Question

Starting at time $t=0$ from the origin with speed $1m{s}^{-1}$, 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 $a_y$ respectuvely. Then

• A. $a_x=1m{s}^{-1}$, implies that when the particle is at the origin $a_y=1m{s}^{-1}$
• B. $a_x=0$ implies $a_y=1m{s}^{-1}$ at all times
• C. at $t=0$, the particles velocity points in the x-direction
• D. $a_x=0$ implies that at $t=1s$, the angle between the particles velocity and the x-axis is $45°$

Answer 1

Answer: (A), (B), (C), (D)

$a_x=0$ implies that at $t=1s$, the angle between the particles velocity and the x-axis is $45°$

Explanation for the correct options:

Finding the values of $a_x$, $a_y$, and $\theta$:

At$t=0,x=0,u=1m/sec$

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

Differentiating the above equation with respect to $t$

$\frac{dy}{dt}=\left(\frac{2x}{2}\right)\frac{dx}{dt}$

Now as we know,

$$\begin{gathered} v_y=x{v}_x \\ \mathrm{where},v_y=\mathrm{velocity}\mathrm{component}\mathrm{of}y;v_x=\mathrm{velocity}\mathrm{component}\mathrm{of}x \end{gathered}$$

And,

$a_y=x{a}_x+v{x}_2$

Option (A):

$$\begin{gathered} \frac{dy}{dx}=x \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=1 \end{gathered}$$

Now,

$$\begin{gathered} R_c=\frac{{\left[1+{\left({\displaystyle \frac{dy}{dx}}\right)}^2\right]}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{d^{2}y}{d{x}^2}}} \\ =\frac{{\left(1+x^2\right)}^{{\displaystyle \frac{3}{2}}}}{1} \\ \mathrm{At}x=0 \\ {R}_c=\frac{{\left(1+0\right)}^{{\displaystyle \frac{3}{2}}}}{1} \\ =1 \end{gathered}$$

Therefore,

$$\begin{gathered} a_y=\frac{v^2}{R_c}=1m{s}^{-2} \\ \mathrm{At}x=0 \\ {a}_y=1m{s}^{-2} \end{gathered}$$

That is independent of $a_x$

Therefore, the correct answer is option (A).

Option (B):

If $a_x=0$

$a_y=0+{v_x}^2=1m{s}^{-2}$

Therefore, the correct answer is option (B).

Option (C):

$v_y=x{v}_x\mathrm{at}x=0\mathrm{and}t=0$

$v_y=0\mathrm{and}{v}_x=1m{s}^{-1}$

Therefore, the correct answer is option (C).

Option (D):

x = 1m

v_y / v_x = tan θ = x = 1

$$\begin{gathered} x=1m \\ \mathrm{So}, \\ \frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}=\mathrm{tan\theta }\left[\because \tan \theta =\frac{\mathrm{component}\mathrm{of}y}{\mathrm{component}\mathrm{of}x}\right] \\ =\mathrm{x}\mathbf{}\left[\because \frac{v_y}{v_x}=x\right] \\ =1 \end{gathered}$$]

So,

$$\begin{gathered} \theta ={\tan }^{-1}\left(1\right) \\ \therefore \theta =45° \end{gathered}$$

Therefore, the correct answer is option (D).

Therefore, the correct answers are options (A), (B), (C), and (D).