Question

Starting at time $t=0$ from the origin with speed $1\text{m/s}$, a particle follows a two-dimensional trajectory in the $\text{x-y plane}$ 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. $a_x=1{\text{m/s}}^2$ implies that when the particle is at origin , $a_y=1{\text{m/s}}^2$
  
• B. $a_x=0$ implies $a_y=1{\text{m/s}}^2$ at all times .
• C. At $t=0$ the particles velocity points in the $\text{x-direction}$.
• D. $a_x=0$ implies that at $t=1\text{s}$, the angle between the particle's velocity and $\text{x-axis}$ is ${45}^{\circ }$.

Answer 1

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

$a_x=0$ implies that at $t=1\text{s}$, the angle between the particle's velocity and $\text{x-axis}$ is ${45}^{\circ }$.

Given, $y=\frac{x^2}{2}.......\left(1\right)$

at $t=0,x=0,y=0$

and initial velocity, $u=1\text{m/s}$.

on differentiating $\left(1\right)$ with respect to time $\left(t\right)$,

$\frac{dy}{dt}=\frac{1}{2}\left(2x\right)\frac{dx}{dt}$

$\Rightarrow v_y=x{v}_x.........\left(2\right)$

On differentiating again, we have

$\Rightarrow \frac{d{v}_y}{dt}=v_x\frac{dx}{dt}+x\frac{d{v}_x}{dt}$

$\Rightarrow a_y=v_x^2+x{a}_x.....\left(3\right)$

At origin, at $t=0,x=0$

So, from $\left(2\right)$, $v_y=0$

It is given, at $t=0,u=1\text{m/s}$.

Thus, $v_x=1\text{m/s}$ at $t=0$.

Hence, particle's velocity points in the $\text{x-direction}$.

option $\left(c\right)$ is correct.

From $\left(3\right)$, at origin,

$a_y=1^2+0\times 1$ $(\text{If}{a}_x=1{\text{m/s}}^2)$

$\Rightarrow a_y=1{\text{m/s}}^2$

option $\left(a\right)$ is correct.

If $a_x=0$, from $\left(3\right)$

$a_y=v_x^2+x\times 0$

$\Rightarrow a_y=1^2=1{\text{m/s}}^2$ $($Always$)$

option (b) is correct.

If $a_x=0$ then $a_y=1{\text{m/s}}^2$.

Given , at $t=0$, $v_x=1$ and $v_y=0$

At $t=1\text{s}$, $(v_x{)}_f=1\text{m/s}$

using, $v=u+at$

and $(v_y{)}_f=0+1\times 1=1$

So , $\tan \theta =\frac{(v_y{)}_f}{(v_x{)}_f}=\frac{1}{1}=1$

$\Rightarrow \theta ={45}^{\circ }$

option $\left(d\right)$ is correct.

Hence, all the options are correct here.