Mathematics of motion
First, let's look at some definitions.
Average (or constant) velocity, v
v = d / t
That is, displacement divided by time.
Another way to compute average velocity:
v = (vi + vf) / 2
where vi is the initial velocity, and vf is the final (or current) velocity.
Strictly speaking, we are talking about speed, unless direction is also specified (in which case "velocity" is the appropriate word).
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)
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).
We usually approximate this number as: 3 x 10^8 m/s, or approximately 186,000 miles/second.
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 this idea (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.
Acceleration, a
a = (change in velocity) / time
a = (vf - vi) / t
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!).
Imagine that a car was accelerating at 2 m/s/s. This would mean:
At the end of: The car would be traveling:
1 second 2 m/s
2 seconds 4 m/s
3 seconds 6 m/s
10 seconds 20 m/s
Got it? The speed increases linearly with time.
If something has a negative acceleration, it is SLOWING DOWN relative to whatever you think of as the forward direction of motion. For example, if a car was initially traveling at 12 m/s, and the driver hits the brakes giving an acceleration of -3 m/s/s:
At the end of: The car would be traveling:
1 second 9 m/s
2 seconds 6 m/s
3 seconds 3 m/s
4 seconds 0 m/s (Stopped)
>
Now we will chat a bit about the equations of motion. There are 5 useful expressions that relate the variables in questions:
vi - initial velocity. Note that the i is a subscript.
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
d = vf t - 0.5 at^2
Note that in each of the 5 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.
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.
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:
- the speed of the car after 7 seconds
- how far the car has traveled after 7 seconds
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!