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ndaceko · Nov 30, 2016, 10:09 PM

Hi everybody, please what is the simplest way to derive the 2nd, 3rd and 4th equations of motion. I don't seems to understand it. Please

ebaines · Dec 1, 2016, 06:39 AM

Please clarify what you mean by 2nd, 3rd, and 4th equations of motion - perhaps you mean these, which apply for the case of constant acceleration:

(1) $v(t) = V_0 + at$
(2) $x(t) = x_0 +V_0t + \frac 1 2 at^2$
(3) $v(x)^2 - v_0^2 = 2ax$

Also, I assume you are a high school student, so not familiar with methods of calculus, correct?

OK, consider an object moving with constant acceleration. If you plot velocity versus time you get the attached graph. The first equation comes from the definition of acceleration: acceleration = change in velocity per unit time, which is equivalent to the slope of the graph:

$a = \frac {\Delta v} {t} = \frac {v-v_0} t$

Rearrange to get $v = V_0 + at$

The second equation comes from the fact that distance traveled is equal to the area under the plot of velocity versus time. Note that the area is a trapezoid, and from basic geometry the area of that trapezoid is equal to one half the sum of the two sides times the base:

$x = (\frac {V_0 + (V_0+at)} 2 ) t$

Simplify and you get:

$x = V_0t + \frac 1 2 a t^2$

The third equation comes from combining the previous two equations. From the first we can solve for t:

$t = \frac {v-V_0} a$

Then sub this value for 't' into the second equation:

$x = V_0 t + \frac 1 2 a t^2 = V_0 (\frac {v-V_0} a) + \frac 1 2 a (\frac {v-V_0} a )^2$

With a bit of manipulation this simplifies to $x = \frac {v^2 - V_0^2} {2a}$

Hope this helps.

Name: v vs t.JPG
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Name: v vs t.JPG
Views: 138
Size: 16.7 KB