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 $y = x^2/2$. The $x$ and $y$ components of its acceleration are denoted by $a_x$ and $a_y$, respectively. Then

Step 1: Differentiate the trajectory equation.

The trajectory of the particle is given by the relation between its coordinates.

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

Differentiate with respect to time $t$ to find the velocity relation.

$$\dot{y} = x\dot{x}$$

Differentiate again with respect to time $t$ to find the acceleration relation.

$$\ddot{y} = \dot{x}^{2} + x\ddot{x}$$

Express this in terms of velocity and acceleration components.

$$a_y = v_x^{2} + x a_x$$

Step 2: Analyze the initial velocity at $t = 0$.

At $t = 0$, the particle is at the origin.

$$x = 0$$

Substitute $x = 0$ into the velocity relation.

$$v_y = 0$$

Since the initial speed is $1\,\text{ms}^{-1}$, the velocity is entirely in the $x$-direction.

$$v_x = 1\,\text{ms}^{-1}$$

Thus, at $t = 0$, the velocity vector is along the $x$-direction (Statement 3 is correct).

Step 3: Determine the acceleration at the origin.

Substitute $x = 0$ and $v_x = 1\,\text{ms}^{-1}$ into the acceleration relation.

$$a_y = 1^{2} + 0 \cdot a_x$$ $$a_y = 1\,\text{ms}^{-2}$$

This value is independent of $a_x$, meaning $a_y = 1\,\text{ms}^{-2}$ at the origin even if $a_x = 1\,\text{ms}^{-2}$ (Statement 1 is correct).

Step 4: Analyze the motion under zero horizontal acceleration.

Assume the horizontal acceleration is zero.

$$a_x = 0$$

The horizontal velocity remains constant at its initial value.

$$v_x = 1\,\text{ms}^{-1}$$

Substitute these values into the acceleration relation.

$$a_y = 1^{2} + x(0)$$ $$a_y = 1\,\text{ms}^{-2}$$

Thus, $a_y = 1\,\text{ms}^{-2}$ for all $t$ (Statement 2 is correct).

Step 5: Calculate the velocity angle at $t = 1\,\text{s}$ for $a_x = 0$.

Find the position $x$ as a function of time.

$$x = t$$

Find the vertical velocity component as a function of time.

$$v_y = t$$

Evaluate the velocity components at $t = 1\,\text{s}$.

$$v_x = 1\,\text{ms}^{-1}$$ $$v_y = 1\,\text{ms}^{-1}$$

Calculate the angle $\theta$ of the velocity vector with the $x$-axis.

$$\tan\theta = \frac{v_y}{v_x} = 1$$ $$\theta = 45^\circ$$

Thus, the velocity vector makes an angle of $45^\circ$ with the $x$-axis (Statement 4 is correct).

Answer: Statements (1), (2), (3), and (4) are all correct.