Example of uniformly accelerated motion

We’re all familiar with the famous tale of an apple falling from a tree, sparking Isaac Newton’s early foundational work theorizing gravity. Newton’s curiosity and drive to understand this seemingly uninteresting falling motion has transformed much of our current understanding of the moving world and universe around us, including the phenomena of uniform acceleration due to gravity happening all around us, all the time.

In this article, we’ll be diving deeper into the definition of uniformly accelerated motion, the relevant formulas to know, how to identify and examine related graphs, and a couple of examples. Let’s get started!

Uniformly Accelerated Motion Definition

Throughout our introduction to kinematics so far, we’ve encountered several new variables and equations to solve problems for motion in one dimension. We’ve paid close attention to displacement and velocity, as well as changes to these quantities, and how different initial conditions affect the overall motion and outcome of a system. But what about acceleration?

Observing and understanding the acceleration of moving objects is just as important in our initial study of mechanics. You may have picked up that so far we’ve primarily been examining systems where acceleration is zero, as well as systems where the acceleration remains constant during some period of time. We call this uniformly accelerated motion.

Uniformly accelerated motion is the motion of an object undergoing constant acceleration that does not change with time.

In other words, the velocity of a moving object uniformly changes with time and the acceleration remains a constant value. Acceleration due to gravity, as seen in the fall of a skydiver, an apple from a tree, or a dropped phone to the floor, is one of the most common forms of uniform acceleration that we observe in our everyday lives. Mathematically, we can express uniform acceleration as:

\begin{align*}a=\mathrm{const.}\end{align*}

Calculus Definition of Acceleration

Recall that we can calculate the acceleration \(a\) of a moving object if we know starting and ending values for both the velocity and time:

\begin{align*}a_{avg}=\frac{\Delta v}{\Delta t}=\frac{v_1-v_0}{t_1-t_0}\end{align*}

where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time. However, this equation gives us the average acceleration over the time period. If we want to determine the instantaneous acceleration instead, we need to remember the calculus definition of acceleration:

\begin{align*}a_{inst}=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\end{align*}

That is, acceleration is mathematically defined as the first derivative of the velocity and the second derivative of position, both with respect to time.

Uniformly Accelerated Motion Formulas

It turns out you already know the formulas for uniformly accelerated motion — these are the kinematics equations we learned for motion in one dimension! When we introduced the core kinematics equations, we assumed that all these formulas accurately describe the motion of an object moving one-dimensionally as long as the acceleration is held constant. Before, this was largely an aspect that we implied and didn’t dig further into.

Let’s rearrange our kinematics equations and isolate the acceleration variable. This way, we can easily use any of our formulas to solve for the value of acceleration, given different initial conditions to start. We’ll start with the formula \(v=v_0+at\).

The value of constant acceleration given the initial velocity, ending velocity, and time is:

\begin{align*}a=\frac{v-v_0}{t}, \\ t \neq 0.\end{align*}

Our next kinematic equation is \(\Delta x=v_0t+\frac{1}{2}at^2\).

The value of constant acceleration given the displacement, initial velocity, and time is:

\begin{align*}a=\frac{2(\Delta x-tv)}{t^2}, \\ t \neq 0.\end{align*}

Our final kinematic equation of interest is \(v^2=v_0^2+2a \Delta x\).

The value of constant acceleration given the displacement, initial velocity, and final velocity is:

\begin{align*}a=\frac{v^2-v_0^2}{2 \Delta x}, \\ \Delta x \neq 0.\end{align*}

You may recall that there is an acceleration independent equation associated with kinematics, but this equation is irrelevant here since the acceleration variable is not included.

Although we’ve isolated the acceleration variable in each kinematic equation here, remember that you can always rearrange your equation to solve for a different unknown — you’ll often be using a known value of acceleration instead of solving for it!

Uniform motion, uniform acceleration — is there really a difference between the two? The answer, perhaps surprisingly, is yes! Let’s clarify what we mean by uniform motion.

Uniform motion is an object undergoing motion with a constant or unchanging velocity.

Although the definitions of uniform motion and uniformly accelerated movement sound similar, there’s a subtle difference here! Recall that for an object moving with a constant velocity, the acceleration must be zero according to the definition of velocity. Therefore, uniform motion does not also imply uniform acceleration, since the acceleration is zero. On the other hand, uniformly accelerated motion means the velocity is not constant but the acceleration itself is.

Graphs for Uniformly Accelerated Motion

We previously looked at a few graphs for motion in one dimension — now, let’s return to uniformly accelerated motion graphs in a bit more detail.

Uniform Motion

We just discussed the difference between uniform motion and uniformly accelerated motion. Here, we have a set of three graphs that visualize three different kinematics variables for an object undergoing uniform motion during some time frame \(\Delta t\):

In the first graph, we observe that the displacement, or change in position from the starting point, linearly increases with time. That motion has a constant velocity throughout time. The velocity curve in the second graph has a slope of zero, held constant to the value of \(v\) at \(t_0\). As for acceleration, this value remains zero throughout the same time period, as we’d expect.

Another important aspect to note is that the area under the velocity-time graph equals the displacement. Take the shaded rectangle in the velocity-time graph above as an example. We can quickly calculate the area under the curve by following the formula for the area of a rectangle, \(a=b \cdot h\). Of course, you can also integrate to find the area under the curve:

\begin{align*}\Delta s = \int_{t_1}^{t_2} v(t)\,\mathrm{d}t\end{align*}

In words, we can integrate the velocity function between a lower and upper limit of time to find the change in displacement that occurred during that time period.

Uniform Acceleration

We can graph the same three types of plots to examine uniformly accelerated motion. Let’s look at a velocity-time graph:

Here, we have a simple velocity function \(v(t)=2t\), plotted from \(t_0=0\,\mathrm{s}\) to \(t_1=5\,\mathrm{s}\). Since the change in velocity is nonzero, we know the acceleration will be nonzero as well. Before we take a look at the acceleration plot, let’s calculate the acceleration ourselves. Given \(v_0=0\, \mathrm{\frac{m}{s}}\), \(v_1=10\, \mathrm{\frac{m}{s}}\), and \(\Delta t=6\, \mathrm{s}\):

\begin{align*} a=\frac{v_1-v_0}{t} \\ a=\mathrm{\frac{10\, \frac{m}{s} - 0\, \frac{m}{s}} {5\, s}} \\ a=\mathrm{2\,\frac{m}{s^2}} \end{align*}

Now, let’s take a look at the acceleration-time graph:

This time, the acceleration-time plot shows a constant, nonzero acceleration value of \(2\,\mathrm{\frac{m}{s}}\). You may have noticed here that the area under the acceleration-time curve is equal to the change in velocity. We can double-check this is true with a quick integral:

\begin{align*} \Delta v = \int_{0}^{5}2\,\mathrm{d}t = 2t \\ \Delta v = 2(5)-2(0) \\ \Delta v = 10\, \mathrm{\frac{m}{s}} \end{align*}

Finally, we can continue working backward to calculate the change in displacement in meters, even though we don’t have a graph for this variable in front of us. Recall the following relationship between displacement, velocity, and acceleration:

\begin{align*} \Delta s = \int v(t)\,\mathrm{d}t = \iint a(t)\,\mathrm{d}t \end{align*}

Although we know functions for both velocity and acceleration, integrating the velocity function is easiest here:

\begin{align*}\Delta s = \int_{0}^{5} 2t\,\mathrm{d}t = \frac{2t^2}{2} = t^2 \\ \Delta s = (5)^2 - (0)^2 \\ \Delta s = 25\, \mathrm{m} \end{align*}

Remember that this calculation gives us the net displacement over the five-second time period as opposed to a general function of displacement. Graphs can tell us quite a lot about an object in motion, especially if we’re given minimal information at the start of a problem!

Now that we’re familiar with the definition and formulas for uniformly accelerated motion, let’s walk through an example problem.

A child drops a ball from a window at a distance of \(11.5\, \mathrm{m}\) from the ground below. Ignoring air resistance, how many seconds does the ball fall in until hitting the ground?

It might seem like we weren’t given enough information here, but we imply the values of some variables in the context of the problem. We’ll have to infer some initial conditions based on the scenario at hand:

• We can assume the child gave no initial velocity when releasing the ball (such as throwing it down), so the initial velocity must be \(v_0=0\, \mathrm{\frac{m}{s}}\).
• Since the ball is undergoing vertical free fall motion due to gravity, we know that the acceleration is a constant value of \(a=9.81\, \mathrm{\frac{m}{s^2}}\).
• We don’t have enough information to determine the final velocity immediately before the ball hits the ground. Since we know displacement, initial velocity, and acceleration, we’ll want to use the kinematic equation \(\Delta y=v_0t+\frac{1}{2}at^2\).

Let’s plug in our known variables and solve for time. Note that of course we don’t want to take the square root of a negative number, which would occur if we use define the acceleration due to gravity following the convention. Instead, we can simply define the downward direction of motion along the y-axis to be positive.

\begin{align*} t^2=\mathrm{\frac{\frac{1}{2}{\Delta y}}{a}} \\ t=\sqrt{\mathrm{\frac{2\Delta y}{a}}} \\ t=\sqrt{\mathrm{\frac{2\cdot11.5\, m}{9.81\, \frac{m}{s^2}}}} \\ t=1.53\, \mathrm{s} \end{align*}

The ball’s journey to the ground lasts \(1.53 \, \mathrm{s}\), uniformly accelerating during this fall.

Before we wrap up our discussion, let’s walk through one more uniformly accelerated motion example, this time applying the kinematics equations we reviewed earlier.

A particle moves according to the velocity function \(v(t)=4.2t-8\). What is the particle’s net displacement after traveling for \(5.0\, \mathrm{s}\)? What is the particle’s acceleration during this time frame?

This problem has two parts. Let’s start with determining the net displacement \(\Delta x\). We know that the value of \(\Delta x\) is related to the velocity function as the area underneath the curve on a graph. The term “area” should remind you that we can integrate the velocity function over the time interval, in this case \(\Delta t=5\, \mathrm{s}\), to calculate the displacement:

\begin{align*} \Delta x=\int_{0}^{5}4.2t-8\, \mathrm{d}t =\frac{21t^2}{10}-8t \\ \Delta x=\frac{21(5)^2}{10}-8(5)-0\\ \Delta x= 12.5\, \mathrm{m} \end{align*}

With calculus, we don’t need to graph our velocity function to have found the displacement, but visualizing the problem can help us check that our answers make sense. Let’s graph \(v(t)\) from (\(t_0=0\, \mathrm{s}\) to (\(t_1=5\, \mathrm{s}\).

We can observe there’s some “negative area” during the first part of its movement. In other words, the particle had a negative velocity and direction of motion during this time. Since the net displacement takes the direction of motion into account, we subtract this area instead of adding it. The velocity is exactly zero at:

\begin{align*}0=4.2t-8 \\ t=1.9\, \mathrm{s} \end{align*}

or more precisely, \(\frac{40}{21}\, \mathrm{s} \). We can quickly double-check our integration above by calculating the area of each triangle by hand:

\begin{align*}\mathrm{A_1=\frac{1}{2}\cdot \frac{40}{21}\, s \cdot -8\, \frac{m}{s} = \frac{-160}{21}\, m} \\ \mathrm{A_2=\frac{1}{2} \cdot (5\, s-\frac{40}{21}\, s) \cdot 13\, \frac{m}{s} = \frac{845}{42} m} \\ \mathrm{A_{net}= \Delta x= \frac{845}{42}\, m-\frac{160}{21}\, m =12.5\, m} \end{align*}

We end up with the same displacement, as expected. Finally, we can calculate the value of acceleration using our kinematics equation with initial velocity, final velocity, and time:

\begin{align*}a=\frac{v-v_0}{t} \\ a=\mathrm{\frac{13\, \frac{m}{s}-(-8\, \frac{m}{s})}{5\, s}} \\ a=4.2\, \mathrm{\frac{m}{s^2}} \end{align*}

The derivative of the velocity equation also confirms this value:

\begin{align*}a=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}(4.2t-8)=4.2\, \mathrm{\frac{m}{s^2}} \end{align*}

Uniformly accelerated motion is a crucial component of our early studies in kinematics and mechanics, the physics of motion that governs much of our everyday experiences. Knowing how to recognize uniform acceleration as well as how to approach these problems is an early step toward bettering your understanding of the universe as a whole!

Uniformly Accelerated Motion - Key takeaways

• Acceleration is mathematically defined as the first derivative of the velocity with respect to time and the second derivative of the position with respect to time.
• Uniform motion is the movement of an object whose velocity is constant and acceleration is zero.
• Uniformly accelerated motion is the movement of an object whose acceleration does not change with the passage of time.
• Downward acceleration due to gravity of falling objects is the most common example of uniformly accelerated motion.
• The area under a velocity-time graph gives us the change in displacement, and the area under an acceleration-time graph gives us the change in velocity.