Question

A point moves according to the law $x=at,y=at(1-\alpha t)$ where $a$ and $\alpha$ are positive constants and $t$ is time. If the moment at which angle between velocity vector and acceleration vector is $\frac{\pi }{4}$ is given as $t=\frac{x}{\alpha }$. Find $x$

Answer 1

$\overrightarrow{v}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}=a\hat{i}+(a-2a\alpha t)\hat{j}$
$f=\frac{dv}{dt}=-2a\alpha \hat{j}$
$\cos \frac{\pi }{4}=\frac{-2a\alpha \hat{j}.[a\hat{i}+(a-2a\alpha t\left)\hat{j}\right]}{2a\alpha \sqrt{a^2+(a-2a\alpha t{)}^2}}$
$\Rightarrow \frac{1}{{}^{\sqrt{2}}}=\frac{-[a-2a\alpha t]}{\sqrt{a^2+(a-2a\alpha t{)}^2}}$ or
$2(1-2\alpha t{)}^2=1+(1-2\alpha t{)}^2$ or
$(1-2\alpha t{)}^2=1$ or $t=\frac{1}{\alpha }$