# Equations in Physics

What is the use of equations in the real world? Why do physicists describe physics in the language of mathematical equations? What is so special about mathematics that we can use it in the world around us? What are some examples of physics equations, what do they tell us, and how do we solve them? Learn about it in this article!

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Contents

## Definition of an equation in physics

There is a definition of an equation (whether in physics or not), but it is very technical and not enlightening at all. What we can say about physics equations is what they do. Equations in physics describe relations between physical quantities. An equation always contains an "equals"-sign, $$=$$.

There is an equation in physics describing the relationship between your speed, the duration of your travel, and the distance you cover. The bigger your speed, the more distance you cover in the same amount of time. The equation will say: $$\text{distance covered}=\text{speed}\times \text{duration of travel}}$$. We see that a doubling in travel time means a doubling in distance covered if the speed remains the same and that a doubling in speed means a doubling in distance covered if the travel time remains the same.

The physics equation we introduced above looks a bit unclear: there are so many letters that the relations (the multiplication- and equal-signs) fade to the background. However, relations are very important in physics. This is why people assign symbols to quantities. It is important to explain to other people what your symbols mean, and it is convenient to use the standard symbols for things that have standard symbols.

To make the equation look clearer, we can give names to the quantities involved in the problem. Let's say that the distance covered is $$d$$ (for distance), our speed is $$v$$ (for velocity), and the duration of our travel is $$t$$ (for time). The equation is now $$d=vt$$.

We omitted the multiplication cross because it is clear that$v$and$t$are two different things. Putting nothing between two quantities means that we multiply them. This equation looks much clearer, and it is not hard to instantly remember that $$d$$ stands for distance, for example. This is good, we just wrote down a physics equation!

Fig. 1: An equation that is always true (also called a tautology).

## Types of physics equations

You may have the impression that there are many different types of physics equations because they all look so different. While it is true that there are many areas of physics that all have their own equations, every equation in physics has the same purpose, namely to describe a relation between quantities. However, we can make a distinction between equations based on how the quantities in the equation present themselves, as the following examples show.

• A linear equation only contains quantities to their first power, e.g. the equation for distance, $$d=vt$$.
• A quadratic equation also contains quantities that are squared, e.g. the equation for kinetic energy, $$E=mv^2/2$$.

We also have differential equations (containing derivatives of quantities), tensor equations (containing quantities with multiple inputs, just like vectors), and many more ways that quantities can appear in equations.

## Examples of physics equations

Fig. 2: Einstein's equations about gravity.

The relation between how much force $$F$$ you exert on an object, the mass $$m$$ of the object, and how fast the object will accelerate (described by the acceleration $$a$$) as a result of the Force you exert is given by the equation

$F=ma$

This is Newton's second law. For example, we see that we need to exert a bigger Force for a bigger acceleration to happen, or to move an object with a bigger mass with the same acceleration.

The relation between the kinetic energy $$E$$ of an object, its mass $$m$$, and its speed $$v$$ is given by the equation

$E=\dfrac{1}{2}mv^2.$

This tells us that a doubling in an object's mass will result in a doubling of its kinetic energy, but a doubling in its speed will result in a quadrupling of its kinetic energy. To see this, assume that the end velocity $$v_e$$ is twice the initial velocity $$v_i$$, so $$v_e=2v_i$$. The end energy dividedby the initial energy is then

$\dfrac{E_e}{E_i}=\dfrac{\frac{1}{2}mv_e^2}{\frac{1}{2}mv_i^2}=\bigg( \dfrac{v_e}{v_i}\bigg)^2=2^2=4.$

We see that, indeed, the energy quadruples if the speed doubles. This is the main cause of why a car accelerates slower at high speeds than at low speeds: it needs to add more energy to gain the same amount of speed because of the square in the equation above!

We have seen a good amount of physics equations, but it is now time to work with them and solve them.

## Solving equations in physics

Fig. 3: Edward Witten writing equations about string theory on a chalkboard.

Solving equations in physics is very similar to solving equations in mathematics, except for two major differences.

• In physics, there is always a context, so a physics equation follows from a little story. We have to unpack the story and convert it to a solvable equation.
• We have to be careful about units. In general, we include units in all calculations.

Let's see how this works with a sample problem.

Q: John always walks at a speed of $$3\,\,\mathrm{mi/h}$$( miles per hour). Last Sunday, he walked around town, and he ended up covering a total of $$9 \,\,\mathrm{ mi}$$. How long did it take him?

A: The first step is to convert this story to a solvable equation. We start by assigning names to the quantities we need. We call John's walking speed $$v$$, his covered distance $$d$$, and how long it took him $$t$$. The facts that it was a Sunday and that he walked around town do not matter to the question. From the story, we know that $$v=3\,\,\mathrm{mi/h}$$, and that $$d=9\,\,\mathrm{mi}$$. We now need to find an equation that relates $$t$$ to $$v$$ and $$d$$. Luckily, we know that $$d=vt$$. Just as with mathematical equations, we isolate $$t$$ by dividing both sides by $$v$$ and we are left with the equation $$t=d/v$$. We know $$d$$ and $$v$$, so this is a solvable equation, so step 1 is done! Let's solve it while being careful with units. We fill in:

$t=\dfrac{9\,\,\mathrm{mi}}{3\,\,\mathrm{mi/h}}=\dfrac{9}{3}\dfrac{\mathrm{mi}}{\frac{\mathrm{mi}}{\mathrm{h}}}=3\,\,\mathrm{h}.$

Now let's make sure to answer the question fully. It took John 3 hours (to walk around town last Sunday).

In the example above, we did not have any problems with the units because the information given to us was in units that "go well together". If this is not the case, we will have to use our knowledge of unit conversions, as shown in the example below.

Q: Anna wants to push a car of which we can assume that it has no friction and that it is not on any slope. She pushes with a force of $$500\,\,\mathrm{N}$$, the mass of the car is $$4000\,\,\mathrm{lbs}$$, and Anna's mass is $$170\,\,\mathrm{lbs}$$. What is the car's acceleration?

A: We call the exerted force $$F=500\,\,\mathrm{N}$$, the mass of the car $$m=4000\,\,\mathrm{lbs}$$, the car's acceleration $$a$$, and Anna's mass $$m_A=170\,\,\mathrm{lbs}$$. We know that $$F=ma$$, so we isolate the acceleration and get the equation $$a=F/m$$. We know $$F$$ and $$m$$, so step 1 is done.

Let's solve it by filling in the data (including the units!):

\begin{aligned}a=&\dfrac{500\,\,\mathrm{N}}{4000\,\,\mathrm{lbs}}=\dfrac{500}{4000}\dfrac{\frac{\mathrm{kg\cdot m}}{\mathrm{s^2}}}{\mathrm{lbs}} \\ =& 0,125\,\,\dfrac{\mathrm{kg}}{\mathrm{lbs}}\cdot \dfrac{\mathrm{m}}{\mathrm{s^2}}=0,125\cdot 2,20\,\,\dfrac{\mathrm{m}}{\mathrm{s^2}}\\ =&0,276\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\end{aligned}.

We see that we needed the conversion between $$\mathrm{kg}$$ and $$\mathrm{lbs}$$, namely that $$1\,\,\mathrm{kg}=2,20\,\,\mathrm{lbs}$$, to answer the question. The answer is that the car's acceleration is $$0,276\,\,\mathrm{m/s^2}$$.

Hopefully, these examples have made it clear how the language of mathematical equations is useful in the real (physical) world.

### Summary: step by step

1. While reading the question, give names to the mentioned quantities and write down their values (with units).
2. Determine what quantity you need to know to answer the question, and write down the physics equation that relates that quantity to quantities that you have written down.
3. Solve the equation (while being careful with units) and answer the question in a complete sentence and with the correct units.

## Equations of motion in physics

Fig. 4: The Schrödinger equation describing quantum mechanics.

In the end, all physics aims to do is predict the motion of objects. Therefore, we could argue that all equations in physics are in some way equations of motion. However, some equations describe only the motion of objects and not the causes of the motion. Let's look at these equations.

We first look at the equation that governs motion with constant speed. For a starting point $$x_0$$, a velocity $$v$$, and a travel duration $$t$$, the location $$x$$ of an object is given by

$x=x_0+vt.$

From earlier, we know that distance covered is $$d=vt$$, so this is logical: the location of an object is its starting location plus the distance it has covered.

An object can also accelerate. For a constant acceleration $$a$$, a starting velocity $$v_0$$, a starting point $$x_0$$, and a travel duration $$t$$, the location$$x$$ of an object is given by

$x=x_0+v_0t + \dfrac{1}{2}at^2.$

We already have some intuition about why the $$x_0$$ is there, but let's examine the rest. The speed gained after a time $$t$$ is $$at$$, so after a time $$t$$, the speed of our object is $$v_0+at$$. This means that the average speed of our object is the average of $$v_0$$ and $$v_0+at$$, which is $$v_0+at/2$$. The distance covered is then

$d=\bigg(v_0+\dfrac{at}{2}\bigg)t=v_0t+\dfrac{1}{2}at^2,$

and we arrive at the equation above! Note that for zero acceleration, we arrive at the equation for a constant speed again, which is what we expect to find.

## Equations in Physics - Key takeaways

• An equation in physics describes a relation between Physical Quantities.
• An equation always contains an "equals"-sign, $$=$$.
• We assign symbols to quantities to make the equations readable.
• Solving equations in physics is largely the same as solving equations in mathematics, with 2 differences:
• Often, we have to set up the equations ourselves from a story. While reading the story, we give names to the relevant quantities and list their values.
• We always include units in our calculations. When the equation is solved, we provide the answer in the correct units.
• The equation of motion for objects that have a constant speed is $$x=x_0+vt$$.
• The equation of motion for objects that have a constant acceleration is $$x=x_0+v_0t+\dfrac{1}{2}at^2$$.

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What is physics equation?

A physics equation is an equation that is used within the field of physics. In nature, a lot of questions or problems involve quantities and numbers, and the answer to the questions can often be found by solving a physics equation.

What are the 3 physics equations?

There are no standard 3 physics equations. Instead, there are thousands of equations in physics, and every subfield has its own set of equations.

What are some examples of physics equations?

Some examples of physics equations are:
F=ma, x=x0+vt, and E=mv2/2. These describe force, constant speed, and kinetic energy respectively.

How to solve physics equations.

Solving a physics equation is the same as solving a "normal" equation: isolate the variable you want to solve for and fill in values for the other variables. However, in physics, it is important to include all units in your calculations and be careful about unit conversions.

What is physics equation of motions?

In physics, equations of motion describe the motion of an object. For example, if an object has a constant speed of 10 miles per hour, its equation of motion would be:
(#miles covered) = 10(#hours traveled).

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