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Derive the three equation of motion by Algebraic method.
Consider a body moving with an initial velocity of 'u' under the influence of constant acceleration 'a'. After time 't', the speed of the body becomes 'v' and displacement becomes 's'.
We know that velocity is defined as the rate of change of displacement. Mathematically it is represented as:
$Velocity=\frac{Displacement}{Time}$
After rearranging it, we get displacement (s), which is the product of the velocity and time period (t), when velocity is constant.
$Displacement=Velocity\times Time$
If the velocity is not constant then we can use average velocity in the place of velocity in the above equation, and rewrite the equation as follows:
$Displacement=Average\ Velocity\times Time$
$s=\frac{u+v}{2}\times t$ $\left(\because Average\ velocity=\frac{u+v}{2}\right)$
Now, from the first equation of motion, we know that $v=u+at$. Putting this value of $v$ in the above equation, we get-
$s=\frac{u+(u+at)}{2}\times t$
$s=\frac{2u+at}{2}\times t$ $s=\left(\frac{2u}{2}+\frac{at}{2}\right)\times t$
$s=\left(u+\frac{1}{2}at\right)\times t$
$s=\left(u\times t\right)+\left(\frac{1}{2}at\times t\right)$ $s=ut+\frac{1}{2}a{t}^{2}$
Hence, the second equation of motion is derived by the algebraic method.
Derivation of Third Equation of Motion- ${v}^{2}={u}^{2}+2as$
We know that displacement is the rate of change of position of an object.
$Displacement=Average\ Velocity\times Time$
$s=\frac{u+v}{2}\times t$ $\left(\because Average\ velocity=\frac{u+v}{2}\right)$
Now, from the first equation of motion, we know that-
$v=u+at$
Rearranging the above formula, we get
$v-u=at$
$t=\frac{v-u}{a}$
Now, substituting this value of 't' in the displacement formula we get the following equation:
$s=\frac{u+v}{2}\times t$
$s=\left(\frac{u+v}{2}\right)\times \left(\frac{v-u}{a}\right)$
$s=\left(\frac{{v}^{2}-{u}^{2}}{2a}\right)$
$2as={v}^{2}-{u}^{2}$
Rearranging the above equation, we got the third equation of motion- ${v}^{2}={u}^{2}+2as$
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