We could be traveling back and forth, but we won't consider changes of the angle of our motion. If the acceleration is constant, then the velocity is changing at a constant rate. Our equation defining the acceleration is then. Since we have selected particular times, everything in this equation is a constant.
But we can be choose one or more of them to turn back into a variable. Then we can see what the velocity looks like as a function of time. If this seems a little weird to you, get used to it. Find the distances necessary to stop a car moving at In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.
Identify the knowns and what we want to solve for. This equation is best because it includes only one unknown, x. We know the values of all the other variables in this equation. There are other equations that would allow us to solve for x , but they require us to know the stopping time, t , which we do not know. We could use them but it would entail additional calculations. Rearrange the equation to solve for x.
This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is —5. The result is. Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. We are looking for x reaction. Identify the best equation to use.
This means the car travels Figure The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at Also shown are the total distances traveled from the point where the driver first sees a light turn red, assuming a 0.
The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements.
But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest. Suppose a car merges into freeway traffic on a m-long ramp. If its initial velocity is Such information might be useful to a traffic engineer.
We are asked to solve for the time t. As before, we identify the known quantities in order to choose a convenient physical relationship that is, an equation with one unknown, t. We need to solve for t. Choose the best equation. We will need to rearrange the equation to solve for t. In this case, it will be easier to plug in the knowns first. Simplify the equation.
The units of meters m cancel because they are in each term. Doing so leaves. Use the quadratic formula to solve for t. This yields two solutions for t , which are. A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that solution. Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable.
The With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. Problem-Solving Basics discusses problem-solving basics and outlines an approach that will help you succeed in this invaluable task. To answer this, choose an equation that allows you to solve for time t , given only a , v 0 , and v.
Rearrange to solve for t. In vertical motion, y is substituted for x. An Olympic-class sprinter starts a race with an acceleration of 4. A well-thrown ball is caught in a well-padded mitt.
If the deceleration of the ball is 2. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.
What is its muzzle velocity that is, its final velocity? How long does it take to reach its top speed of How long does it take to come to a stop from its top speed? While entering a freeway, a car accelerates from rest at a rate of 2. To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. Solve for this unknown in the same manner as in part c , showing all steps explicitly.
At the end of a race, a runner decelerates from a velocity of 9. Does it make sense? Professional Application: Blood is accelerated from rest to These units may seem a little awkward to a beginning physics student. Yet they are very reasonable units when you begin to consider the definition and equation for acceleration. The reason for the units becomes obvious upon examination of the acceleration equation.
Since acceleration is a vector quantity , it has a direction associated with it. The direction of the acceleration vector depends on two things:. The general principle for determining the acceleation is:. This general principle can be applied to determine whether the sign of the acceleration of an object is positive or negative, right or left, up or down, etc. Consider the two data tables below. In each case, the acceleration of the object is in the positive direction. In Example A, the object is moving in the positive direction i.
When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction i. According to our general principle , when an object is slowing down, the acceleration is in the opposite direction as the velocity. Thus, this object also has a positive acceleration. This same general principle can be applied to the motion of the objects represented in the two data tables below.
In each case, the acceleration of the object is in the negative direction. In Example C, the object is moving in the positive direction i. According to our principle , when an object is slowing down, the acceleration is in the opposite direction as the velocity. Thus, this object has a negative acceleration. In Example D, the object is moving in the negative direction i.
Thus, this object also has a negative acceleration. Observe the use of positive and negative as used in the discussion above Examples A - D. In physics, the use of positive and negative always has a physical meaning.
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