get a better understanding of why the world works the way it does. That’s pretty important stuff! Time to travel back to basic algebra for a quick refresher.
You need to be able to isolate different variables. For instance, the following equation tells you the distance, s, that an object travels if it starts from rest and accelerates at rate of a for a time, t:
Now suppose the problem actually tells you the time the object is in motion and the distance it travels and asks you to calculate the object’s acceleration. By rearranging the equation algebraically, you can solve for the acceleration:
In this case, you’ve multiplied both sides by 2 and divided both sides by t2 to isolate the acceleration, a, on one side of the equation.
What if you have to solve for the time, t? By moving the number and variables around, you get the following equation:
Do you need to memorize all three of these variations on the same equation? Certainly not. You just memorize one equation that relates these three items – distance, acceleration, and time – and then rearrange the equation as needed. (If you need a review of algebra, get a copy of Algebra I For Dummies, by Mary Jane Sterling [Wiley].)
Example
Q. The equation for final speed, vf – where the initial speed is vi, the acceleration is a, and the time is t – is vf = vi + at. Solve for acceleration.
A. The correct answer is a = (vf – vi)/t. To solve for a, divide both sides of the equation by time, t.
Practice Questions
1. The equation for potential energy, PE, for a mass m at height h, where the acceleration due to gravity is g, is PE = mgh. Solve for h.
2. The equation relating final speed, vf , to initial speed, vi , in terms of acceleration a and distance s is vf 2 = vi 2 + 2as. Solve for s.
3. The equation relating distance s to acceleration a, time t, and speed v is
4. The equation for kinetic energy is
Practice Answers
1. h = PE/mg. Divide both sides by mg to get your answer.
2.
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Divide both sides by t to get your answer.
4.
Take the square root to get your answer.
Tackling a Little Trig
You need to know a little trigonometry, including the sine, cosine, and tangent functions, for physics problems. To find these values, start with a simple right triangle. Take a look at Figure 2-1, which displays a right triangle in all its glory, complete with labels I’ve provided for the sake of explanation. Note in particular the angle
© John Wiley & Sons, Inc.
Figure 2-1: A labeled triangle that you can use to find trig values.
Remember: To find the trigonometric values of the triangle in Figure 2-1, you divide one side by another. Here are the definitions of sine, cosine, and tangent:
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If you’re given the measure of one angle and one side of the triangle, you can find all the other sides. Here are some other forms of the trig relationships – they’ll probably become distressingly familiar before you finish any physics course, but you don’t need to memorize them. If you know the preceding sine, cosine, and tangent equations, you can derive the following ones as needed:
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To find the angle
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And here’s one more equation, the Pythagorean theorem. It gives you the length of the hypotenuse when you plug in the other two sides:
If you need a more in-depth refresher, check out Trigonometry For Dummies, by Mary Jane Sterling (Wiley).
