapter, we will return to the case in which the acceleration is not constant. The above relationship is our first kinematic relationship.
The velocity-time graph shows a line with a positive upward slope meaning that there is a positive acceleration ; the line is located in the negative region of the graph corresponding to a negative velocity.
The acceleration-time graph shows a horizontal line in the positive region of the graph meaning a positive acceleration. For more information on physical descriptions of motion, visit The Physics Classroom Tutorial.
Detailed information is available there on the following topics:.
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