SAT Physics Subject Test
Chapter 2 Kinematics
UNIFORMLY ACCELERATED MOTION and THE BIG FIVE
The simplest type of motion to analyze is motion in which the acceleration is constant (possibly equal to zero). Although true uniform acceleration rarely occurs in the real world, many common motions exhibit approximately constant acceleration and, in these cases, the kinematics of uniformly accelerated motion provide a pretty good description of what’s happening. Notice that if the acceleration is constant, then taking an average yields nothing new, so ā = a.
Another thing that makes our discussion easier is that we’ll only consider motion that takes place along a straight line. In these cases, there are only two possible directions of motion—one is positive, and the opposite direction is negative. Most of the quantities we’ve been dealing with—displacement, velocity, and acceleration—are vectors, which means that they include both a magnitude and a direction. With straight-line motion, we can show direction simply by attaching a plus or minus sign to the magnitude of the quantity; therefore we will drop the standard vector notation.
Keep in mind that all of the
quantities in the Big Five
(except t) are vector quantities
(that is, they can be
positive or negative).
Fundamental Quantities: A Quick Review
The fundamental quantities are displacement (∆ s), velocity (v), and acceleration (a). Acceleration is a change in velocity, from an initial velocity (vi or v0) to a final velocity (vf or simply v—with no subscript). And, finally, the motion takes place during some elapsed time interval, ∆t. Therefore, we have five kinematics quantities: ∆s, v0, v, a, and ∆t. Since time usually begins at zero, we will replace ∆t with t.
These five quantities are related by a group of five equations that we call the Big Five. They work in cases where acceleration is uniform, which are the cases we’re considering.
In Big Five #1, the average velocity is simply the average of the initial velocity and the final velocity: = (v0 + v). This is true because the acceleration is constant.
Each of the Big Five equations is missing one of the five kinematic quantities. The way you decide which equation to use when solving a problem is to determine which of the kinematic quantities is missing from the problem—that is, which quantity is neither given nor asked for—and then use the equation that doesn’t contain that variable. For example, if the problem never mentions the final velocity—v is neither given nor asked for—you should use the equation that’s missing v: Big Five #2.
The first part of Big Five #1 and Big Five #4 are simply the definitions of v and ā written in forms that don’t involve fractions. The other Big Five equations can be derived from these two definitions and the equation = (v0 + v), using a little algebra. Although the derivations are irrelevant for the purposes of SAT Physics, to remember the equations, you may wish to think of the first three as resembling d = vt and the next two as resembling each other.
8. An object with an initial velocity of 4 m/s moves along a straight axis under constant acceleration. Three seconds later, its velocity is 14 m/s. How far did it travel during this time?
Here’s How to Crack It
We’re given v0, t, and v, and we’re asked for ∆s. So a is missing; it isn’t given and it isn’t asked for, so we use Big Five #1.
9. A car that’s initially traveling at 10 m/s accelerates uniformly for 4 seconds at a rate of 2 m/s2 in a straight line. How far does the car travel during this time?
Here’s How to Crack It
We’re given v0, t, and a, and we’re asked for s. So v is missing; it isn’t given and it isn’t asked for, so we use Big Five #2.
10. A rock is dropped off a cliff that’s 80 m high. If it strikes the ground with an impact velocity of 40 m/s, what acceleration did it experience during its descent?
Here’s How to Crack It
If something is dropped, then that means it has no initial velocity: v0 = 0. So, we’re given v0, ∆s, and v, and we’re asked for a. Since t is missing, we use Big Five #5.
Those Pesky Signs
Make sure you are clear
which direction is positive.
If an object ends more in
the negative direction than
it started, ∆s is negative.
Notice that since a has the same sign as ∆s, the acceleration vector points in the same direction as the displacement vector. This makes sense, because the object moves downward and the acceleration it experiences is due to gravity, which also points downward.