# 02. Mechanics 1: Position, Velocity & Acceleration

## Mechanics

### Introduction

Mechanics is the part of physics that is all about motion. We’ll talk about the motion of things that you can touch — a ball, a bicycle, a car, etc. We’ll learn about how these things move, how fast they move (velocity) and how their velocity changes with time or direction (acceleration).

Also, as I like to say, the language of Physics is math, and you can’t understand the subject if you don’t speak the language. You must be proficient with algebra. We’ll use a fair bit of trigonometry, and we’ll also need a little bit of calculus. If you’ve taken (or are taking) calculus, this should be easy. If not, I’ll explain what you need to know as we go along.

So get used to reading equations — you’ll need to develop a “feel” for what they are telling you. The good news is that physics will make you more proficient with algebra!

### Position, Velocity, Acceleration

Knowing where your object is — we call that

*position.*To represent it, we use position variables, such as the 3 Cartesisan coordinates: \(x,\ y,\ z.\) We’ll start with just one dimension at a time, and typically we’ll choose \(x\) for horizontal motion or \(y\) for vertical motion.

The measure of how an object’s position changes with time is called *velocity.* We write it like so

$$v = \frac{\Delta x}{\Delta t}

\qquad {\rm or} \qquad

v = \frac{\Delta y}{\Delta t}

$$

where we are using a Greek letter \(\Delta\) (delta) to mean “a change in.” The measure of how an object’s velocity changes with time is called *acceleration.* We can write it like this

$$a = \frac{\Delta v}{\Delta t}$$

First, we’ll write a description of an object’s position as a function of time, like so:

$$x(t) = kt$$

so that at 1 second it’s \(k\) meters from the start, at 2 seconds it’s \(2k\) meters, etc.

So the velocity is \(\Delta x /\Delta t\) (read “change in x over change in t”), which is the slope of that line, equal to \(k.\) Notice that the formula for velocity is similar to a slope — rise over run. As we just showed, velocity is the slope of the position graph.

### Average Velocity

Now let’s calculate an average velocity. Say we drive 120 km in 1.5 hours. We know that velocity is distance divided by time, giving us a velocity of \(120 / 1.5 = 80\) kph (kilometers per hour). Next, we stop for 1 hour for lunch. Then we drive another 90 km in 1.5 hours. The velocity for the last part is \(90 / 1.5 = 60\) kph. So our trip has three pieces that have velocities 80, 0, and 60 kph. We can calculate an average velocity for the entire trip; just divide total distance by total time, which is \(210 / 4 = 52.5\) kph.

Here is a graph of the trip, plotted as distance (sometimes called *displacement*) vs. time

where the individual trip segments are shown in black and the average is in red.

### Instantaneous Velocity

We’ve been talking about calculating velocity as distance over “some” period of time (\(\Delta t\)). Instantaneous velocity is the answer to “what’s my velocity

*right now?*” You are familiar with this from a car’s speedometer. What if your position-vs-time graph looked like this:

We could still calculate an average velocity (slope in red), but the slopes of each individual segment vary wildly, and even go backwards. We could calculate the slope of each little segment. But what if the time period was made smaller and smaller? As this happens (we say, “as \(\Delta t\) goes to 0”) the simple formula \(\Delta x / \Delta t\) becomes something called a *derivative.*

### What is a derivative?

Here’s the first place where calculus comes in. What if we let the \(\Delta t\) part get smaller and smaller? As \(\Delta t\) approaches zero, we call that a

*derivative,*and we write it like this:

$$v = \frac{dx}{dt}$$

which is read “the derivative of \(x\) with respect to \(t.\)” If you’ve had calculus, you understand this. If not, you just need to understand the definition (above) and the recipe for how to calculate it (below).

Just as in the previous sections (\(\Delta x /\Delta t\)), a derivative is a slope. But because it is calculated over a very small interval (mathematicians call that interval an “infinitesimal”), a derivative can calculate the slope of any function — a straight line, a curve, a sine wave — at any point along that curve. So if you want the slope of a curve at a particular point, think of it as drawing a tangent line to the curve at that point, then calculating rise over run of that tangent line.

To calculate a derivative “with respect to \(t,\)” we need to use the exponent of \(t\) (the power that it’s raised to). So \(t\) (really \(t^1\); the 1 is assumed) has an exponent of 1, \(t^2\) has an exponent of 2, \(t^3\) has an exponent of 3, and so on. So, we write down the exponent (power) of the \(t\) variable, then write down the \(t\) variable with its exponent reduced by 1. So for example, the derivative of \(t^2\) is \(2t^1\) or just \(2t.\)

The first few powers of \(t\) and their derivatives are:

\(f(t)\) | \(\frac{df}{dt}\) |

\(t^0\) | 0 |

\(t^1\) | \((1) t^0 = 1\) |

\(t^2\) | \(2t^1 = 2t\) |

\(t^3\) | \(3t^2\) |

\(t^4\) | \(4t^3\) |

and so on. This “recipe” only works for polynomials — other functions like sines and cosines we’ll get to later.

OK, so let’s use our new knowledge. Above, we had \(x(t) = kt,\) which can be written \(kt^1.\) Note that the \(k\) is just a multiplied constant and as far as the derivative is concerned, it just “comes along for the ride.” Since the exponent of \(t\) is 1, so we get:

$$\frac{dx}{dt} = (1) k t^0$$

and since \(t^0 = 1,\) we have \(dx/dt = k,\) which is the slope of the line.

### Speed, Velocity and Acceleration

We sometimes (inaccurately) use the words speed and velocity. What’s the difference? To answer that question, we need to introduce a tiny bit more math. We won’t go too deeply right now, just a definition — more later.

What we need here is called a *vector.* A vector is a quantity that has a *magnitude* and a *direction.* I’ll explain more by example. Say we’re traveling on a highway at 65 mph. That’s our *speed.* Just a number. But if we said we were traveling 65 mph east, that’s our *velocity.* We’ve just defined a vector — magnitude and direction. So, speed is just a number (called a *scalar*); velocity is a vector.

So remember that we said acceleration is defined as change in velocity over time. This means either a change in magnitude or direction (or both). A change in magnitude would be speeding up or slowing down; a change in direction would be turning a corner. Either is acceleration.

So if we have an equation for position vs. time, we can calculate the velocity by taking the derivative of the position function. We can then calculate the acceleration by taking the derivative of the velocity function. Position, velocity and acceleration are related by derivatives. The progression looks like this:

$${\rm position} \stackrel{\rm derivative}{\longrightarrow} \ {\rm velocity} \stackrel{\rm derivative}{\longrightarrow}\ {\rm acceleration}$$

### Putting it all Together — With Numbers and Pictures!

Let’s go through an example using simple graphs and simple numbers to illustrate. What if we were moving in a car at some constant velocity, say 10 m/s. A constant velocity of 10 m/s means that for every second, we travel another 10 m. So after 1 s, we’ve traveled 10 m, then after 2 s, we’ve moved 20 m, 3 s is 30 m, and so on.

What would that look like on a graph? Similar to above, but now we have a number for the slope:

Above, we said that the velocity is the slope of the position graph. We can calculate that as rise/run and in this case, we get 10. We can also write a simple position equation, \(x(t),\) which is read, “x as a function of time,”

$$x(t) = 10t$$

Then, the velocity as a function of time, \(v(t) = dx/dt.\) Remember that \(t\) has an exponent of 1, and the “recipe” for taking a derivative is: multiply by the current exponent (1) and reduce the power by one, so we get

$$\frac{d(10t^1)}{dt} = (1) 10 t^0 = 10$$

What does that look like as a graph?

This graph shows the slope of the position graph, which is a constant number, 10. Then, the acceleration is the slope of this graph. The slope of a horizontal line is zero, which tells us that there is no acceleration in this case. Also, we just learned that the derivative of a constant is zero. Remember that acceleration is the change in velocity, and we said that our car is moving with constant velocity (change = 0). We could graph the acceleration as a horizontal line at zero (not shown).

So to summarize, the position graph is a line with a slope, velocity is a flat line with zero slope, and acceleration is zero. Constant velocity means zero acceleration, and vice versa.

### Examples

- Say we traveled 100 km in 2 hrs. What would be our average velocity?

Simply calculate \(\Delta x / \Delta t,\) which gives 100 / 2 = 50 kph.

- If we had a projectile whose motion was described by \(x(t) = 4t^2,\) what would be its instantaneous velocity?

We can answer this by taking the derivative, like so:

$$ v(t) = \frac{dx}{dt} = \frac{d}{dt} \big(4t^2\big) = 8t$$

So at any moment in time, the instantaneous velocity is \(v(t) = 8t.\) We can go one step further and calculate the acceleration, like so:

$$ a(t) = \frac{dv}{dt} = \frac{d}{dt} \big(8t\big) = 8$$

So at any moment in time, the acceleration is 8 m/s\(^2\). This is called*constant*acceleration — it doesn’t change with time. - If we had a boat whose motion was described by \(x(t) = 6t^2 – 3t + 2,\) what would be its instantaneous velocity and acceleration?
We can answer this by taking the derivative, like so:

$$ v(t) = \frac{dx}{dt} = \frac{d}{dt} \big(6t^2 – 3t + 2\big) = 12t – 3$$

because the derivative of a constant is zero.

So at any moment in time, the instantaneous velocity is \(v(t) = (12t – 3)\) m/s. Now calculate the acceleration,

$$ a(t) = \frac{dv}{dt} = \frac{d}{dt} \big(12t – 3\big) = 12$$

So at any moment in time, the acceleration is 12 m/s^{2}. Again, constant acceleration. - Say we had a vehicle whose position was described by \(y(t) = 3t^3 – 2t^2 + 7t.\) What are the velocity and acceleration of this vehicle, as a function of time?

Again, take the derivatives, like this for velocity:

$$v(t) = \frac{dy}{dt} = \frac{d}{dt} \big(3t^3 – 2t^2 + 7t\big) = 9t^2 – 4t + 7$$

and this for acceleration

$$a(t) = \frac{dv}{dt} = \frac{d}{dt} \big(9t^2 – 4t + 7\big) = 18t – 4$$

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Had two stumbling blocks: “Notice that the formulas for velocity and acceleration are both similar to a slope…” graph for the velcoty was clear yet did not see one for acceleration to see similarity.

the example of v=dx/dt=d(4t^2)/dt=8t…where did 4t^2 come from? maybe a picture would help?

I hope these are the comments you are looking for, but other than that this has been GREAT so far.

Answers:

(1) Acceleration would be similar to velocity. Since velocity is a straight line, the acceleration would be a single number, so I didn’t graph it.

(2) the 4t^2 comes from the problem statement: “If we had a projectile whose motion was described by x = 4t^2 …”

Hope this helps. YES! these are exactly the comments I’m looking for! Please tell me more about what is not clear. Thanks for the comments and the kind words.