Linear Motion Equations | Kinematics | Linear Motion Formulas

What: “Kinematics” deals with objects in motion. Linear motion means that the objects move in a straight line, which simplifies the mathematics.

Why: If you know some information about the movement of an object you can figure out other details.

How: You use two basic linear motion equations. They include terms where you can use calculus for more complicated situations (velocity and acceleration)

1.Two fundamental equations

v = v0 + at

x = x0 + v0t + (1/2)at2

What:

These are two very useful equations that you can use for early linear motion problems.

Why: If you understand these equations and have them memorized, you can solve many problems without memorizing a lot more equations. In general, I prefer to memorize less if possible.

How: Certain details of a situation, such as starting at “zero” or having a zero acceleration will simplify these equations. And you can also combine these two equations to do more.

**Wait…..** I thought I needed 10 equations, not 2. If you understand all the parts of these two equations and can combine them, you will generally not need more equations for many problems. If you

2. Parts of the equation

v: velocity (later on, after v0)

v0: starting velocity, “v naught”

a: acceleration

t: time

x: position

x0: starting position

How: you see some of the variables show up more than once in these equations. In the second equation, you do see the time is squared in the final term. Sometimes gravity is the acceleration that you will use.

Note: you will want to use SI units,

meters and seconds

If you do that, the outputs will also be in SI units

3. Examples

Simple example:

A person runs at a velocity of 5 m/s, how long will it take to run 20 m?

Use

x = x0 + v0t + (1/2)at2

You can take the starting position to be 0 and there is no acceleration mentioned, so you take that to also be 0.

The equation simplifies to

x = v0t

“How long” asks for the time, so you want to solve the equation for time

x = v0t

Divide both sides by v0

x/v0 = t

20 m/(5 m/s) = 4 s

Harder example:

A rocket ascends vertically (from rest) with an acceleration of 4.00 m/s^2 until it runs out of fuel at an altitude of 1000 m (1.00 x 10^3 m). At that point, its new acceleration is due to gravity, downward.

a) What is the velocity of the rocket when it runs out of fuel?

v = v0 + at

x = x0 + v0t + (1/2)at2

For many problems that are a bit more complicated, you will use the two kinematic equations.

Let’s start with the second equation

1000 m = 0 + 0 + (1/2)at2

1000 m = (1/2)at^2

2000 m = (4 m/s^2)t2

5000 s^2 = t^2

t = 500^(1/2) s

~22.4 s

Note! Keep this stored in memory, you don’t want to round until the end.

We can put that result into the first equation to get the velocity at this point

v = 4 m/s^2 * 500^(1/2) s

v ~ 89.4 m/s

b) What is the maximum altitude of the rocket?

Probably need to use both equations, definitely need the equation with distance

To keep track carefully, it can be useful to use subscripts for variables, especially since there are two stages of motion here (with fuel and without)

red and blue!

x = x0 + v0t + (1/2)at2

x = 1000 m + v0t – (1/2) 9.8 t2

v = v0 + at

0 m/s = 89.4 m/s – 9.8t

-89.4 m/s = – 9.8t

(-89.4 m/s)/(-9.8 m/s^2) = t

t ~ 9.1 s

x = x0 + v0t + (1/2)at^2

= 1000 m + v0 + (1/2)at^2

1000 m + 89.4(9.1) + 0.5 * -9.8 * 9.1^2

~ m

Let me know what questions you have, I could make a followup video

saif says

2000 m = (4 m/s^2)t2

5000 s^2 = t^2

t = 5000^(1/2) s

~70.7 s

that sounds wrong because (1000*2)/4=500

and square root of that is 22.3s

Neal says

Thanks for catching that.