Starting at time from the origin with speed , a particle follows a two-dimensional trajectory in the plane so that its coordinates are related by the equation . The and components of its acceleration are denoted by and , 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 $y = \frac{x^{2}}{2}$ .

The $x$ and $y$ components of its acceleration are denoted by $a_{x}$ and $a_{y}$ , respectively. Then [2020]

Smart Achievers · Accepted Answer

(1, 2, 3, 4)

According to question, equation

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

$\begin{matrix} x = 0, y = 0 \\ At t = 0, \\ u = 1 \end{matrix}} Given$

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

$\frac{d y}{d t} = \frac{1}{2} \cdot 2 x \frac{d x}{d t} = x \frac{d x}{d t} \Rightarrow v_{y} = x v_{x}$

$Now differentiate w.r.t. time$

$\frac{d^{2} y}{d t^{2}} = x \frac{d^{2} x}{d t^{2}} + {(\frac{d x}{d t})}^{2}$

$a_{y} = \frac{d x}{d t} \cdot v_{x} + x a_{x}$

$a_{y} = v_{x}^{2} + x a_{x}$

$(a) If a_{x} = 1 and particle is at origin (x = 0, y = 0)$

$a_{y} = v_{x}^{2}$

$a_{y} = 1^{2} = 1 {m/s}^{2}$

$(b) a_{x} = 0$

$a_{y} = v_{x}^{2} + x a_{x} \Rightarrow a_{y} = v_{x}^{2}$

$If a_{x} = 0, v_{x} = constant = 1 \Rightarrow a_{y} = 1^{2} = 1$

$(c) At t = 0, x = 0, v_{y} = x v_{x}$

$speed = 1; v_{y} = 0 \Rightarrow v_{x} = 1$

$(d) a_{x} = 0 implies that at t = 1 s$

$a_{y} = v_{x}^{2} + x a_{x} \Rightarrow v_{y} = x v_{x} \Rightarrow a_{y} = v_{x}^{2}$

$If a_{x} = 0 \Rightarrow v_{x} = constant initially (v_{x} = 1) \Rightarrow a_{y} = 1^{2} = 1$

$At t = 1 s$

$v_{y} = 0 + a_{y} t = 1 \cdot 1 = 1$

$\tan θ = \frac{v_{y}}{v_{x}} = x$ ( $θ \to$ angle with $x$ axis)

$∴$ $\tan θ = \frac{v_{y}}{v_{x}} = \frac{1}{1} = 1$ $\Rightarrow θ = 45 °$

Solution

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

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

3 $at t = 0, the particle's velocity points in the x -direction$

4 $a_{x} = 0 implies that at t = 1 s, the angle between the particle's velocity and the x -axis is 45 °$

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Solution

Q. 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=x22. The x and y components of its acceleration are denoted by ax and ay, respectively. Then [2020]

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

2 ax=0 implies ay=1 ms-2 at all times

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

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

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

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

3 $at t = 0, the particle's velocity points in the x -direction$

4 $a_{x} = 0 implies that at t = 1 s, the angle between the particle's velocity and the x -axis is 45 °$