OR
Prove the following relations by calculus method:OR
$\therefore$ instantaneous velocity, $\text{v}=\frac{\text{dx}}{\text{dt}}$ or dx = vdt.
dx = (u + at)dt $(\because$ v = u + at)
let x0 and x be the displacements of the obhect at time 'zero' and 't'.
$\int\limits^{\text{x}}_{\text{x}_0}\text{dx}=\int\limits^{\text{t}}_0(\text{u}+\text{at})\text{dt}$
$=\text{u}\int\limits^{\text{t}}_0\text{dt}+\text{a}\int\limits^{\text{t}}_0\text{t dt}$
$(\text{x})^{\text{x}}_{\text{x}_0}=\text{u}(\text{t})^\text{t}_0+\text{a}\Big(\frac{\text{t}^2}{2}\Big)^\text{t}_0$
$\text{x}-\text{x}_0=\text{ut}+\frac{1}{2}\text{at}^2$
If x = x0 = s (distance) covered by the object.
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
Then, $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{dv}}{\text{dx}}\times\frac{\text{dx}}{\text{dt}}=\frac{\text{dv}}{\text{dx}}\times\text{v}$
$\text{a dx}=\text{v dv}$
Integrating,
$\int\limits^{\text{x}}_{\text{x}_0}\text{a dx}=\int\limits^{\text{V}}_\text{u}\text{v dv}$
$\Rightarrow\ \text{a}|\text{x}|^{\text{x}}_{\text{x}_0}=\Big[\frac{\text{v}^2}{2}\Big]^\text{V}_\text{u}$
$\text{a}(\text{x}-\text{x}_0)=\frac{\text{v}^2}{2}-\frac{\text{u}^2}{2}$
$\text{v}^2-\text{u}^2=2\text{a}(\text{x}-\text{x}_0)$
$\text{v}^2-\text{u}^2=2\text{a}\text{s.}$
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