The study of motion with relationship to time considers primarily how the position or velocity of an object changes with respect to time. In exploring these changes, definitions and formulas emerge:

displacement: the magnitude of the change in position of an object.velocity: the rate at which the position of an object changes with respect to time.acceleration: the rate at which the velocity of an object changes with respect to time.df = di + vtwhere df is final (or total) displacement, di is initial displacement, v is average velocity, and t is elapsed time.vf = vi + atwhere vf is final velocity, vi is initial velocity, a is acceleration, and t is elapsed time.v(ave) = (vf+vi)/2where v(ave) is average velocity, vf is final velocity, and vi is initial velocity.

Using the formulas above, it is possible to directly solve problems that ask about any combination of displacements, velocity and time or that ask about velocities, acceleration and time. However, to solve problems that ask about displacements, acceleration, and time, three steps were needed, if only the above formulas were used.

To solve the problems that that ask about displacements, acceleration, and time directly, a fourth formula is needed. This formula can be derived from the other three, but to get started, here it is:

df = di + vit + 1/2at

^{2}*(t*^{2}means t•t or "t squared")
This equation is much more fun with di and vi are zero! But, to break it down as is first

df is the final, total displacement.

di is the initial displacement. How far from whatever point of reference is the object when the thing starts accelerating?

vit accounts for the motion of the object based on its starting velocity. It keeps covering distance at the initial rate.

The last term tells how much MORE distance is covered based on the acceleration. Since any di and vi are covered in the first term, the last term can be analyzed for the zero case, but first two examples.

EXAMPLE 1 (the zero case)

How far will a rocket travel while it accelerates from rest at a rate of 4 (m/s)/s for 5 seconds?di = 0vi = 0vf = not given, not asked fordf = what you are looking fora = 4t = 5df = di + vit + 1/2at^{2}df = 0 + 0 + 1/2 • 4 • 5•5PEMDOSdf = 1/2 • 4 • 25df = 2 • 25df = 50

EXAMPLE 2

How far will a rocket travel while it accelerates from 20 m/s at a rate of 6 (m/s)/s for 3 seconds?di = 0vi = 20vf = not given, not asked fordf = what you are looking fora = 6t = 3df = di + vit + 1/2at^{2}df = 0 + 20•3 + 1/2•6 • 3•3PEMDOSdf = 60 + 1/2 • 6 • 9df = 60 + 3 • 9df = 60 + 27df = 87

CONCLUSION:

The above examples can be checked using the three-step method. The equation used above can be used to solve for any of the variables,

*but does require dealing with squares and square roots*. However, if that level of math does not present a problem,df = di + vit + 1/2at^{2}

becomes the only equation needed for all cases of motion related to displacement. Combined with

vf = vi + at

all of the motion problems can be solved.

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Example 3 (Find a when given distance and time.)

Working through the three-step method to find acceleration for the case of di and vi = 0 results in a quick and easy calculation method that requires

Step 1: Divide df by t

Step 2: Double it

Step 3: Divide by t again.

Using the new equation, this emerges like this:

df = 1/2 a t^{2}df = 1/2 a • t • t2 • df = a • t • t2df/t = a • t(2df/t)/t = a

(Same as the 3 step method! Wooo!)

To more correctly write the above equation results in:

2•(df/t^{2}) = a

And that is called physics!

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WHO LOVES SOME MATH?

Okay, don't answer that.

It is not hard to come up with the df = di + vit + 1/2at

^{2}formula. Starting with the basic, simple case formulas, combining them quickly results indf = 1/2at^{2}

How is that done? Easily! Begin with

df = vt

where v is the average velocity.

Remember that

vf = at

where vf is the final velocity. In the case of vi = 0, the formula for average velocity easily reduces:

v(ave) = (vf + vi)/2v(ave) = (vf + 0)/2v(ave) = (vf)/2

Now, back to vf = at! Plug it in!

So…v(ave) = (vf)/2 becomesv(ave) = (at)/2

Now, back to df = vt! Plug in (at)/2 for

*v*:df = (at)/2 • t

Clean up time:

df = a • 1/2 • t • t

df = 1/2 • a • t • tdf = 1/2at^{2}

Bam! That just happened!

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