## What is linear motion?

**Linear motion **is any motion along a straight line. In other words, the object in motion does not change its direction. It can be mathematically described as using only one dimension. According to Newton’s first law of motion, an object will keep moving in a straight line until an external force changes its direction. Under normal circumstances, an object will be affected by many external forces, and thus will not move in a straight line.

**Displacement **is the distance between a moving object’s starting and finishing point. For linear motion, the displacement is always equal to the distance the object has traveled.

## Types of linear motion

So, to summarize, two requirements need to be fulfilled: the object needs to move in a straight line, and the acceleration needs to be constant. Based on the second one, we have two types of linear motion.

### Completely uniform linear motion

The first one is **completely uniform linear motion**. In this type, other than the direction, the velocity is also constant. In other words, there is no acceleration (a=0). This is the most primitive type of motion. However, you will almost never encounter it, unless in a controlled environment.

### Accelerated motion with uniform acceleration

The second type is **accelerated motion**, with **uniform acceleration**. This time, the velocity is constantly changing, but at a consistent rate (a≠0). This type of motion can be achieved during a free fall in a vacuum.

Although rare, there are examples of linear motion in our daily lives, such as elevators. We can split their motion into three stages. At first, they start moving from a standstill. They accelerate at a high rate until they achieve a certain velocity. From that point, they move without acceleration, until they almost reach the end. Before the end of their intended path, they decelerate at a high rate, and stop right at the finish line. As we can see, the first stage is with uniform acceleration, the second with no acceleration, and the third with uniform deceleration. If you want to learn more about acceleration, check out our Acceleration Calculator!

## Motion under force of gravity

Any motion where the only thing acting upon an object is gravity is called a free fall. During a free fall in a vacuum, the object accelerates at a rate of 9.80664 m/s^{2} or 32.17405 ft/s^{2}. This fulfills all the requirements in order to be classified as linear motion.

If the object is not in a vacuum, however, the acceleration will not be linear. Thus, the motion can’t be classified as linear.

## Calculations for linear motion

When talking about linear motion, we need to consider a few values. Those are: the initial velocity (**v _{0}**), the final velocity (

**v**), the displacement (

**s**), the acceleration (

**a**) and the time (

**t**).

First, I’m going to give you all the formulas and then we’re going to do some examples.

v=v_{0}+a*tAlternatively, if the time is unknown, we can use:

v^{2}=v_{0}^{2}+2*a*sIt is important to keep track of all your basic units, because that’s what we use when calculating. For velocity, the basic unit is m/s. For acceleration, it’s m/s^{2}, for displacement it’s meters, and for time it’s seconds.

Alternatively, if the acceleration is unknown, we can use:

s=\frac {(v_{0}+v)*t}{2}Now that we know all the formulas, we can go over some examples.

## Linear motion examples

Let’s assume a particle started linear motion from a standstill, and is accelerating at a rate of *5 m/s ^{2}*. After

*10 seconds*, what will its velocity be, and how much distance will it have passed?

Keep in mind, since it started moving from a standstill, v_{0} will be 0. This means we can eliminate it from the equation:

For the displacement, we can use either formula, but we are going to use the simpler one. Once again, because v_{0} is 0, we can remove it from the equation:

Let’s do one that’s more complex. This time, we’re going to assume a particle moving at a consistent velocity of *5 m/s*. After *8 seconds*, it starts accelerating at a consistent rate of* 4 m/s ^{2}*.

*10 seconds*after that, it suddenly stops moving. What was its peak velocity, and what distance did it travel?

We can split this calculation in two segments: pre-acceleration and post-acceleration. For pre-acceleration, we need to find the displacement. Because this is completely uniform linear motion (no acceleration), we can use this formula for displacement:

s= v*tWith this in mind, we can calculate:

s_{1}=5 \frac {m}{s}*8s=40mNow, for the post-acceleration segment. We need to calculate the displacement and the peak velocity:

v=5 \frac {m}{s}+4 \frac {m}{s^{2}}*10s=45 \frac {m}{s}Now, the only thing left to calculate is the displacement:

s_{2}=5 \frac {m}{s}*10s+ \frac {4 \frac{m}{s^{2}}*(10s)^{2}}{2}=250mNow we just need to add the two displacement calculations up, and we will have the final value:

s= s_{1}+s_{2} \rightarrow s= 40m + 250m= 290mWe hope this helps you better understand how to use our calculator!

## FAQ

**Linear motion definition**

**Linear motion **is any motion along a straight line.

**How to convert rotary motion to linear motion?**

The simplest way is to use a slider-crank mechanism.

**What are the two types of linear motion?**

The first type is completely uniform linear motion, with no acceleration, and the second one is linear motion with uniform.