# Kinematic equations - What do they mean?

1. Oct 24, 2009

### marshall104

I'm in a physics 201 course and very confused about the kinematic equations. Is there anybody out there who can describe what each one means?

Thanks,

Mike

2. Oct 24, 2009

### Staff: Mentor

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3. Oct 24, 2009

### marshall104

Yes those are the ones. What is it that I should be looking for when applying these equations? Is it just a matter of feasibility?

Thanks,

Mike

4. Oct 24, 2009

### Staff: Mentor

It's hard to give generic advice without seeing the kind of problem that's bugging you. Like anything else, it depends on what you're given and what you're trying to find. Often you'll have to use several kinematic relationships to solve a problem.

5. Oct 24, 2009

### Staff: Mentor

Note that "Physics 201" tells us diddly-squat about what the course is about. Different colleges and universities use different systems of course numbers. Where I teach, we don't have a Physics 201. The closest number we've ever had is 203 which was our Introductory Astronomy course.

It sounds like the course you're taking is similar to our Physics 121 (General Physics I, which doesn't use calculus) or perhaps Physics 214 (Physics with Calculus I), which are introductory courses for non-majors and physics majors respectively. What those numbers mean at your school is something else again...

6. Oct 24, 2009

### Staff: Mentor

To get a better feel for the kinematics equations, I suggest you learn how they are derived. That will make you more comfortable with using them.

7. Oct 24, 2009

### Andy Resnick

They simply describe motion of point masses for the special case of constant acceleration (and no air resistance, etc). Using the definition of velocity v = (change in position)/(time interval) and acceleration a = (change of velocity)/(time interval), one can derive 3 of the 4 kinematic equations:

vf = vi + a*t
xf = x1 + vi*t + 1/2 a*t^2
v_average = (vi+vf)/2

The fourth one, vf^2 = vi^2 + 2a(xf - xi), I confess is a bit of a mystery to me. I think one needs calculus to derive it, and it is also special in that 'time' is not a part of the expression.

8. Oct 24, 2009

### mikelepore

There's a usual pattern for solving kinematics problems that most people use. You have five variables (V_i, V_f, a, t and s). You have four formulas for which it is true that each formula mentions four of the five variables, and doesn't mention one of the five variables. A word problem that you must solve also mentions four variables, in the form of giving values for three of them and asking you to find one of them, and the word problem doesn't mention one of the variables, in that it doesn't give a value for it and also doesn't ask you to find it. Getting accustomed to that pattern immediately tells you which formula to use. Given a word problem, you just ask yourself which of the formulas mentions the same four variables, and omits any mention of the same one variable, that the problem does.

When I say that a problem "gives" a value for a variable, it might be encoded in words, for example, the phrase "dropped from rest" means V_i = 0, "comes to rest" means v_f=0, "projectile reaches its maximum height" means v_f=0, "falls near the surface of the earth" means a=-g, etc.

9. Oct 24, 2009

### ideasrule

Not at all. It can easily be derived from:

vf = vi + a*t
xf = x1 + vi*t + 1/2 a*t^2

Just isolate t in the first equation and substitute into the second.

10. Oct 24, 2009

### Andy Resnick

thank you!

11. Oct 25, 2009

### johnnyies

in your basic kinematic problems, there are four variables: position, velocity, acceleration, and time. You just need to know which equation to use with the variables you have.