For the one dimensional motion, described by `x=t-sint`

Options

A. `x(t) gt 0` for all t gt 0
B. `nu(t) gt 0` for all `t gt 0`
C. `a(t) gt 0` for all `t gt 0`
D. all of these

Explanation

To solve the problem of one-dimensional motion described by the equation \( x = t - \sin t \), we will analyze the position, velocity, and acceleration step by step.

### Step 1: Analyze the Position Function
The position function is given by:
\[ x(t) = t - \sin t \]
We will evaluate this function for various values of \( t \).

- For \( t = 0 \):
\[ x(0) = 0 - \sin(0) = 0 \]

- For \( t = \frac{\pi}{2} \):
\[ x\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - 1 \]
Since \( \frac{\pi}{2} \approx 1.57 \), \( \frac{\pi}{2} - 1 > 0 \).

- For \( t = \pi \):
\[ x(\pi) = \pi - \sin(\pi) = \pi - 0 = \pi > 0 \]

- For \( t = 2\pi \):
\[ x(2\pi) = 2\pi - \sin(2\pi) = 2\pi - 0 = 2\pi > 0 \]

From the evaluations, we can see that \( x(t) \) is greater than 0 for \( t > 0 \).

### Step 2: Calculate the Velocity
Velocity is the derivative of the position function:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t - \sin t) = 1 - \cos t \]

Now, we will evaluate the velocity for various values of \( t \):

- At \( t = 0 \):
\[ v(0) = 1 - \cos(0) = 1 - 1 = 0 \]

- At \( t = \frac{\pi}{2} \):
\[ v\left(\frac{\pi}{2}\right) = 1 - \cos\left(\frac{\pi}{2}\right) = 1 - 0 = 1 \]

- At \( t = \pi \):
\[ v(\pi) = 1 - \cos(\pi) = 1 - (-1) = 2 \]

- At \( t = 2\pi \):
\[ v(2\pi) = 1 - \cos(2\pi) = 1 - 1 = 0 \]

The velocity \( v(t) \) varies between 0 and 2, indicating that it is not always greater than 0 for \( t > 0 \).

### Step 3: Calculate the Acceleration
Acceleration is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(1 - \cos t) = \sin t \]

Now, we will evaluate the acceleration for various values of \( t \):

- At \( t = 0 \):
\[ a(0) = \sin(0) = 0 \]

- At \( t = \frac{\pi}{2} \):
\[ a\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

- At \( t = \pi \):
\[ a(\pi) = \sin(\pi) = 0 \]

- At \( t = 2\pi \):
\[ a(2\pi) = \sin(2\pi) = 0 \]

The acceleration \( a(t) \) varies between -1 and 1, and is not always 0 for \( t > 0 \).

### Conclusion
Based on the evaluations:
1. The position \( x(t) \) is greater than 0 for all \( t > 0 \).
2. The velocity \( v(t) \) is not always greater than 0 for \( t > 0 \).
3. The acceleration \( a(t) \) is not always 0 for \( t > 0 \).

Thus, the only correct option is that the position is always greater than 0 for all \( t > 0 \).