Question

A particle moves in $XY$ - plane with $x$ and $y$ varying with time $t$ as $x(t)=5t,y(t)=5t\left(27-t^2\right)$. At what time in seconds, the direction of velocity and acceleration will be perpendicular to each other?

• A. $5\sqrt{\frac{27}{2}}$
• B. 5
• C. $5\sqrt{12}$
• D. 3

Answer 1

Answer: (D)

Velocity in $x$ -direction is
$v_x=\frac{d}{dt}(x)=\frac{d}{dt}(5t)=5m{s}^{-1}$
Velocity in $y$ -direction is
$v_y=\frac{d}{dt}(y)=\frac{d}{dt}\left(5\times 27t-5t^3\right)=5\times 27-5\times 3t^2$
As, $v_x=$ constant and $v_y=$ time dependent.
$\therefore$ This is a case similar to projectile motion and velocity and acceleration are perpendicular, when $v_y=0.$
$\Rightarrow 5\times 27-5\times 3t^2=0$
$\Rightarrow t^2=9\Rightarrow t=3s$