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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

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a_x = 0 implies a_y = 1ms^-2 at all times

a_x = 1ms^-2 implies that when the particle is at the origin, a_y = 1ms^-2

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 45⁰

answer is A, B, C, D.

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Detailed Solution

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

$v_{y} {= x v}_{x}$

$a_{y} {=xa}_{x} {+v}_{x}^{2}$

$x = 0 \Rightarrow a_{y} {=v}_{x}^{2} {= 1m/s}^{2}$

For any value of a_x

$a_{x} = 0 \Rightarrow a_{y} {= v}_{x}^{2} =1$

$tan θ= \frac{V_{y}}{V_{x}} = x =1$

x = v_x t = 1

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