Kinematic equations - What do they mean?

In summary: sometimes the problem will give you a value for one of the variables, and you need to find the equation that gives you that value.other times, the problem might not give you a value for one of the variables, but it will give you an equation that describes how one of the variables is related to the other three. you just need to use that equation to solve for the variable.
  • #1
marshall104
13
0
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
 
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  • #3
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
marshall104 said:
What is it that I should be looking for when applying these equations?
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
marshall104 said:
I'm in a physics 201 course

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. :smile:

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
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
marshall104 said:
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

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
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
Andy Resnick said:
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.

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
ideasrule said:
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.

thank you!
 
  • #11
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.
 

Related to Kinematic equations - What do they mean?

What are kinematic equations?

Kinematic equations are mathematical equations that describe the motion of objects. They are based on the principles of motion, such as velocity, acceleration, and displacement.

How many kinematic equations are there?

There are five kinematic equations, which are also known as the "Big Five" equations. These include the equations for displacement, velocity, acceleration, time, and initial velocity.

What do these equations represent?

These equations represent the relationship between an object's motion and its physical properties, such as position, velocity, and acceleration. They can be used to solve for unknown variables in a given motion problem.

What is the difference between kinematic equations and dynamics equations?

Kinematic equations only describe the motion of an object, while dynamics equations also take into account the forces acting on the object. Kinematic equations are used to solve for motion variables, while dynamics equations are used to solve for force and motion variables.

How can I use kinematic equations?

Kinematic equations can be used to solve a variety of problems involving motion, such as finding the velocity of an object, calculating the distance traveled, or determining the acceleration of an object. They are commonly used in physics, engineering, and other scientific fields.

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