Question

A particle moves in $X Y$ - plane with $x$ and $y$ varying with time $t$ as $x(t)=5 t, y(t)=5 t\left(27-t^{2}\right)$. 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
$v_{x}=\frac{d}{d t}(x)=\frac{d}{d t}(5 t)=5\, ms ^{-1}$
Velocity in $y$ -direction is
$v_{y}=\frac{d}{d t}(y)=\frac{d}{d t}\left(5 \times 27 t-5 t^{3}\right)=5 \times 27-5 \times 3 t^{2}$
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 $v_{y}=0.$
$\Rightarrow 5 \times 27-5 \times 3 t^{2}=0$
$\Rightarrow t^{2}=9 \Rightarrow t=3\, s$