Motion

Given:

At time $ t = 0 $, the particle starts from the origin $ (0, 0) $ with a speed of $ v = 1\, \text{m/s} $.

The particle moves along the path defined by $ y = \frac{x^2}{2} $.

Understanding the Particle's Motion

Since the particle follows the path $ y = \frac{x^2}{2} $, we can find relationships between its velocity and acceleration components.

1. Velocity Components

Let $ x(t) $ and $ y(t) $ be the particle's position as functions of time. Then:

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

Differentiate both sides with respect to time $ t $:

$ \frac{dy}{dt} = x \frac{dx}{dt} $

The velocity components are:

$ v_x = \frac{dx}{dt} $

$ v_y = \frac{dy}{dt} = x v_x $

2. Acceleration Components

The acceleration components are:

$ a_x = \frac{dv_x}{dt} $

$ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(x v_x) = v_x^2 + x a_x $

Analyzing Each Option

Option A

Statement: $ a_x = 1\, \text{m/s}^2 $ implies that when the particle is at the origin, $ a_y = 1\, \text{m/s}^2 $.

Verification:

At the origin $ x = 0 $, and from initial conditions, $ v_x = 1\, \text{m/s} $.

Using the expression for $ a_y $:

$ a_y = v_x^2 + x a_x = (1\, \text{m/s})^2 + 0 \times a_x = 1\, \text{m/s}^2 $

Conclusion: Option A is true.

Option B

Statement: $ a_x = 0 $ implies $ a_y = 1\, \text{m/s}^2 $ at all times.

Verification:

If $ a_x = 0 $, then $ v_x = \text{constant} = 1\, \text{m/s} $.

Position along $ x $: $ x = v_x t = t $.

Then $ v_y = x v_x = t \times 1\, \text{m/s} = t\, \text{m/s} $.

Therefore, $ a_y = \frac{dv_y}{dt} = 1\, \text{m/s}^2 $.

Conclusion: Option B is true.

Option C

Statement: At $ t = 0 $, the particle's velocity points in the $ x $-direction.

Verification:

The slope of the path is $ \frac{dy}{dx} = x $.

At $ x = 0 $, $ \frac{dy}{dx} = 0 $, meaning the path is horizontal.

Thus, $ v_y = 0 $ at $ t = 0 $, and $ v_x = 1\, \text{m/s} $.

Conclusion: Option C is true.

Option D

Statement: $ a_x = 0 $ implies that at $ t = 1\, \text{s} $, the angle between the particle's velocity and the $ x $-axis is $ 45^\circ $.

Verification:

With $ a_x = 0 $, $ v_x = 1\, \text{m/s} $.

At $ t = 1\, \text{s} $, $ x = 1\, \text{m} $.

Then $ v_y = x v_x = 1\, \text{m} \times 1\, \text{m/s} = 1\, \text{m/s} $.

The angle $ \theta $ between $ \vec{v} $ and the $ x $-axis is:

$ \tan \theta = \frac{v_y}{v_x} = \frac{1\, \text{m/s}}{1\, \text{m/s}} = 1 \implies \theta = 45^\circ $

Conclusion: Option D is true.

Final Answer

All options A, B, C, and D are true.