The position function `x(t)` of a particle moving along an `x` axis is `x=4.0-6.0t^(2)`, where `x` in meters and t in seconds. (a) At what time and (b) where does the particle ( momentarily) stop ? At what (c) negative time and (d) positive time does the particle pass through the origin ?

Answer

To solve the problem step by step, we will analyze the position function of the particle and derive the necessary information.

### Given:
The position function of the particle is given by:
\[ x(t) = 4.0 - 6.0t^2 \]
where \( x \) is in meters and \( t \) is in seconds.

### (a) At what time does the particle momentarily stop?
To find when the particle stops, we need to determine when the velocity is zero. The velocity \( v(t) \) is the derivative of the position function \( x(t) \).

1. **Differentiate the position function:**
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(4.0 - 6.0t^2) = -12.0t \]

2. **Set the velocity to zero:**
\[ -12.0t = 0 \]
Solving for \( t \):
\[ t = 0 \, \text{seconds} \]

### (b) Where does the particle momentarily stop?
Now, we will find the position of the particle at \( t = 0 \).

1. **Substitute \( t = 0 \) into the position function:**
\[ x(0) = 4.0 - 6.0(0)^2 = 4.0 \, \text{meters} \]

### (c) At what negative time does the particle pass through the origin?
To find when the particle passes through the origin, we set \( x(t) = 0 \).

1. **Set the position function to zero:**
\[ 0 = 4.0 - 6.0t^2 \]

2. **Rearranging gives:**
\[ 6.0t^2 = 4.0 \implies t^2 = \frac{4.0}{6.0} = \frac{2}{3} \]

3. **Taking the square root:**
\[ t = \pm \sqrt{\frac{2}{3}} \approx \pm 0.816 \, \text{seconds} \]

Since we are looking for the negative time:
\[ t = -\sqrt{\frac{2}{3}} \approx -0.816 \, \text{seconds} \]

### (d) At what positive time does the particle pass through the origin?
From the previous calculation, the positive time is:
\[ t = \sqrt{\frac{2}{3}} \approx 0.816 \, \text{seconds} \]

### Summary of Answers:
(a) The particle momentarily stops at \( t = 0 \, \text{seconds} \).
(b) The position where it stops is \( x = 4.0 \, \text{meters} \).
(c) The particle passes through the origin at \( t \approx -0.816 \, \text{seconds} \).
(d) The particle passes through the origin at \( t \approx 0.816 \, \text{seconds} \).