1. 3.3 Working with Equations1

2. Chapter Objectives
Calculate time, distance, or speed when given two of the three values.
Solve an equation for any of its variables.
Use and interpret positive and negative values for velocity and position.
Describe the relationship between three-dimensional and one-dimensional systems.
Draw and interpret graphs of experimental data, including velocity versus position, and speed versus time.
Use a graphical model to make predictions that can be tested by experiments.
Derive an algebraic model from a graphical model and vice versa.
Determine velocity from the slope of a position versus time graph.
Determine distance from the area under a velocity versus time graph.

3. Chapter Vocabulary average speed constant speed coordinates
coordinate system
displacement
instantaneous speed
instantaneous velocity
origin
position
rate
slope
time
vector
velocity

4. Inv 3.3 Equations of Motion
Investigation Key Question:
How are equations used in physics? 4

5. 3.3 Working with Equations
An equation is a much more powerful form of model than a graph.
While graphs are limited to two variables, equations can have many variables and can be used over a wide range of values.

6. 3.3 Working with Equations
Equations can also be rearranged to show how any one variable depends on all the others.

7. Calculating time from speed and distance
How far do you go if you drive for 2 h at a speed of 100 km/h?
You are asked for distance.
You are given time in h and speed in km/h.
Use d = vt.
Solve. d = 2 h × 100 km/h = 200 km

8. 3.3 Solving an equation
To “solve” means to get a desired variable by itself on one side of an equals sign.
Whatever you do to the left of the equals sign you must do exactly the same to the right.
Get in the habit of solving an equation before you plug in numbers.
More complex problems require you to substitute whole equations for single variables.

9. 3.3 Solving an equation To solve this equation for distance (d):
Multiply both sides of the equation by “t”.
Multiplying by “t”on both sides of the equation allows you to cancel a t from the numerator and the denominator on the right side of the equation.

10. 3.3 Position vs. time equation
The equation says your position, x, is equal to the position you started at, x0, plus the additional amount you traveled, vt.

11. Calculating time from speed and distance
A car moving in a straight line at constant velocity starts at a position of 10 meters and finishes at 30 meters in five seconds. What is the velocity of the car?
You are asked for velocity.
You are given that the motion is at constant velocity, two positions, and the time.
Use x = x0 + vt, solve for v.
x – x0 = vt
x – x0 = v t
Substitute numbers for variables: v = 30 m – 10 m = 4 m/s
5 s

12. 3.3 Relating equations and graphs
In science and engineering, any two variables can be used in the equation for a line, not just x and y.

13. 3.3 Relating equations and graphs
The y corresponds to x, the position at any time;
the x corresponds to time “t” ;
the slope, m, corresponds to the velocity, v;
the y-intercept, b, corresponds to the initial position, x0.

15. 3.3 Scientific process
The process of developing a model or theory in science starts with actual experiments and data, and produces a validated model in the form of an equation.

16. 3.3 How to solve physics problems
Step 1
Identify clearly what the problem is asking.
Step 2
Identify the information you are given.
Step 3
Identify relationships.
Step 4
Combine the given information and the relationships.

17. Calculating distance from time and speed
A space shuttle is traveling at a speed of 7,700 meters per second.
How far in kilometers does the shuttle travel in one hour?
At an altitude of 300 kilometers, the circumference of the shuttle’s orbit is 42 million meters.
How long does it take the shuttle to go around Earth one time?

18. Calculating distance from time and speed
This is a two-part problem asking for distance in kilometers and time in hours.
You are given a speed and time for the first part, and a speed and distance for the second.
d = vt, and t = d ÷ v 1 h = 3,600 s 1 km = 1,000 m
Part 1:
d = (7,700 m/s)(3,600 s) = 27,720,000 m
Convert to kilometers: 27,720,000 m ÷ 1,000 km/m = 27,720 km
Part 2:
t = 42 × 106 m ÷ 7,700 m/s = 5,455 s
Convert to minutes: = 5,455s ÷ 60 s/min = 90.9 minutes

19. Slow Motion Photography
A video camera does not photograph moving images.
It takes a sequence of still images called frames and changes them fast enough that your brain perceives a moving image.
You can use an ordinary video camera to analyze motion in laboratory experiments.