Constant Acceleration

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This article is about constant acceleration when applied to one dimensional motion.

Constant Acceleration and Gravity

Objects with a constant acceleration undergo an unchanging acceleration. This means that no matter how much time goes by or what the velocity is, the acceleration remains the same. This means that objects under constant acceleration are effected by an unchanging force. This force includes gravity. Gravity is the constant acceleration that all objects feel everywhere. Gravity pulls all objects towards its center, meaning that all of us are subject to that force. If gravity did not exist, then the universe is impossible. On different planets and stars, the acceleration due to gravity g is different. On Earth, g is -9.8 m/s2. The acceleration is negative because the force goes down, which is a negative direction. On the moon, g is -1.6 m/s2. So gravity is different on different celestial bodies. This force, g on earth, effects us in many different ways. The gravity on earth also effects all objects in various ways, such as the cliche, "What goes up must come down". There are four core equations for dealing with constant acceleration in one dimension. In this article, we will discuss them all.

The Formulas

There are four core formulas for constant acceleration each with different uses and purposes. The variables used are x, displacement, vf, final velocity, v0, initial velocity, a, the constant acceleration, and t, time. The four equations include and disregard certain variables in order to have many solve able features. You will see this after observing the equations.

x,a, and t Relation

The first formula includes x, a, and t. It disregards vf and v0. The equation is a relation of displacement to constant acceleration and time. Here is the equation:

(1)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} x = {1 \over 2}at^2 \end{align}

This equation says that the displacement of an object is one half the constant acceleration times time squared. This equation can answer questions such as:
How far did the object fall? with given time and acceleration
How long did it take the object to fall? with given displacement and acceleration
What acceleration did the object experience? with given time and displacement
The equation is simple, and any of the three variables in it can be solved for. Most problems, unless stated, will refer the constant acceleration as the acceleration due to gravity, which remember, is -9.8 m/s2. Now, let's put this equation to action.
Question 1
If a ball is dropped off a cliff and hits the ground in 10 seconds, how tall was the cliff?
First, identify what you know. The problem tells us the time t as 10 seconds. The problem does not specify the acceleration, so we can assume that it is -9.8 m/s2. What we don't know is the displacement, x. Our problem is using the variables x, a, and t, so we know to use our first formula. Plug in what you know into the equation. $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} x = {1 \over 2}(-9.8)(10)^2$ Begin to solve using PEMDAS. The 10 needs to be squared, and in the meantime, you can simplify the 1/2(-9.8). Our new equation after some simplification is $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} x = -4.9(100)$. When solved, we get x=-490 meters. Wait, the cliff can't be -490 meters tall! Actually, the x represents the displacement of the object, and since in went down, the answer was negative. One of the largest things in physics is keeping your signs correct. The cliff is actually 490 meters tall.
Question 2
An object on the moon takes how long to fall 3 meters?
Again, we must identify what we know. We know our displacement x, 3 meters. Also, we know our acceleration. This time, the problem refers to gravity on the moon. If you don't know that specific number, then find it on the Internet. Otherwise, the acceleration due to gravity on the moon is -1.6 m/s2. What we don't know is our time, t. Again, the problem calls for the three variables x, a, and t meaning we will use the first equation. Substitute known values into the equation. $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} -3 = {1 \over 2}(-1.6)(t)^2$ Uh oh, this time we must use algebra. Simplify the equation so it looks simpler. $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} -3 = -0.8t^2$ Now, using algebra, divide by -0.8 on both sides. We are then left with $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} 3.75 = t^2$. To make the t2 into just t, we must take the inverse of squaring, which is square rooting. Square root both sides of the equation to get $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \sqrt{3.75} = t$. The square root of 3.75 is about 1.936. So, it takes approximately 1.936 seconds for an object to fall 3 meters on the moon.

Here are variants of the equation solved for the other variables.

$\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} t = \sqrt{{2x \over a}}$ for t

$\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} a = {2x \over t^2}$ for a

vf, v0, a, and x Relation

The next formula uses vf, v0, a, and x, and it disregards t. This equation relates the final velocity to the object's initial velocity, acceleration, and displacement regardless of time. Here is the equation:

(2)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} v_f^2 = v_0^2+2ax \end{align}

The equation states that the final velocity squared is equal to its initial velocity squared plus two times the acceleration times the displacement. This equation can answer questions such as:
What was the object's final velocity? with given initial velocity, acceleration, and displacement
What was the object's beginning velocity? with given final velocity, acceleration, and displacement
How far did the object travel? with given final and initial velocity, and acceleration
What acceleration did the object experience? with given final and initial velocity, and displacement
This equation can be cumbersome, since we have two squares. This equation will often require finding square roots for an answer. Let's see what happens when you apply this problem to everyday life.
Question 1
An object is thrown down a 600 meter high cliff with a velocity of 3 meters per second. What was the object's speed right before it hit the ground?
First, identify what you know and what you want to find out. Again, remember, that if a problem doesn't state an alternate acceleration and it is referring to falling objects, use -9.8 m/s2 for a. Our displacement, x, is 600 meters, but wait! Since the object is falling, it is going in the negative direction. Remember that if an object is going down, its displacement is negative. The actual distance is positive, but since displacement is a vector, it needs direction. So our x is -600 meters. We also know our initial velocity, v0, as 3 meters per second. Note once again what the problem is saying. The object was thrown down, so the initial velocity is also negative. If the problem said just thrown, or thrown up, the velocity would be positive. So, we have a=-9.8, x=-600, and v0=-3. We do not know vf, which is what we are trying to find. So, begin plugging in what you know into the equation. $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} v_f^2 = (-3)^2+2(-9.8)(-600)$ Simplify the (-3)2 to 9, and the 2(-9.8)(-600) to 11760. You get this after simplifying: $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} v_f^2 = 11769$. You can then find the square root in order to get a singular vf. The equation is not solved if you have vf2, so find the square root if you get this. Also note that if our x was not negative, we would be finding the square root of -11769, which can't be done. Keep those signs right! When we solve, we get a final answer of 108.49 m/s. So the velocity right before landfall was 108.49 meters per second.
Question 2
A ball is thrown at someone 10 meters away. When the person was hit, the ball was going 6 meters a second. At what velocity was the ball thrown?
Question 3
A ball is thrown with a velocity of 10 meters per second. How far did the ball travel when it is going 5 meters per second?
Here are variants of the equation solved for the other variables.

$\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} v_f = \sqrt{v_0^2+2ax}$ for vf
$\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} v_0 = \sqrt{v_f^2-2ax}$ for v0
$\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} a = {v_f^2-v_0^2 \over 2x}$ for a
$\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} x = {v_f^2-v_0^2 \over 2a}$ for x

vf, v0, a, and t Relation

The third equation uses time, final and initial velocity, and of course constant acceleration. Displacement is not a concern in this figure. Here is the equation:

_c
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