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In Grade 10 physics, Australia, I have been given these formulas without how they are achieved, which is something that frustrates me in math. I have done some investigation and while I know they work, I cant figure out how they work, in a sense, and I usually find in a situation like this understanding how these formulas are discovered helps.

$s=ut+ \frac12 at^2$

Displacement = Initial Velocity * Time + Half Acceleration * Time squared.

If you were asked a question that needed this formula without knowing it, what would be the thought process one goes through in order to find this formula and use it.

Apply the same to $v^2=u^2+2as$

Velocity Squared = Initial Velocity Squared + 2 * Acceleration * Displacement.

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2 Answers 2

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Your textbook wants to introduce you to the Kinematic equations, which are helpful in solving basic physics problems in one or two dimensional motion. Did you notice that you don't have to work with problems including changing accelerations at your grade?

The reason is, you haven't been introduced to Calculus yet.

These equations are derived from the principle of Area under a specific curve. I'll give you some examples that suits your level.

Were you introduced to the Velocity-Time graphs? If you were, you know that the area under the curve equals to the distance traveled. $s = vt$

You also know that $a = v/t = s/t^2$ Thus $at^2 = s$

Now you know that $[S] = [vt] = [at^2]$. In words, these are dimensionally same and thus can be used in arithmetic operations.

Kinematic equations are made of adding the total area under the specific curve. For example, if you draw the graph for motion of a certain object, it's area under the curve will be made up by adding the elements of the related Kinematic equation.

If you draw the graph for the Displacement of the object, the area under your curve will be made up from Initial Velocity * Time + Half * Acceleration * Time squared.

Please note that my explanation is an oversimplified one. My intention is to let you know why you weren't told calculus and were told a few equations out of nowhere to remember instead. You'll learn the correct definitions once you learn Calculus.

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  • $\begingroup$ Thanks. So what i seem to be getting at, with math, physics and everything else, i will be given formulas and the likes that i struggle to explain, and they always link back to calculus. Cant wait to learn it then. $\endgroup$
    – user559905
    Commented May 10, 2018 at 0:16
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Short of deriving the formulas, we might veify them. If we take the derivative of both sides of $$\tag1s=ut+\frac12at^2,$$ it becomes $$\tag2 v=\dot s=u+at$$ i.e., the current velocity is the initial velocity plus time times acceleration. This is surely correct at $t=0$ and otherwise it is more or less the definition of acceleration - or differentiate again: $$ \tag3\dot v=\ddot s=a$$ which now really is the definition of acceleration.

In order to derive instead of verify, one would start from $(3)$ and then get to $(2)$ and $(1)$ by the reverse process, integration. At each integration, one would have to notice the need of an integration constant (essentially the value of the left hand side at $t=0$), thus introducing the initial speed $u$ - and for comleteness an initial displacement $s_0$ so that inrtead of $(1)$ we rather find $$ s=s_0+ut+\frac12 at^2.$$ So strictly speaking, $(1)$ is only correct if we define the initial position of the object as our coordinate origin.


From the fundamental result above, we find other equations by simple algrabriac transfomratins,e.g., $$v^2=(u+at)^2=u^2+2uat+a^2t^2=u^2+2a(ut+\tfrac12at^2)=u^2+2as $$

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