One-Dimensional Motion with Uniform Acceleration
One-Dimensional Motion with Uniform Acceleration
Many introductory physics problems can be simplified to the special case of uniform motion in one dimension with constant acceleration. That is, most problems will involve objects moving in a straight line whose acceleration doesn’t change over time. For such problems, there are five variables that are potentially relevant: the object’s position, x; the object’s initial velocity, ; the object’s final velocity, v; the object’s acceleration, a; and the elapsed time, t. If you know any three of these variables, you can solve for a fourth. Here are the five kinematic equations that you should memorize and hold dear to your heart:
The variable represents the object’s position at t = 0. Usually, = 0.
You’ll notice there are five equations, each of which contain four of the five variables we mentioned above. In the first equation, a is missing; in the second, x is missing; in the third, v is missing; in the fourth, is missing; and in the fifth, t is missing. You’ll find that in any kinematics problem, you will know three of the five variables, you’ll have to solve for a fourth, and the fifth will play no role in the problem. That means you’ll have to choose the equation that doesn’t contain the variable that is irrelavent to the problem.
Learning to Read Verbal Clues
Problems will often give you variables like t or x, and then give you verbal clues regarding velocity and acceleration. You have to learn to translate such phrases into kinematics-equation-speak:
When They Say . . . They Mean . . .
“. . . starts from rest . . .”
“. . . moves at a constant velocity . . .” a = 0
“. . . comes to rest . . . ” v = 0
Very often, problems in kinematics on SAT II Physics will involve a body falling under the influence of gravity. You’ll find people throwing balls over their heads, at targets, and even off the Leaning Tower of Pisa. Gravitational motion is uniformly accelerated motion: the only acceleration involved is the constant pull of gravity, –9.8 m/s2 toward the center of the Earth. When dealing with this constant, called g, it is often convenient to round it off to –10 m/s2.
A student throws a ball up in the air with an initial velocity of 12 m/s and then catches it as it comes back down to him. What is the ball’s velocity when he catches it? How high does the ball travel? How long does it take the ball to reach its highest point?
Before we start writing down equations and plugging in numbers, we need to choose a coordinate system. This is usually not difficult, but it is vitally important. Let’s make the origin of the system the point where the ball is released from the student’s hand and begins its upward journey, and take the up direction to be positive and the down direction to be negative.
We could have chosen other coordinate systems—for instance, we could have made the origin the ground on which the student is standing—but our choice of coordinate system is convenient because in it, = 0, so we won’t have to worry about plugging a value for into our equation. It’s usually possible, and a good idea, to choose a coordinate system that eliminates . Choosing the up direction as positive is simply more intuitive, and thus less likely to lead us astray. It’s generally wise also to choose your coordinate system so that more variables will be positive numbers than negative ones, simply because positive numbers are easier to deal with.
What is the ball’s velocity when he catches it?
We can determine the answer to this question without any math at all. We know the initial velocity, m/s, and the acceleration due to gravity, m/s2, and we know that the displacement is x = 0 since the ball’s final position is back in the student’s hand where it started. We need to know the ball’s final velocity, v, so we should look at the kinematic equation that leaves out time, t:
Because both x and are zero, the equation comes out to But don’t be hasty and give the answer as 12 m/s: remember that we devised our coordinate system in such a way that the down direction is negative, so the ball’s final velocity is –12 m/s.
How high does the ball travel?
We know that at the top of the ball’s trajectory its velocity is zero. That means that we know that = 12 m/s, v = 0, and m/s2, and we need to solve for x:
How long does it take the ball to reach its highest point?
Having solved for x at the highest point in the trajectory, we now know all four of the other variables related to this point, and can choose any one of the five equations to solve for t. Let’s choose the one that leaves out x:
Note that there are certain convenient points in the ball’s trajectory where we can extract a third variable that isn’t mentioned explicitly in the question: we know that x = 0 when the ball is at the level of the student’s hand, and we know that v = 0 at the top of the ball’s trajectory.
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