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When learning kinematics, I have learned these three formulae

$s=1/2(u+v)t$

$s=ut+1/2at^2$

$v^2=u^2+2as$

  • s = displacement
  • u = initial velocity
  • v = final velocity
  • t = time
  • a = acceleration

My teacher then say that we can get the second and third formulae from the first formulae and the formulae of acceleration

Proof for $s=ut+1/2at^2$

From the formulae $a=(v-u)/t$ and arrange you get $v=u+at$

Then substitute $v=u+at$ into $s=1/2(u+v)t$

$s=1/2(u+v)t$

$s=1/2(u+u+at)t$

$s=1/2(2u+at)t$

$s=1/2(2ut+at^2)$

$s=ut+1/2at^2$

Proof for $v^2=u^2+2as$

From the formulae $a=(v-u)/t$ and arrange you get $t=(v-u)/a$

Then substitute $t=(v-u)/a$ into $s=1/2(u+v)t$

$s=1/2(u+v)t$

$s=1/2(u+v)*(v-u)/a$

$s=(v^2-u^2)/2a$

$2as=v^2-u^2$

$v^2=u^2+2as$

So, my questions is if we can get the second and third formulae from the first formulae, how do we get the first formulae?

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  • $\begingroup$ The first formula only makes sense if we assume that the acceleration is constant, in which case the effective velocity is the average velocity. To be very honest with you, learning these formulae doesn't teach anything about physics. It's better to write everything in differential form $\Delta s = v\Delta t$ and $\Delta v = a\Delta t$, which is not only valid for constant a and v and large $\Delta t$, but also for infinitesimal quantities. Trouble is... you probably didn't have pre-calculus, yet, right? Without some understanding of how to manipulate small quantities correctly, you are stuck. $\endgroup$ – CuriousOne May 1 '15 at 8:17
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The first one is basically derived from the velocity-time graph. As the acceleration is constant the area under the graph represents displacement (s).

Thus by the formula to get the area under a trapezium. $$\text{Area} =\frac{1}{2} ( \text{sum of the parallel sides}) \times ( \text{vertical height})$$.

We have $$S = \frac{1}{2} ( v + u)\cdot t$$.

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Suppose you have a constant acceleration $a$. Acceleration is the derivative of velocity with respect to time, so we have:

$$ \frac{dv}{dt} = a $$

and if we integrate this we get:

$$ v = at + C $$

where $C$ is the constant of integration. To find $C$ we note that when $t = 0$ the velocity is equal to $u$ (the initial velocity) so that means $C = u$ and the full equation is:

$$ v = u + at $$

Now, velocity is the derivative of distance with time, so we have:

$$ \frac{ds}{dt} = u + at $$

Again, we can integrate this to get:

$$ s = ut + \tfrac{1}{2}at^2 + C $$

where again $C$ is a constant of integration. This time we use the condition that at $t = 0$ the distance $s$ is zero, and this means $C = 0$. So we have derived your second equation:

$$ s = ut + \tfrac{1}{2}at^2 $$

We then get the first and third equations from the second one.

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