Starting time t = 0 from the origin with speed 1 $m s^{- 1}$ , a particle follows a two-dimensional trajectory in the x-y plane so that it 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

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$a_{x} = 1 m s^{- 2}$ implies that when the particle is 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 $45^{\circ}$ .

answer is A, B, C, D.

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Detailed Solution

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

$\frac{d y}{d t} = \frac{x d x}{d t}$

$a y = {(v_{x})}^{2} + x a_{x}$

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