How to Solve An Academic problem (6 steps)
By Jason Guidos 10/17/2008
Let’s face it; solving academic problems can be hard. What is the scariest problem you can remember: Algebra, Chemistry, Physics, or Calculus? I would argue that it is most likely a word problem found in just about any subject. Yes, I said word problem. Remember those? Word problems, when they count, as a grade for your student, can be the scariest problem they have ever faced. Not only is there pressure to get it right, but they also feel unprepared and overwhelmed. Try to remember yourself sitting in that position, clock ticking, test unfinished, another test that day, high expectations of self and of parents weighing down on you, and your hand unable to move because you do not know what to write. You see the words, the numbers, the units, the punctuation, everything; but it is too much. You ask yourself, “Are some of these numbers a trick?” “Do I need to use them all?” “How is this going to help me in life?” “I knew I should have been sick today.” That is what is going on in their head. But fortunately there is a solution. Look at this example problem:
An object is thrown upward at an angle of 30o with respect to the ground with an initial velocity of 75 m/s. The acceleration of gravity is constant at 9.8 m/s2. What is the velocity in the x and y direction at t = ?
(Note: Vy = Vyo + gt & Vx = Vxo)
Where do you start? In this example, I have even given you the equations and the value of the constant for gravity. In most cases, these will have to be known and/or memorized. Although not part of the six steps, the first thing you MUST do is reeelaaax!!! “How can I relax? I am already freaking out!” I know, I know, but relaxation starts BEFORE the test. You have to go into it relaxed and maintain it throughout. (How to Take a Test is another article I have written.)
STEP 1
What is the question asking? Read the problem in full and make a note of the desired answer. Do NOT assume you know what it is asking by simply looking at the problem. The directions will tell you what you need to do to solve the problem. This will prevent unnecessary work, wasting time, and wrong answers. Example: The question is asking for velocity in two (2) directions referencing the x-axis and y-axis of a graph. Done.
STEP 2
What information do you have? Look at the given information and write it down. The given information within the problem has a tendency to blend together. Make it easy on yourself and remove it from the rest of the question. Draw a picture or pictures if necessary; sometimes you'll be able to see things easier in this form. Example: Vo = 75 m/s g = -9.8 m/s2 (negative because up is assumed positive) t= Vy = Need to solve for Vyo = UNK Vx = Need to solve for Vxo = UNK Done.
STEP 3
Establish relationships. Looking at the given information and the desired response, you should be able to see a relationship. This will usually be in the form of some known equation. Knowing what you need to solve for, you can also eliminate unnecessary information. Remember, not all given information is needed to solve the problem. Example: In this example, I have given you the equations, so that makes it easier, but I did leave out two things. Vyo which is unknown is equal to Vo * sin (30o) Vxo which is unknown is equal to Vo * cos (30o) Done.
STEP 4
Solve. Solve the problem. Don't forget about unit agreement!! Take your time and DO NOT skip steps. Too many mistakes are made when skipping. You think you are saving time, but when the answer does not work out, it ends up wasting more time than it was worth. If you write all the steps, it does not take that much longer. Once you get an answer, remember, you are NOT done. Example: Vy = Vyo + gt, substituting in the known values yields: = [75 m/s * sin (30o)] + (-9.8 m/s2 * s) = 37.5 m/s – 30.77 m/s = 6.71 m/s Vx = Vxo = 75 * cos (30o) = 64.95 m/s Done (with this step only).
STEP 5
Check. The key step in all of this is right here. You have to check your answer. This usually takes less time than you think and will increase your confidence in your answer to about 100%. Even if you check you answer and it works, can you still be wrong? Yes, BUT, 99% of the time you will be right. If you are wrong, it's usually because you did not answer the question being asked or you only gave part of the answer. Example: In our problem, we need to ensure that the cosine goes with the xvalue, and it does. The sine needs to go with the y-value, and it does. Did we use negative 9.8 for acceleration? Yes. Did we use the right equation? Yes. Did our units agree? Yes. Did we put units in the answer? Yes. Did we answer the question that was being asked? Yes. Looks good. Done.
STEP 6
Does the answer make sense? Finally, you are almost done. Last step is asking yourself if the answer you just got makes sense. What does this mean? Well, if you were asked to find the time it took a rock to fall from a height of 50 feet and your answer came out to 150 seconds, it would not make sense, because you know that it would only take a few seconds. The answer would not make sense and you would have to find your mistake. This is where most of the damage is done. When grading homework, quizzes, or tests, if the answer does not make sense, partial credit will not count. So if the problem is worth 4 points and the student answered 150 seconds, no matter how much work was done and/or shown, they will receive 0/4. You might think this is unfair, but application and understanding of what the student is doing is necessary for real learning to be occurring. Anyone can “plug-and-chug,” but the students that stand out are those that can see the problem, solve it, and have a basic understanding of what is going on. Example: Now the fun part. Now we get to try and understand our answer. Don’t be lazy now; all the hard work is over. Since the initial velocity was 75 m/s at 30 degrees, common sense should apply and you should realize that neither component of the initial velocity can be greater than 75 m/s. Our initial velocity was broken down into an x and y component both yielding answers less than 75 m/s, so that makes sense. Our final velocity in the y-direction was quite a bit less and also less than the initial y-velocity of 37.5 m/s, which should also make sense. Because projectile motion never changes the velocity of an object in the x-direction, (as the equation would also suggest), it makes sense that no matter what time you choose over its flight path, this velocity will be the same. Both our answers make sense. It doesn’t mean we are exactly right, but at least they make sense. Done.