# Newton’s Laws, Part 3

What I would like to eventually get to, is to develop the equation of motion of a rocket. An equation of motion is just an equation that calculates an object’s position given a time. I did this without a lot of detail, in my Springy Thingy posts back in February. For this set of posts, I would like to add a bit more development.

Let’s go back to our basic equation that describes motion: F = ma. Let’s look at the a (acceleration) part. Acceleration is a rate of change of velocity. If a car goes from 0 to 100 km/s in 10 seconds, its acceleration is the difference in velocity divided by the time interval:

${a}\hspace{0.33em}{=}\hspace{0.33em}\frac{{100}\hspace{0.33em}{\mathrm{k}}{\mathrm{m}}{/}{\mathrm{s}}}{{10}\hspace{0.33em}{\mathrm{s}}}\hspace{0.33em}{=}\hspace{0.33em}{10}\hspace{0.33em}\frac{\mathrm{k}\mathrm{m}/\mathrm{s}}{\mathrm{s}}\hspace{0.33em}{=}\hspace{0.33em}{10}\hspace{0.33em}{\mathrm{k}\mathrm{m}/\mathrm{s}}^{2}$

This means the car increases its velocity 10 km/s each second. So acceleration is the rate that velocity changes per unit of time. However, velocity is also a rate of change measurement. Velocity is the rate of change of position (or distance) per unit of time. An equation of motion is finding the position of an object given a time.

So acceleration is the rate of change of velocity which is the rate of change of position, and it’s position that we want. How do we get there? This is where calculus comes in.

Calculus essentially deals with rate of change equations. It can find the rate of change of position, that is velocity, given a position equation. It can also go backwards and find a position equation, given a velocity equation. In our case, it can take the rate of change of velocity (acceleration) and find the velocity equation and then take the velocity equation and find the position equation, that is the equation of motion.

This ability to take a simple equation like F = ma, and from it, describe the motion of objects is one of the many reasons I love maths.

With this as a background, next time, let’s launch a rocket, then cut off it’s motors and answer questions like:

How high is the rocket when the motors cut off?

How fast is the rocket going when the rockets cut off?

How high does the rocket go?

When will the rocket hit the ground (or will it hit the ground)?

Posted on Categories Physics, Pre-VCE