NWP Learning Commons: Physics: Linear Motion

Some of the content of this guide was modeled after a guide published by OpenStax and has been adapted for the NWP Learning Commons in January 2022. This work is licensed under a Creative Commons BY 4.0 International License.

Displacement

Linear motion is movement along a straight line or one-dimensional movement. For these examples, we will be using the x-axis from the cartesian coordinate plane, but you can just as easily assume the y-axis or any other arbitrary straight line plane.

Displacement is the difference between the end point and the starting point along a straight line. The diagram below shows a dog starting at point ${x_1}=10m$ and ending at point ${x_2}=80m$. The displacement would then be calculated as $\Delta{x}={x_2}-{x_1}$ where the delta symbol represents the change of quantity.

$$\Delta{x}={x_2-x_1}$$

$$𝐹_1=10𝑁$$

Velocity

When we combine this change in linear displacement with time, we can calculate an average velocity over that interval.

This average velocity is a good estimate of the general travel, but if the change in time is reduced, the average velocity becomes more accurate to the velocity in that shorter interval in time.

As these intervals become smaller, the average velocity approaches what is known as the instantaneous velocity.

Instantaneous velocity is represented by reducing the delta t component of the average velocity equation to 0. In calculus, this is also defined as the derivative of the distance over time.

Acceleration

Following in the same fashion as velocity, acceleration is the change from one velocity at a point in time to a second velocity at a second point in time. As that change in time decreases to zero, the average acceleration becomes an instantaneous acceleration.

Average Acceleration

$$a_{avg-x}=\frac{\Delta{v_x}}{\Delta{t}}=\frac{v_{x_2}-v_{x_1}}{t_{2}-t_{1}}$$

Instantaneous Acceleration

$$a_{x}=\lim_{\Delta{t}\to{0}}\frac{\Delta{v_x}}{\Delta{t}}=\frac{dv_{x}}{dt}$$

Instantaneous Acceleration in terms of distance

$$a_{x}=\frac{dv_{x}}{dt}=\frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{d^2x}{dt^2}$$

Constant Velocity or Acceleration

When acceleration is constant, the velocity of an object can be found by taking the initial velocity and adding the constant acceleration multiplied by time.

\[v_x={v_0}_x+a_xt\] \[(eq.\,1)\]

Using this new velocity, we can create an equation for distance with respect to the constant acceleration as well. Starting with the standard calculation of average velocity as the change in distance over the change in time, set the initial time to zero and rearrange the equation for the variable x:

\[{v_{avg}}_x=\frac{x-x_0}{t-0}\longrightarrow x=x_0+{v_{avg}}_xt\] \[(eq.\,2)\]

The average velocity can also be described as the average of the velocity at the initial point in time and the velocity at the final point in time:

\[{v_{avg}}_x=\frac{{v_0}_x+v_x}{2}\] \[(eq.\,3)\]

Input equation 3 into equation 2:

\[x=x_0+\frac{{v_0}_x+v_x}{2}t\] \[(eq.\,4)\]

Separate your terms:

\[x=x_0+\frac{{v_0}_x}{2}t+\frac{v_x}{2}t\] \[(eq.\,5)\]

Input equation 1 into equation 5:

\[x=x_0+\frac{{v_0}_x}{2}t+\frac{{{v_0}_x}+a_xt}{2}t\] \[(eq.\,6)\]

Separate your terms:

\[x=x_0+\frac{{v_0}_x}{2}t+\frac{{v_0}_x}{2}t+\frac{a_xt}{2}t\] \[(eq.\,7)\]

Simplify the equation. This is the final form to find distance x with respect to time for a constant acceleration a, initial distance x, and initial velocity v:

\[x=x_0+{v_0}_xt+\frac{1}{2}a_xt^2\] \[(eq.\,8)\]

Another equation that can be rearranged from these steps removes the need to calculate time beforehand and can be very handy when looking to solve certain problems:

\[{v_x}^2={{v_0}_x}^2+2a_x(x-x_0)\] \[(eq.\,9)\]