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Solving for any variable...
Rearranging Equations - Practice Problems

Practice rearranging equations below using the "rules" that you have just learned. Answers are provided (but try doing them on your own before peeking!) You can download a sheet with all the questions (Acrobat (PDF) 127kB Jul25 09) so you can print them out to try on your own.

Velocity, distance and time

Many problems in the geosciences deal with the concept of velocity, from stream flow to plate motion. Being able to manipulate the equation for velocity (v = d/t) gives you a powerful tool for understanding the world around you. And there are two other equations that you can get by rearranging the equation that you probably already know for velocity (also rate)!

Problem 1:Generally, we know the equation for velocity (a rate) to be:
equation for velocity
Where v = velocity, d = distance and t = time.

This equation can be rearranged so that you have an equation for distance (d) and time (t).
  1. Rearrange the velocity equation to create an equation for distance (d).

    Remember to first ask yourself: Which variable do I want to isolate? In this case, we want to isolate d. So we begin by working on moving all other variables to the other side:

    1. Multiply both sides by t:
      Eq1-1-1
    2. Cancel t's where appropriate:
      Eq1-1-2
      The two steps above are exactly what you do when you cross-multiply. If you are more familiar (or comfortable) with doing it this way, please feel free to do so.
    3. and you end with an equation for d:
      Eq1-1-3
      If you are more comfortable with the unknown being on the left, you can also rearrange it to read:
      d = v x t.
  2. Rearrange the velocity equation to create an equation for time (t).

    Remember to first ask yourself: Which variable do I want to isolate? In this case, you want to isolate t. But t is on the bottom of the right side (in the denominator of the fraction). We want to get it out from under that d. The first three steps of the process are exactly the same as the question above, if you start with the original equation.

    1. Multiply both sides by t:
      Eq1-1-1
    2. Cancel t's where appropriate:
      Eq1-1-2
      The two steps above are exactly what you do when you cross-multiply. If you are more familiar (or comfortable) with doing it this way, please feel free to do so.
    3. and you end up with an equation for d:
      Eq1-1-3
    4. Now we need to isolate t by dividing both sides of the equation by v :
      Eq1-2-1
    5. cancel v's where appropriate:
      Eq1-2-2
    6. and we end up with an equation for t:
      Eq1-2-3
Use the equations that you manipulated above to solve the following problems:

cartoon of seismic profiling
Cartoon showing Woods Hole Science Center Seismic Profiling scheme photo from Woods Hole Sea Floor Mapping Project.
Problem 2:A wave traveling downward from the surface of the ocean at 1.5 km/sec takes six seconds to reflect off the ocean floor. How deep is the ocean at that site?
  1. Depth is a distance, so for this problem, you will use the equation that you rearranged in the first part of Problem 1:
    an equation for distance rearranged from the velocity equation
  2. Now, plug in the numbers you have. Remember to keep track of units (so that you can say what units the distance is in)!
    solving for depth of ocean
  3. This equation has units that can be canceled (we want to end up with some measurement of distance (yards, feet, meters, miles, etc.).
    equation for depth of ocean with units canceled
  4. If you plug the numbers into your calculator (or do the math in your head) you should end up with a depth of 9 km!

Problem 3:Imagine that you are working with Ms. Homeowner to understand the groundwater flow in her area. She is particularly interested in an underground tank that is located 2.6 km from her home. You have measured the velocity of the groundwater to be 0.033 km/day. About how long will it take any contaminants leaking from the tank to reach Ms. Homeowner's well?
  1. Because we want to solve for time (t), we need to use the equation that you rearranged in the second part of question 1:
    the equation for t from velocity
  2. Because you've got an equation that allows you to solve for t (without rearranging), you can simply plug in the numbers (v = 0.032 km/day from the above problem, and d = 2.6 km) and do the math.
    rearranged velocity equation for time with numbers for sample problems
  3. If you do the math, you end up with:

    time calculated from equation
  4. Check the units. Did you end up with a unit of time? Seems like that's a strange set of units, can we cancel any of them? Well, it turns out that the units are in time, you just have to remember the rules for dividing by a fraction - it's the same as multiplying by the inverse of the fraction:
    showing how to get days from units
    So finally, you end up with units that "days" a unit of time.
So, the answer is that it would take about 79 days for the contaminant to get to Ms. Homeowner's well.


Density

Density plays an important role in our understanding of the physical properties of Earth materials. The equation for density is similar to that for velocity and, as such, it can be manipulated so that you can solve for any of the variables involved. The next few problems utilize the equation for density:

diagram of Earth's layers
Diagram showing the layers of the Earth and their thicknesses. photo from USGS - Science of Earthquakes.

equation for density
which can be shortened to:
equation for density (variables)
Problem 1: Manipulate (rearrange) the density equation (above) to create an equation for mass.
Begin with the density equation:

equation for density (variables)
Multiply both sides by v (to isolate m):
isolating mass in the density equation
Cancel where appropriate:
cancel volume to solve for mass
And you end up with an equation for mass:
solving for mass

Problem 2:Manipulate the density equation to create an equation for volume.
Begin with the density equation:
equation for density (variables)
Multiply both sides by v:
isolating mass in the density equation
Cancel where appropriate:
cancel volume to solve for mass
And you end up with the equation for mass that we rearranged above:
solving for mass
Now, divide both sides by density (ρ):
manipulating density (div by density)
Cancel like terms (ρ):
canceling like terms to sove for V
To end up with the equation for volume (when you know density and mass):
density rearranged to solve for v

Use the equations that you manipulated above to solve the following problems:

Problem 3:
Continental crust has a density of about 2.75 g/cm3. What volume would 1000 g (1 kg) of continental crust occupy?
  1. First, you should start with the equation you came up with in Problem 2 because we want to solve for volume:
    density rearranged to solve for v
  2. Because you already have an equation for volume, no need to manipulate the equation, just plug in the numbers:
    calculating continental crust volume
  3. Use your calculator to perform the operation (divide) and check the units:
    solving for continental crust volume
    If you would like more complete description of how to figure out these seemingly complex units, please see the answer to Problem 3 in the velocity portion of these problems!
  4. 1000 g of continental crust would occupy 364 cm3

Problem 4:The outer core has a density of about 10.5 g/cm3. What volume would 1000 g (1 kg) of the outer core occupy?
  1. First, you should start with the equation you came up with in Problem 2 because we want to calculate volume:
    density rearranged to solve for v
  2. Because you already have an equation for volume, no need to manipulate the equation, just plug in the numbers:
    calculating core volume
  3. Use your calculator to perform the operation (divide) and check the units:
    solving core volume
    If you would like more complete description of how to figure out these seemingly complex units, please see the answer to Problem 3 in the velocity portion of these problems!
  4. 1000 g of the outer core would occupy 95.2 cm3. This a significantly smaller volume (by almost 4 times) than the volume occupied by the same mass of continental crust!



Isostasy

Isostasy is an important concept in the geosciences that is related to density. The concept of isostasy explains why continental crust sits so much higher than oceanic crust. It involves properties of an object like height and density. It also explains why only a portion of an iceberg is visible above water. The following problems involve an equation that is important to understanding the way that continental crust/oceanic crust behave on the Earth. If you need help visualizing the problem, click on the illustration below to see a larger version.

Illustration of isostatic equilibrium for two stylized blocks with densities and heights labeled. Illustration by Jennifer Wenner.
The equation for calculating the height of an object above the "fluid" in which it is floating is:

equation for calculating height above

where:


Htotal = total height of the object
ρobject = density of the object of interest (e.g., iceberg, continental crust, etc.)
ρfluid = density of the fluid in which the object is floating (e.g., seawater, mantle, etc.)
Habove = height of the object visible above the fluid.

Problem 1: Rearrange the equation above to solve for Htotal.

To isolate Htotal, we divide both sides by density division :

Step 1: total height
Then we can cancel like terms:
cancel like terms:  Iso Prob 1
And we end up with Htotal isolated on one side of the equation:
final equation Iso problem 1


Problem 2: Use the equation that you generated in Problem 1 to calculate the thickness (Htotal) of continental crust if:

Now that you have an equation for Htotal, you can plug in the numbers and calculate the thickness of the crust. We start with:
final equation Iso problem 1
Plug in the numbers:
Step 2: Problem 2
Divide densities (ρ) to get a unitless number:
Problem 2 Step 3
Perform the subtraction step in the denominator:
Problem 2 Step 4
and, finally, divide to get the thickness of the continental crust:
Problem 2 Final


Problem 3: Use the information from Problem 2 to calculate the thickness of crust that lies below the mantle (e.g., Hbelow).

This is a simple subtraction problem but takes a little thinking. Remember that the total of something (e.g., the height of the crust) is the sum of the parts (the height below a line + the height above a line):

Problem 3 Final

Next Steps


I think I've mastered manipulating equations! Let me try the quiz!
(If this is not how you feel, see the links below for more practice!)

Still need more practice?

There are many web sites and books that walk you through the algebraic manipulation of equations. You can use these to get more practice with rearranging equations. You may need to go to the library to find some of these.


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