When is constant acceleration

And we will often fix a variable at a specific value in order to have an equation apply to a specific situation. If we want to figure out what the average velocity in that time interval is, we have to find the constant line so that the areas under the true velocity curve equals the area under the average velocity line a constant.

We do this by adjusting the average velocity line so that the light pink area the area no longer included and the light blue area the extra area now included are equal. It's pretty clear both from thinking about how the velocity changes and from looking at the graph, that the average velocity is going to be halfway between the endpoints; that is, if a is constant, then.

Since we know that the definition of the average velocity is "that velocity which, if you go constantly with it, will produce the same displacement", we can see that if the acceleration is constant:. The airplane lands with an initial velocity of Note that the acceleration is negative because its direction is opposite to its velocity, which is positive. From it we can see, for example, that. All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.

An intercontinental ballistic missile ICBM has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight actual ICBM burn times are classified—short-burn-time missiles are more difficult for an enemy to destroy. But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does.

The Space Shuttle does this by accelerating for a longer time. We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with. Adding v 0 to each side of this equation and dividing by 2 gives. Dragsters can achieve average accelerations of Suppose such a dragster accelerates from rest at this rate for 5.

How far does it travel in this time? Figure 6. William Thurmond. Photo Courtesy of U. We are asked to find displacement, which is x if we take x 0 to be zero.

Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it. Plug the known values into the equation to solve for the unknown x :. Substituting the identified values of a and t gives. If we convert m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing.

So the answer is reasonable. This is an impressive displacement in only 5. We see that:. Identify the known values. To get v , we take the square root:. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration. In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques.

The box below provides easy reference to the equations needed. On dry concrete, a car can decelerate at a rate of 7. 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 Physics 2D Motion Relative Motion.

Prashant S. Apr 15, See below. Explanation: Constant acceleration refers to the motion where the speed of the object increases in the same amount in per unit time. Of course, they require consistent systems of units to be used. If the values of three of the variables are known, then the remaining values can be found by using two of the equations. More simply:. We can derive this equation using the fact that the displacement is equal to the signed area under the velocity—time graph.

For the velocity—time graph on the left, the region under the graph is a trapezium. So the second equation of motion holds in this case. One of the triangles has positive signed area and the other has negative signed area. Finding the displacement of a particle from the velocity—time graph using integration will be discussed in a later section of this module.

Substituting this into the second equation gives. Derive the third and fifth equations of motion from a velocity—time graph. Motion due to gravity is a good context in which to demonstrate the use of the constant-acceleration formulas.

As discussed earlier, our two directions in vertical motion are up and down, and a decision has to be made as to which of the two directions is positive. Use this value in the following exercise.