Any calculation of time (

*t*) is relatively easy in the cases where either*vi*or*a*was zero. In the cases where neither are zero, the math results in a second degree polynomial equation such as:
0 = x2 + 3x - 12

Where this occurs in physics of motion is in the full distance equation:

df = di + (vi)(t) + (1/2)(a)(t2)

Although this does not exactly match the expected form for a quadratic equation, it can easily be rearranged as such:

df = di + (vi)(t) + (1/2)(a)(t2)

0 = di - df + (vi)(t) + (1/2)(a)(t2)

0 = (1/2)(a)(t2) + (vi)(t) + di - df

Keep in mind that di - df will yield a number when the values are plugged in and simplified.

To solve these problems, the steps are the same for any problem in science, BUT the algebra becomes harder.

After you have plugged in the numbers, combine like terms and simplify. Suppose the following:

df = 89

di = 50

vi = 4

a = 6

and you need to find t

df = di + (vi)(t) + (1/2)(a)(t2)

89 = 50 + (4)(t) + 1/2(6)(t2)

0 = -39 + 4t + 3t2

(Put it in normal quadratic form.)

0 = 3t2 + 4t - 39

Now you can either factor or use the quadratic formula to find the values for t:

0 = ( t - 3 )( 3t + 13)

0 = t - 3 AND 0 = 3t + 13

3 = t AND -13/3 = t

Since time cannot be negative within the context of classical physics, only t = 3 is a valid answer.

Using the quadratic equation will yield the same results.

While it is far easier to find

*t*when either*a*or*vi*is zero, the math to find t when that is not the case is not beyond the skills of a student taking an introductory physics class.
For another look at this process, check out this video:

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