SAT Physics Subject Test

Chapter 2 Kinematics


The simplest real-life example of motion under almost constant acceleration is the motion of objects in the earth’s gravitational field, near the surface of the earth and ignoring any effects due to the air (mainly air resistance). With these effects ignored, an object can fall freely, that is, it can fall experiencing only acceleration due to gravity. Near the surface of the earth, the gravitational acceleration has a constant magnitude of about 9.8 m/s2 (or, for our purposes, about 10 m/s2); this quantity is denoted g (for gravitational acceleration). And, of course, the gravitational acceleration vector, g, points downward.

Hammers and Feathers

At a given location on the
earth and in the absence
of air resistance, all
objects fall with the same
uniform acceleration.
Thus, two objects of
different sizes and
weights, such as hammers
and feathers, dropped
from the same height will
hit the ground at the same

Since the acceleration is constant, we can use the Big Five with a replaced by +g or –g. To decide which of these two values to use for a, make a decision at the beginning of your calculations whether to call “down” the positive direction or the negative direction. If you call “down” the positive direction, then a = +g. If you call “down” the negative direction, then a = –g. Just to make things easier, you should default to referring to the direction of the object’s displacement as positive.

In each of the following examples, we’ll ignore effects due to air resistance.

13. A rock is dropped from an 80-meter cliff. How long does it take to reach the ground?

Here’s How to Crack It

Since the rock’s displacement is down, we call down the positive direction, so a = +g. We’re given v0s, and a, and asked for t. So v is missing; it isn’t given and it isn’t asked for, and we use Big Five #3.

14. One second after being thrown straight down, an object is falling with a speed of 20 m/s. How fast will it be falling 2 seconds later?

Here’s How to Crack It

Call down the positive direction, so a = +g and v0 = +20 m/s. We’re given v0a, and t, and asked for v. Since s is missing, we use Big Five #4.

v = v0 + at = (+20 m/s) + (+10 m/s2)(2 s) = 40 m/s

15. If an object is thrown straight upward with an initial speed of 8 m/s and takes 3 seconds to strike the ground, from what height was the object thrown?

Here’s How to Crack It

The figure below shows that the displacement is down, so we call down the positive direction. Therefore, a = +g and v0 = –8 m/s (because up is the negative direction). We’re given av0, and t, and we need to find ∆s. Since v is missing, we use Big Five #2.