Reading homework for Tuesday.

If you haven't had a chance to play with Logger Pro or the graphing program of your choice, please do that ASAP.  Then read the material below, most of which is review from class.

Intro to the mathematics of motion

Today, we are going to talk about how we think about speed and the rate of change in speed (usually called acceleration).  It is a bit math-y, but don't panic - we'll summarize things nicely in a couple of simple-to-use equations.

First, let's look at some definitions.

Average (or constant) velocity, v

v = d / t

That is, distance divided by time.  The SI units are meters per second (m/s).

* Strictly speaking, we are talking about speed, unless the distance is a straight-line and the direction is also specified (in which case "velocity" is the appropriate word).  However, we'll often use the words speed and velocity interchangeably if the motion is all in one direction (1D).

Some velocities to ponder....

Approximately....

Keep in mind that 1 m/s is approximately 2 miles/hour.

Your walking speed to class - 1-2 m/s
Running speed - 5-7 m/s
Car speed (highway) - 30 m/s
Professional baseball throwing speed - 45 m/s
Terminal velocity of skydiver - 55 m/s
Speed skiing - 60 m/s
Speed of sound (in air) - 340 m/s
Bullet speed (typical) - 900 m/s
Satellite speed (in orbit) - 6200 m/s
Escape velocity of Earth - 11,200 m/s
(That's around 7 miles per second, or 11.2 km/s)


What about.....
The Speed of light

Speed of light (in a vacuum) -

c = 299,792,458 m/s

This number is a physical constant, believed to be true everywhere in the universe. The letter c is used to represent the value being of constant celerity (speed).

By the way, it's hard to remember this exact number, and I wouldn't expect you to.  However, here are some approximations that may make it easier to keep it in mind.  The speed of light is approximately:

- 300,000,000 meters/sec

- 186,000 miles/sec

- 7 times around the Earth's equator in 1 second

- Out to the Moon in around 1 second (1.3 seconds is closer) - so, the Moon is approximately 1.3 "light seconds" away (on average)

- To the Sun in about 8 minutes - so, the Sun is approximately 8 "light-minutes" away (on average)

- To Mars in about 13 minutes, though this varies depending on the relative locations of Earth and Mars in their respective orbits.  (Consider that this is, under the best circumstances, about a 9 month one-way trip at current spaceship speeds.)

- To the nearest (non-Sun) star, Proxima Centauri B (actually part of a 3-star system, and home to a recently-discovered Earth-like planet!), in 4.24 years.  Yes, YEARS.  So, that 3-star system is around 4.3 "light years" away from us.  And that's our closest neighbors!!  See why we don't get too far in space travel?

Instantaneous Velocity

Average velocity should be distinguished from instantaneous velocity (what you get, more or less, from a speedometer):

v(inst) = d / t, where t is a very, very, very tiny time interval. There's more to be said about this sort of thing, and that's where calculus begins.

Now the idea of velocity is pretty useful if you care about the velocity at a specific time OR the average velocity for a trip. However, if you care about the details of velocity, if and when it changes, then we need to introduce a new concept: acceleration.

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Acceleration, a

a = (change in velocity) / time

a = (vf - vi) / t

Note that the i and f are subscripts.  The units here are m/s^2, or m/s/s.

Acceleration is a measure of how quickly you change your speed - that is, it's a measure of 'change in speed' per time. Imagine if you got in a car and floored it, then could watch your speedometer. Imagine now that you get up to 10 miles/hr (MPH) after 1 second, 20 MPH by the 2nd second, 30 MPH by the 3rd second, and so on. This would give you an acceleration of:

10 MPH per second. That's not a super convenient unit, but you get the idea (I hope!).

The unit we usually use for acceleration is the m/s/s or m/s^2.

EXAMPLE PROBLEMS:

1.  What is the acceleration of a car that travels from rest to 30 m/s in 4 seconds?

2.  If a falling object is subject to an acceleration of 10 m/s/s, how fast will it be traveling after 3.5 seconds, assuming that it starts from rest?   (A little algebra may be needed here.)

Read the next section if you want to be ahead for our next discussion.  If you don't have time, that's ok, too.

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The equations of motion

Recall v = d/t.  That's usually how we calculate average velocity.  However, there is another way to compute average velocity:

v = (vi + vf) / 2

where vi is the initial velocity, and vf is the final (or current) velocity.  This is the same as taking the average of two numbers, in this case, the initial and final velocities.

Knowing these equations for average velocity, as well as the definition for acceleration, allows you to relate (or calculate) the interesting things about an object's motion:  initial velocity, final velocity, displacement, acceleration, and time.

As it happens, you can do a bunch of algebra to put the equations together into more convenient forms.  I will do this for you and summarize the most useful equations.  In general, there are 2 or 3 really useful equations for accelerated motion.

Today we will chat about the equations of motion. There are 4 very useful expressions that relate the variables in questions:

vi - initial velocity

vf - velocity after some period of time

a - acceleration

t - time

d - displacement

Now these equations are a little tricky to come up with - we can derive them in class, if you like. (Remember, never drink and derive. But anyway....)

We start with 3 definitions, two of which are for average velocity:

v (avg) = d / t

v (avg) = (vi + vf) / 2

and the definition of acceleration:

a = (change in v) / t or

a = (vf - vi) / t

Through the miracle of algebra, these can be manipulated (details shown, if you like) to come up with:

vf = vi + at

d = 0.5 (vi + vf) t

d = vi t + 0.5 at^2

vf^2 = vi^2 + 2ad

Note that in each of the 4 equations, one main variable is absent. Each equation is true - indeed, they are the logical result of our definitions - however, each is not always helpful or relevant. The expression you use will depend on the situation.

By the way, there is a 5th equation of motion  (d = vf t - 0.5 at^2) that is sometimes useful.  We won't need it in this class.)

In general, I find these most useful:

vf = vi + at

d = 0.5 (vi + vf) t

d = vi t + 0.5 at^2

By the way, note that the 2nd equation above is the SAME THING as saying distance equals average velocity [0.5 (vi + vf)] multiplied by time.

Also, if the initial velocity is zero (as it usually is in our problems), the equations become even simpler:

vf = at

d = 0.5 (vf) t

d = 0.5 at^2

Please keep THESE final 3 at your fingertips.  Really, the first and the third are the ones that get used the most.

Let's look at a sample problem:

Consider a car, starting from rest. It accelerates uniformly (meaning that the acceleration remains a constant value) at 1.5 m/s^2 for 7 seconds. Find the following:

Part 1

- the speed of the car after 7 seconds

- how far the car has traveled after 7 seconds

To start this problem, ask yourself:  "What do I know in this problem, and how can I represent these things as symbols?"

For example, "starting from rest" indicates that the initial velocity (vi) is zero.    "7 seconds" is the time, and "1.5 m/s^2" is the acceleration.

See if you can solve the problem from here.

Part 2

Then, the driver applies the brakes and brings the car to a halt in 3 seconds. Find:

- the acceleration of the car in this time

- the distance that the car travels during this time

Got it? Hurray!

Physics - YAY!