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=(x2/2). The x and y components of its acceleration are denoted by a x and a y, respectively. Then

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=(x2/2). 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°

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f25607e978a3145117f8b7921df27a3c-.png" /> y=(x2/2) vy=x vx ay=x ax+vx2 x=0 ⇒ ay=ax2=1 m / s 2, for any value of ax ax=0 ⇒ ay=vx2=1 ax=0 tan θ=(vx/vy)=x=1(x=vx t=1)

$a_{x} = 1 m s^{- 2}$ implies that when the particle is at the origin, $a_{y} = 1 m s^{- 2}$

$a_{x} = 0$ implies $a_{y} = 1 m s^{- 2}$ at all times

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

$a_{x} = 0$ implies that at $t = 1 s$ , the angle between the particle's velocity and the $x$ axis is $4 5^{\circ}$

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

Q. Starting at time t=0 from the origin with speed 1ms−1, a particle follows a two-dimensional trajectory in the x−y plane so that its coordinates are related by the equation y=2x2​. The x and y components of its acceleration are denoted by ax​ and ay​, respectively. Then

ax​=1ms−2 implies that when the particle is at the origin, ay​=1ms−2

ax​=0 implies ay​=1ms−2 at all times

at t=0, the particle's velocity points in the x-direction

ax​=0 implies that at t=1s, the angle between the particle's velocity and the x axis is 45∘

y=2x2​ vy​=xvx​ ay​=xax​+vx2​ x=0⇒ay​=ax2​=1m/s2, for any value of ax​ ax​=0⇒ay​=vx2​=1 ax​=0 tanθ=vy​vx​​=x=1(x=vx​t=1)

$a_{x} = 1 m s^{- 2}$ implies that when the particle is at the origin, $a_{y} = 1 m s^{- 2}$

$a_{x} = 0$ implies $a_{y} = 1 m s^{- 2}$ at all times

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

$a_{x} = 0$ implies that at $t = 1 s$ , the angle between the particle's velocity and the $x$ axis is $4 5^{\circ}$

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