Deriving Kinematics Equation without Calculus / First Principle

Question

How do I derive the Kinematics equation $v=v_0+at$ from the equation $x = x_0+v_0t+\frac12at^2$?

This is how I have tried :

$$x - x_0=v_0t+\frac12at^2$$

$$\Delta x= t(v_0+\frac12at)$$

$$\dfrac{\Delta x}{t}= v_0+\frac12at$$

$$v = v_0+at$$

I'm stuck with the $\frac12$. How to get rid of that ? Please Help !

Answer 1

It is entirely possible to derive the equation without calculus, here is how:

Firstly, you are wrong in substituting $\Delta x/t$ with $v$. The ratio $\Delta x/t$ is, the average velocity, and not the velocity at time $t$. Since this is a case of uniform acceleration, we have $$\frac{\Delta x}{t}=v_{\text{avg}}=\dfrac{v+v_0}{2}$$ implying $$\dfrac v2+\dfrac{v_0}{2}=v_0+\dfrac12at$$ Simplifying will give you the required equation.

Hope this helps. Ask anything if not clear :)

Answer 2

To derive it the way you want you would need some basic calculus (derivatives), but the equation $v = v_{0} + at$ can be understood without that. The definition of acceleration ($a$) is change in velocity ($v$) per second. This equation only works when the acceleration is constant so the change in velocity over a period of ($t$) seconds is $at$ (literally change per second times number of seconds). Then it is clear that the equation says "final velocity = initial velocity + change in velocity".

If you want to get the basics of calculus, here is a great series on it: https://www.youtube.com/playlist?list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr

Answer 3

$$ x = v_0t + ½at²$$ $$\frac{dx}{dt}=v_0+\frac{2}{2}at$$ ~ Differentiate respect to time. $$v=v_0+at$$

You can learn calculus from here. I like these tutorial.

Answer 4

Hmm how about this:

$$x(t+\Delta t)=x_0 + v_0( t + \Delta t)+ \frac12a(t+ \Delta t)^2$$

We know that at each and every point the velocity is different since there is an acceleration, but let's take a small time interval $(t,t+\Delta t)$, then change in position for this interval of time is given as::

$$ x(t + \Delta t) - x(t) = v_0 \Delta t + \frac12 a( 2t \Delta t + \Delta t^2)$$

Now, if to find velocity we divide by the time gap:

$$ v(t) = \frac{x( t+ \Delta t) - x(t)}{\Delta t} = v_0 + \frac12 a( 2t+ \Delta t)$$

But remember, each point has a different velocity due to acceleration, so to get the best idea for rate of change of position for the motion at a point, let us take the time gap to be really small then we find that all terms of $\Delta t$ on RHS die off: $$ v(t) = v_0 + at$$

And that's it.