## Motion Equations for Constant Acceleration in One Dimension

Read this text for more examples and practice on how to solve motion equations for constant acceleration.

### Notation, Solving for Displacement and Final Position

Figure 2.25 Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England.

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.

### Notation: t, x, v, a

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is $\Delta t=t_{\mathrm{f}}-t_{0}$, taking $t_{0}=0$ means that $\Delta t=t_{\mathrm{f}}$, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity.

That is, $x_{0}$ is the initial position and $v_{0}$ is the initial velocity. We put no subscripts on the final values. That is, $t$ is the final time, $x$ is the final position, and $v$ is the final velocity. This gives a simpler expression for elapsed time-now, $\Delta t=t$. It also simplifies the expression for displacement, which is now $\Delta x=x-x_{0}$. Also, it simplifies the expression for change in velocity, which is now $\Delta v=v-v_{0}$. To summarize, using the simplified notation, with the initial time taken to be zero,

$\left.\begin{array}{l} \Delta t=t \\ \Delta x=x-x_{0} \\ \Delta v=v-v_{0} \end{array}\right\}$

where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.

We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is,

$\bar{a}=\, a\, = \, constant$

so we use the symbol $a$ for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant acceleration.

#### Solving for Displacement $(\Delta x)$ and Final Position $(x)$ from Average Velocity When Acceleration (a) is Constant

To get our first two new equations, we start with the definition of average velocity:

$\bar{v}=\frac{\Delta x}{\Delta t}.$

Substituting the simplified notation for $\Delta x$ and $\Delta t$ yields

$\bar{v}=\frac{x-x_{0}}{t}.$

Solving for $x$ yields

$x=x_{0}+\bar{v} t,$

where the average velocity is

$\bar{v}=\frac{v_{0}+v}{2}(\text { constant } a).$

The equation $\bar{v}=\frac{v_{0}+v}{2}$ reflects the fact that, when acceleration is constant, $v$ is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to $60 \mathrm{~km} / \mathrm{h}$, then your average velocity during this steady increase is $45 \mathrm{~km} / \mathrm{h}$. Using the equation $\bar{v}=\frac{v_{0}+v}{2}$ to check this, we see that

$\bar{v}=\frac{v_{0}+v}{2}=\frac{30 \mathrm{~km} / \mathrm{h}+60 \mathrm{~km} / \mathrm{h}}{2}=45 \mathrm{~km} / \mathrm{h},$

which seems logical.

#### Example 2.8: Calculating Displacement: How Far does the Jogger Run?

A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?

#### Strategy

Draw a sketch.

Figure 2.26

The final position $x$ is given by the equation

$x=x_{0}+\bar{v} t$

To find $x$, we identify the values of $x_{0}, \bar{v}$, and $t$ from the statement of the problem and substitute them into the equation.

#### Solution

1. Identify the knowns. $\bar{v}=4.00 \mathrm{~m} / \mathrm{s}, \Delta t=2.00 \mathrm{~min}$, and $x_{0}=0 \mathrm{~m}$

2. Enter the known values into the equation.

$x=x_{0}+\bar{v} t=0+(4.00 \mathrm{~m} / \mathrm{s})(120 \mathrm{~s})=480 \mathrm{~m}$

#### Discussion

Velocity and final displacement are both positive, which means they are in the same direction.

Source: Rice University, https://openstax.org/books/college-physics/pages/2-5-motion-equations-for-constant-acceleration-in-one-dimension