Question

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

• A. $1/a$
• B. $1/b$
• C. $1/a+1/b$
• D. $(a+b)/(a^2+b^2)$

Answer 1

Answer: (B)

$1/b$

$x=at$

$y=at(1-bt)$

We have to find the time when the angle between acceleration vector and velocity vector will be $\pi /4$

Now,

$v_x=\frac{dx}{dt}=a$

$v_y=\frac{dy}{dt}=a(1-2bt)$

So, the velocity vector will be $\overrightarrow{v}=a\hat{i}+a(1-2bt)\hat{j}$

the acceleration vector will be $\overrightarrow{A}=\frac{d\overrightarrow{v}}{dt}=-2ab\hat{j}$

So, as you can see the picture acceleration $\overrightarrow{A}$ is in $-Y$ direction and velocity $\overrightarrow{v}$ makes an angle $\pi /4$

$So,tan\left(\pi /4\right)=\frac{a}{|a(1-2bt)|}$

$or,|1-2bt|=1$

$or,1-2bt=\pm 1$

$or,t=0or\frac{1}{b}$

But, when $t=0$, the Y-component of velocity will be in positive direction. So the condition will not work.

Hence, our answer is $t=\frac{1}{b}$

The correct answer is $Option:B$