Animated Solution for Physics - Kinematics: Starting at time $t=0$ from the origin with speed $1{\text{ ms}}^{-1}$, 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

The problem presents a fascinating scenario of a particle moving along a parabolic trajectory, $y=\frac{x^2}{2}$. We are given its initial speed and asked to deduce the relationships between its velocity and acceleration components under various conditions. This is a classic test of your ability to link geometry with kinematics using calculus.

We know the particle starts at the origin $(0,0)$ at time $t=0$ with a speed of $1{\text{ ms}}^{-1}$. The path it follows is a parabola opening upwards.

To understand how the particle moves, we need to relate its position to its velocity. We do this by differentiating the trajectory equation with respect to time $t$.

Using the chain rule, differentiating $y=\frac{x^2}{2}$ gives:

This is our first master equation. It tells us exactly how the y-velocity depends on the x-position and the x-velocity.

Let's see what happens at the very beginning. At $t=0$, the particle is at the origin, so $x=0$.

Plugging this into our velocity equation:

This means initially, the particle has no vertical velocity. Since we are given that the total initial speed is $1{\text{ ms}}^{-1}$, and speed is $\sqrt{v_x^2+v_y^2}$, it must be that $v_x=1{\text{ ms}}^{-1}$ (assuming it moves in the positive x-direction).

Therefore, at $t=0$, the velocity vector is $\vec{v}=1\hat{i}$, which points purely in the x-direction. This confirms that Option (C) is correct.

To evaluate the other options, we need to bring acceleration into the picture. We do this by differentiating our velocity equation, $v_y=x{v}_x$, with respect to time.

Applying the product rule on the right side:

This is our second master equation, linking the acceleration components.

Now, let's systematically check the remaining options using our master equations.

Checking Option (A): The option asks what happens if $a_x=1{\text{ ms}}^{-2}$ when the particle is at the origin. At the origin, $x=0$. Substituting this into our acceleration equation: We already established that at the origin, $v_x=1{\text{ ms}}^{-1}$. This perfectly matches the statement. Thus, Option (A) is correct.

Checking Option (B): What if $a_x=0$ at all times? If the x-acceleration is zero, it means the x-velocity ($v_x$) is constant. Since it started at $1{\text{ ms}}^{-1}$, it will remain $1{\text{ ms}}^{-1}$ forever. Now, look at our acceleration equation again: Since $v_x$ is always $1$, $a_y$ will always be $1^2=1{\text{ ms}}^{-2}$. This confirms that Option (B) is correct.

Checking Option (D): This option builds on the condition that $a_x=0$. As we just saw, this implies $v_x=1{\text{ ms}}^{-1}$ (constant) and $a_y=1{\text{ ms}}^{-2}$ (constant). We need to find the angle of velocity at $t=1\text{ s}$. Since $a_y$ is constant, we can use the first equation of motion for the y-direction: Initially, $u_y=0$. So at $t=1\text{ s}$: The velocity vector at $t=1\text{ s}$ is $\vec{v}=1\hat{i}+1\hat{j}$. The angle $\theta$ it makes with the x-axis is given by: This proves that Option (D) is correct.

In conclusion, all four options are correct. This problem beautifully demonstrates the power of calculus in unraveling the kinematics of curvilinear motion!