Acceleration!

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!).

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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!