Long distance runners try to maintain constant velocity with very little acceleration or deceleration to conserve energy.

### Displacement During Constant Acceleration

When acceleration is constant, there are three equations that relate displacement to two of the other three quantities we use to describe motion – time, velocity, and acceleration. These equations only work when acceleration is constant, but there are, fortunately, quite a few cases of motion where the acceleration is constant. One of the most common, if we ignore air resistance, are objects falling due to gravity.

When an object is moving with constant velocity, the displacement can be found by multiplying the velocity by the time interval, as shown in the equation below

**\begin{align*}d = vt\end{align*} d=vt**

If the object is moving with constant acceleration, but not a constant velocity, we can use a derivation of this equation. Instead of using *v*, as velocity, we must calculate and use the average velocity using this equation:

**\begin{align*}v_{\text{ave}} = \frac{1}{2} (v_f + v_i)\end{align*} vave=12(vf+vi)**

The distance, then, for uniformly accelerating motion can be found by multiplying the average velocity by the time.

**\begin{align*}d = \frac{1}{2}(v_f + v_i)(t) \qquad (\text{Equation} \ 1)\end{align*} d=12(vf+vi)(t)(Equation 1)**

We know that the final velocity for constantly accelerated motion can be found by multiplying the acceleration times time and adding the result to the initial velocity, **\begin{align*}v_f = v_i + at\end{align*} vf=vi+at. **

The second equation that relates displacement, time, initial velocity, and final velocity is generated by substituting this equation into equation 1.

Start by distributing the 1/2 in equation 1 through:

\begin{align*}d &= \frac{1}{2} (v_f + v_i)(t) = \frac{1}{2} v_ft + \frac{1}{2} v_it \\ \end{align*}

We know that **\begin{align*}v_f = v_i + at\end{align*} vf=vi+at**. Therefore:

\begin{align*}\frac{1}{2} v_it + \frac{1}{2}(t)(v_i + at)
\end{align*}

\begin{align*}= \frac{1}{2} v_it + \frac{1}{2} v_it + \frac{1}{2} at^2\\
\end{align*}

\begin{align*}= v_it + \frac{1}{2} at^2 \qquad (\text{Equation} \ 2)\end{align*}

The third equation is formed by combining \begin{align*}v_f = v_i + at\end{align*}

\begin{align*}d=\left(\frac{1}{2}\right) (v_f+v_i) \left(\frac{v_f-v_i}{a}\right)=\left(\frac{1}{2}\right) \left(\frac{v^2_f-v^2_i}{a}\right)\end{align*}

And solving for \begin{align*}v{_f}^2\end{align*} yields

**\begin{align*}v{_f}^2 = v{_i}^2 + 2ad \qquad (\text{Equation} \ 3)\end{align*}**

Keep in mind that these three equations are only valid when acceleration is constant. In many cases, the initial velocity can be set to zero and that simplifies the three equations considerably. When acceleration is constant and the initial velocity is zero, the equations can be simplified to:

\begin{align*}d &= \frac{1}{2} \ v_f \ t\\ d &= \frac{1}{2} \ at^2 \ \text{and}\\ v{_f}^2 &= 2ad.\end{align*}

**Example: ** Suppose a planner is designing an airport for small airplanes. Such planes must reach a speed of 56 m/s before takeoff and can accelerate at 12.0 m/s^{2}. What is the minimum length for the runway of this airport?

**Solution: **The acceleration in this problem is constant and the initial velocity of the airplane is zero. Therefore, we can use the equation \begin{align*}v{_f}^2 = 2ad\end{align*} and solve for \begin{align*}d\end{align*}.

\begin{align*}d=\frac{v{_f}^2}{2a}=\frac{(56 \ \text{m/s})^2}{(2)(12.0 \ \text{m/s}^2)}=130 \ \text{m}\end{align*}

**Example: ** How long does it take a car to travel 30.0 m if it accelerates from rest at a rate of 2.00 m/s^{2}?

**Solution: ** The acceleration in this problem is constant and the initial velocity is zero, therefore, we can use \begin{align*}d = \frac{1}{2} at^2\end{align*} solved for \begin{align*}t\end{align*}.

\begin{align*}t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{(2)(30.0 \ \text{m})}{2.00 \ \text{m/s}^2}}=5.48 \ \text{s}\end{align*}

**Example: ** A baseball pitcher throws a fastball with a speed of 30.0 m/s. Assume the acceleration is uniform and the distance through which the ball is accelerated is 3.50 m. What is the acceleration?

**Solution: ** Since the acceleration is uniform and the initial velocity is zero, we can use \begin{align*}v{_f}^2 = 2ad\end{align*} solve for \begin{align*}a\end{align*}.

\begin{align*}a=\frac{v_f^2}{2d}=\frac{(30.0 \ \text{m/s})^2}{(2)(3.50 \ \text{m})}=\frac{900. \ \text{m}^2/\text{s}^2}{7.00 \ \text{m}}=129 \ \text{m/s}^2\end{align*}

Suppose we plot the velocity versus time graph for an object undergoing uniform acceleration. In this first case, we will assume the object started from rest.

If the object has a uniform acceleration of 6.0 m/s^{2} and started from rest, then each succeeding second, the velocity will increase by 6.0 m/s. Here is the table of values and the graph.

In displacement versus time graphs, the slope of the line is the velocity of the object. In this case of a velocity versus time graph, the slope of the line is the acceleration. If you take any segment of this line and determine the \begin{align*}\Delta y\end{align*} to \begin{align*}\Delta x\end{align*} ratio, you will get 6.0 m/s^{2} which we know to be the constant acceleration of this object.

We know from geometry that the area of a triangle is calculated by multiplying one-half the base times the height. The area under the curve in the image above is the area of the triangle shown below. The area of this triangle would be calculated by \begin{align*}\text{area} = \left(\frac{1}{2}\right)(6.0 \ \text{s})(36 \ \text{m/s}) = 108 \ \text{m}\end{align*}.

By going back to equation 2, we know that \begin{align*}\text{displacement}= \frac{1}{2} at^2\end{align*}. Using this equation, we can determine that the displacement of this object in the first 6 seconds of travel is \begin{align*}\text{displacement}= \left(\frac{1}{2}\right)(6.0 \ \text{m/s}^2)(6.0 \ \text{s})^2 = 108 \ \text{m}\end{align*}.

It is not coincidental that this number is the same as the area of the triangle. In fact, the area underneath the curve in a velocity versus time graph is always equal to the displacement that occurs during that time interval.

#### Summary

- There are three equations we can use when acceleration is constant to relate displacement to two of the other three quantities we use to describe motion – time, velocity, and acceleration:
- \begin{align*}d = \left(\frac{1}{2}\right)(v_f + v_i)(t)\end{align*} (Equation 1)
- \begin{align*}d = v_it + \frac{1}{2} at^2\end{align*} (Equation 2)
- \begin{align*}v{_f}^2 = v{_i}^2 + 2ad\end{align*} (Equation 3)

- When the initial velocity of the object is zero, these three equations become:
- \begin{align*}d = \left(\frac{1}{2} \right)(v_f)(t)\end{align*} (Equation 1’)
- \begin{align*}d = \frac{1}{2} at^2\end{align*} (Equation 2’)
- \begin{align*}v{_f}^2 = 2ad\end{align*} (Equation 3’)

- The slope of a velocity versus time graph is the acceleration of the object.
- The area under the curve of a velocity versus time graph is the displacement that occurs during the given time interval.

#### Practice

*Questions*

Use this resource to answer the questions that follow.

https://www.youtube.com/watch?v=d-_eqgj5-K8

- For the example in the video, what acceleration is used?
- What time period is used in the example?
- What does the slope of the line in the graph represent?
- What does the area under the curve of the line represent?

#### Review

*Questions*

- An airplane accelerates with a constant rate of 3.0 m/s
^{2}starting at a velocity of 21 m/s. If the distance traveled during this acceleration was 535 m, what is the final velocity? - An car is brought to rest in a distance of 484 m using a constant acceleration of -8.0 m/s
^{2}. What was the velocity of the car when the acceleration first began? - An airplane starts from rest and accelerates at a constant 3.00 m/s
^{2}for 20.0 s. What is its displacement in this time? - A driver brings a car to a full stop in 2.0 s.
- If the car was initially traveling at 22 m/s, what was the acceleration?
- How far did the car travel during braking?