Starting at time t=0 from the origin with speed 1 ms ^ -1 , a par… - Vidyadip

MCQ

Starting at time $t=0$ from the origin with speed $1 ms ^{-1}$, 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

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

$(B)$ $a_x=0$ implies $a_y=1 ms ^{-2}$ at all times

$(C)$ at $t=0$, the particle's velocity points in the $x$-direction

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

A
$A,B,C$
✓
$A,B,C,D$
C
$A,B,D$
D
$A,B$

✓

Answer

Correct option: B.

$A,B,C,D$

b
$y =\frac{ x ^2}{2}$

at $\left.t=0, \begin{array}{l}x=0, y=0 \\ u=1\end{array}\right\}$ given

$y=\frac{x^2}{2}$

$\frac{d y}{d t}=\frac{1}{2} \cdot 2 x \frac{ dx }{ dt }$

$\Rightarrow v _{ Y }= xv _{ x }$

difference wit time

$a_y=\frac{d x}{d t} \cdot V_x+x a_x$

$a_y=v_x^2+x a_x$

Option

$(A)$ If $a _{ x }=1$ and particle is at origin

$(x=0, y=0)$

$a_y=v_x^2$

$z_y=1^2=1$

At origin, at $t=0 sec$

speed $=1$ given

$(B)$ Option

$a_y=v_s^2+\pi a_s$

given in option $B, a _{ x }=0$

$\Rightarrow a _{ s }= v _{ s }^2$

If $a _{ x }=0, v_{ s }=$ constant $=1$. (all the time)

$\Rightarrow a_{ y }=1^2=1 \text { (all the tume) }$

$(C)$

$at t =0, x =0 \quad v_{ y }=\pi v_{ s }$

$\text { speed }=1$

$v_{ y }=0$

$v_{ s }=1$

$(D)$

$a _{ y }=v_{ x }^2+ xa _{ x }$

$v_y=\pi v_x$

$a _2=0 \text { (given m D option) }$

$\Rightarrow a _{ y }= v _{ x }^2$

$\text { If } a _{ a }=0 \Rightarrow V _{ x }=\text { comstant initially }\left( v _{ x }=1\right)$

$\Rightarrow a _{ y }=1^2=1$

$a t t=1 sec$

$v_y=0+3, \times t=1 \times 1=1$

$\tan \theta=\frac{v_y}{v_s}= x$

$(\theta \rightarrow \text { angle with } x \text { axis) }$

$\tan \theta=\frac{v_s}{v_s}=\frac{1}{1}=1$

$\theta=45^{\circ}$

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