Question

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 $45^{\circ}$

Answer 1

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

$y=\frac{x^{2}}{2}$
$v_{y}=x v_{x}$
$a_{y}=x a_{x}+v_{x}^{2}$
$x=0 \Rightarrow a_{y}=a_{x}^{2}=1 m / s ^{2}$,
for any value of $a_{x}$
$a_{x}=0 \Rightarrow a_{y}=v_{x}^{2}=1$
$a_{x}=0$
$\tan \theta=\frac{v_{x}}{v_{y}}=x=1\left(x=v_{x} t=1\right)$