Starting at 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 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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$a_{x} = 1 m s^{- 2}$ 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 $45 °$ .

answer is A, B, C, D.

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Detailed Solution

The equation of trajectory,
$y = \frac{x^{2}}{2} ... (i)$
Question Image
At $t = 0, \begin{matrix} x = 0, y = 0 \\ u = 1 m / s \end{matrix}}$ given
Differentiating Eq. (i) w.r.t. time
$\frac{d y}{d t} = \frac{1}{2} \times 2 x \frac{d x}{d t}$
$v_{y} = x v_{x} ... (ii)$
Again differentiating Eq. (ii) w.r.t. time
$\frac{d v_{y}}{d t} = \frac{d x}{d t} \times \frac{d x}{d t} + x (\frac{d^{2} v_{x}}{d t^{2}})$
$a_{y} = v_{x}^{2} + x a_{x} ... (iii)$
If $a_{x} = 1 m s^{- 2}$ and particle is at origin $(x = 0, y = 0)$
$a_{y} = v_{x}^{2}$
$a_{y} = 1^{2} = 1 m s^{- 2}$
Hence option (A) is correct.
If $a_{x} = 0$ , from Eq. (iii), $a_{y} = v_{x}^{2}$
$a_{y} = 1^{2} = 1 m s^{- 2}$ (all the time)
Hence option (B) is correct.
At $t = 0, x = 0$ from Eq. (ii), $v_{y} = 0$
But speed of the particle at $t = 0$ is 1 m/s.
Hence , $v_{x} = 1 m / s$ Hence option (C) is correct.
If $a_{x} = 0$ , from Eq. (iii)
$a_{y} = v_{x}^{2} = 1^{2} = 1 m s^{- 2}$ (constant)
At $t = 1 \sec$
$v_{y} = 0 + a_{y} \times t = 1 \times 1 = 1 m s^{- 1}$
Also, if $a_{x} = 0 \Rightarrow v_{x} =$ constant $= 1 m s^{- 1}$
Then angle made by velocity vector with x-axis
$\tan θ = \frac{v_{y}}{v_{x}} = \frac{1}{1} = 1 \Rightarrow θ = 45 °$
Hence, option (D) is correct.