Question
What is the position at any time, for a body starting from rest, with an acceleration $a = at^2$?
✓
Answer
$\text{a}=\alpha\text{t}^2$ is a variable acceleration.
$\therefore \text{dv}=\alpha\text{t}^2\text{dt}$ $|\text{v}|^\text{v}_0=\Big|\frac{\alpha\text{t}^3}{3}\Big|^\text{t}_0$
$\Rightarrow \text{v}=\frac{\alpha\text{t}^3}{3}$ Also $\text{v}=\frac{\text{dx}}{\text{dt}}$
$\therefore \text{dx}=\frac{\alpha\text{t}^3}{3}\text{dt}$
$|\text{x}|^{{\text{x}}_\text{f}}_{\text{x}_\text{i}}=\frac{\alpha\text{t}^4}{12}$
$\therefore \text{x}_\text{f}-\text{x}_\text{i}=\frac{\alpha\text{t}^4}{12}$
$\text{x}_\text{f}=\text{x}_\text{i}+\frac{\alpha\text{t}^4}{12}$ is the final position at time 't' seconds.
$\therefore \text{dv}=\alpha\text{t}^2\text{dt}$ $|\text{v}|^\text{v}_0=\Big|\frac{\alpha\text{t}^3}{3}\Big|^\text{t}_0$
$\Rightarrow \text{v}=\frac{\alpha\text{t}^3}{3}$ Also $\text{v}=\frac{\text{dx}}{\text{dt}}$
$\therefore \text{dx}=\frac{\alpha\text{t}^3}{3}\text{dt}$
$|\text{x}|^{{\text{x}}_\text{f}}_{\text{x}_\text{i}}=\frac{\alpha\text{t}^4}{12}$
$\therefore \text{x}_\text{f}-\text{x}_\text{i}=\frac{\alpha\text{t}^4}{12}$
$\text{x}_\text{f}=\text{x}_\text{i}+\frac{\alpha\text{t}^4}{12}$ is the final position at time 't' seconds.
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