Question

A point in xy-plane according to equation $x=at$, $x=at\left(1-bt\right)$ where a and b are positive constants and t is time. The instant at which velocity vector is at $\frac{\pi }{4}$ with acceleration vector is given by-

• A. $\frac{1}{a}$
• B. $\frac{1}{b}$
• C. $\frac{1}{a}+\frac{1}{b}$
• D. $\left(a+b\right)\left(a^2+b^2\right)$

Answer 1

The correct option is B $\frac{1}{b}$

$x=aty=at-\frac{1}{2}b{t}^2$

$v_x=\frac{dx}{dt}=a{v}_y=\frac{dy}{dt}=a-ab\times 2t$

$a_x=0a_y=\frac{d{v}_y}{dt}=-2ab$

$\overrightarrow{a}=-2ab\hat{i}$

$\left|\overrightarrow{a}\right|=2ab$

At t, $v_x=a$

$v_y=a-2abt$

$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}-a\hat{i}-2abt\hat{j}$

Now, $\overrightarrow{a.}\overrightarrow{v}=\left|\overline{a}\right|.\left|\overline{b}\right|\cos \theta$

$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

$=a\hat{i}+\left(a-2bt\right)\hat{j}$

$\overrightarrow{a}=-2ab\hat{j}$

$\overline{v}.\overline{a}=\left|\overline{v}\right|.\left|\overline{a}\right|\cos \frac{\pi }{4}$

Or, $0+\left(a-2abt\right)\left(-2ab\right)=\sqrt{a^2+{\left(a-2abt\right)}^2}.2ab\frac{1}{\sqrt{2}}$

Or, ${\left(\left(-a+2abt\right)\sqrt{2}\right)}^2=a^2+{\left(a-2abt\right)}^2$

Or, $2\left[{\left(a-2abt\right)}^2\right]-{\left(a-2ab\right)}^2=a^2$

Or, ${\left(a-2abt\right)}^2=a^2$

$a-2abt=-a$

$2a=2abt$

$t=\frac{1}{b}$