Question

Starting at time $t=0$ from the origin with speed $1{\text{ms}}^{-1}$, a particle follows a two-dimensional trajectory in the $x-y$ plane so that its coordinates are related by the equation $y=\frac{x^2}{2}$. The $x$ and $y$ components of its acceleration are denoted by $a_x$ and $a_y$, respectively. Then

• A. $a_x=0$ implies that at $t=1\text{s}$, the angle between the particle’s velocity and the $x$ axis is ${45}^{\circ }$
• B. $a_x=0$ implies $a_y=1{\text{ms}}^{-2}$ at all times
• C. at $t=0$, the particle’s velocity points in the $x-$direction
• D. $a_x$ = $1{\text{ms}}^{-2}$ implies that when the particle is at the origin, $a_y=1{\text{ms}}^{-2}$

Answer 1

Answer: (A), (B), (C), (D)

$a_x$ = $1{\text{ms}}^{-2}$ implies that when the particle is at the origin, $a_y=1{\text{ms}}^{-2}$

$y=\frac{x^2}{2}$
at $t=0$, ${}_{u=1}^{x=0,y=0}\}\text{given}$
equation of trajectory,
$y=\frac{x^2}{2}$
On differentiating both sides,
$\frac{dy}{dt}=\frac{1}{2}.2x\frac{dx}{dt}$
$\Rightarrow v_y=x{v}_x$
Again differentiating wrt time,
$a_y=\frac{dx}{dt}.v_x+x{a}_x$
$\Rightarrow a_y=v_x^2+x{a}_x....\left(1\right)$

Option
$\left(A\right)$
If $a_x=1$and particle is at origin i.e $(x=0,y=0)$
$a_y=V_x^2$
$a_y=1^2=1$
At origin, at $t=0$ the speed $=1\text{m/s}$ is given .

Option $\left(B\right)$:
$a_y=v_x^2+x{a}_x$
given that, $a_x=0$
$\Rightarrow a_y=v_x^2$
If $a_x=0$ , $v_x=\text{constant}=1$, all the time
$\Rightarrow a_y=1^2=1$ (all the time)

Option $\left(C\right)$:
at $t=0,x=0$
$\Rightarrow v_y=x{v}_x=0$
$\because \text{speed}=1\text{m/s}$
$\Rightarrow v_x=1\text{m/s}$

Option $\left(D\right)$:
$a_y=v_x^2+x{a}_x$
$v_y=x{v}_x$
$\because a_x=0$, (given in option 'D')
$\Rightarrow a_y=v_x^2$
If $a_x=0\Rightarrow v_x=\text{constant}$
initially $(v_x=1\text{m/s})$
$\Rightarrow a_y=1^2=1$
at $t=1\text{s}$
$\Rightarrow v_y=0+a_y\times t=1\times 1=1\text{m/s}$
$\tan \theta =\frac{v_y}{v_x}$
($\theta \to$ angle made with $x-$ axis)
$\tan \theta =\frac{v_y}{v_x}=\frac{1}{1}=1$
$\therefore \theta ={45}^{\circ }$