Chapter 2 Part 4 Vertical Motion and Motion Graphs

Prof. Clements Notes/Hints for Physics Courses

Chapter 2 Open Stax College Physics or Most Introductory Physics Courses

Vertical Motion and Graphs of Motion

I highly recommend that you go to the web site https://openstax.org/details/books/college-physics to access the free OpenStax College Physics textbook and resources.

YouTube videos of my (usually short) lectures and example problems are indexed at http://www.physics.gpclements.com/ . There are also a few videos that give review of basic math tools that are used in an introductory physics course.

Terms you should know at the end of this unit: acceleration due to gravity, free-fall, tangent line, slope

Falling Objects

Free fall is vertical motion with only the force of gravity acting on the object. (i.e. ignore air resistance and don’t allow any propulsion)

“g” is the symbol for the acceleration due to gravity Near the earth’s surface the magnitude of g is about 9.80 m/s² and the direction is towards the center of the Earth. (The value of g changes slightly based on various conditions of the location that we will ignore.) g tells us how much the velocity of an object in free fall will change every second (9.8 meters/second). g is not gravity. Gravity is a force, not an acceleration. The force of gravity produces an acceleration.

For vertical motion I will usually use the symbol Y for the position of the object. I will try to usually use X for position when the motion of the object is in the horizontal direction. We will assume that the motion is occurring near the surface of the Earth. In this case, the value of “g” will be constant. We will usually ignore air resistance. The four kinematic equations may be used for vertical motion because the acceleration of the object is constant.

Write the four kinematic equation with these changes. Wherever X appears use Y. Wherever “a” appears use “g.”

Equation 1 V = V_o + g t

Equation 2 Y = Y_o + [ 0.5 (V + V_o) ] * t or S = [ 0.5 (V + V_o) ] * t

Equation 3 Y = Y_o + V_o * t + 0.5 * a * t² or S = V_o * t + 0.5 * a * t²

Equation 4 V² = V_o² + 2 * a * (Y – Y_o) or V² = V_o² + 2 * a * S

with S = Y - Y_o

Note: I will work problems with the upward direction as the positive direction. This means that I will use g = - 9.8 m/s² .

Note: Some people have a misconception that the value of the free-fall acceleration depends on the weight of the object. In reality, all objects near the earth have a free-fall acceleration of g (9.8 m/s² ) if air resistance is ignored. This fact will proven when force and Newton’s laws are discussed.

Suppose that an object is dropped from rest and is in free fall from a height of 400 meters. Calculate the position of the object for every second before it hits the ground.

Here are some questions for you to consider: Where will you put Y = 0 meters? Is Y positive in the upward direction or downward? (These questions do not have one right answer!) You are in control of labeling your coordinate system. The analysis will give identical physical results for an infinite number of choices for the 0 point and + or – direction on the axes.

Solution:

1. Make a sketch and write the values for quantities that are known. You should make your own sketch right now. I will use upward as the positive direction.

I will label the ground as Y = 0 meters and assign the upward direction as +.

Y_o = 400 m V_o = 0 m/s g = - 9.8 m/s²

Y = Y_o + V_o * t + 0.5 * a * t²

Y = 400 + 0 * t + 0.5 * (-9.8) * t²

for t = 0 seconds Y_o= 400m for t = 1 second Y₁ = 395m

For the rest of the list the subscript for Y is the time value in seconds:

Y₂ = 380m Y₃ = 356m Y₄ = 322m Y₅ = 278m Y₆ = 224m Y₇ = 160m Y₈ = 86.4m Y₉ = 3.1m

Y₁₀ = -90m ...This value will not be used since it is below ground level.

If you would plot the Y value vs. time the graph would be a parabola.

The velocity at any particular time can be calculated with

V = V_o + at

For this problem V = -9.8 (m/s) * t V_o = 0 m/s V₁ = -9.8 m/s

The graph of velocity vs. time is a straight line with a slope of -9.8 m/s².

The value of the slope on the V vs time graph is equal to the acceleration value.

The graph of acceleration vs time is a horizontal line that has a value of -9.8 m/s² .

You should workother sample problems until you are comfortable with free-fall problems. Let your instructor know if you have questions on the example problems.

Graphical Analysis of Linear Motion

A straight line can be represented by the equation Y = m * X + b. The English language (and other languages) does not have enough symbols for all the variables that might be discussed. The Y and X symbols here are NOT vertical and horizontal position. They are just the coordinates of a Cartesian coordinate system.

“m” is the slope of the line. It is common to calculate the slope by dividing the “rise” by the “run.” The “rise” is the change in the vertical variable value. The “run” is the change in the horizontal variable value.

“b” is the Y-intercept. For the graph represented by Y = 5X + 3, the value of the Y intercept is 3.

When X = 0 the value of Y is 3. When X = 1 the value of Y is 8.

The slope of the line between these two points is (8 – 3) / (1-0) or 5. Note that 5 is the coefficient of X in the equation above.

The slope of the graph at a particular time can be found by drawing a straight line tangent to the graph. The tangent line touches the graph at the time of interest and has equal spacing between the tangent line and the graph on both sides of the touch point.  The graph below is a parabola.  The tangent lines for the times of 1 and 3 seconds and their slopes are shown.  The slopes are calculated by selecting two widely separated points on the line and dividing the change in the vertical (on the graph axis) coordinate by the change in the horizontal coordinate.

The slope of the tangent line on a graph of X vs. time gives information about the instantaneous velocity at that time. The slope of the tangent line on a graph of V vs. time gives information about the instantaneous acceleration at that time. You should be able to draw a tangent line on a graph and find the slope value.

The area “under the graph” of velocity vs. time gives information on the change in position of the object. The area “under the graph” of acceleration vs. time gives information on the change in velocity of the object. “Under the graph” means find the area between the line and the time axis of the graph (where “V” or “a” is zero).

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