A particle moves along x-axis such that its position veries with time as `x=50t-5t^2`. Select the correct alternative (s).

Options

A. The particle has travelled 130m distance by `t=6s`.
B. At `t=4s` and `t=12s`, the particle is at 120 m distance from starting point.
C. the particle will never be at a distance of 130 m from starting point.
D. During the whole motion, the particle does not have same velocity at 2 different instants.

Explanation

To solve the problem, we need to analyze the motion of a particle along the x-axis, given its position as a function of time:

\[ x(t) = 50t - 5t^2 \]

### Step 1: Determine the position of the particle at \( t = 6 \) seconds.

We substitute \( t = 6 \) into the position equation:

\[ x(6) = 50(6) - 5(6^2) \]

Calculating this:

\[ x(6) = 300 - 5(36) = 300 - 180 = 120 \text{ meters} \]

### Step 2: Find the maximum position of the particle.

To find when the particle comes to rest (maximum position), we need to find the velocity by differentiating the position function with respect to time:

\[ v(t) = \frac{dx}{dt} = 50 - 10t \]

Setting the velocity to zero to find the time at which the particle stops:

\[ 50 - 10t = 0 \implies t = 5 \text{ seconds} \]

### Step 3: Calculate the position at \( t = 5 \) seconds.

Now, we substitute \( t = 5 \) into the position equation:

\[ x(5) = 50(5) - 5(5^2) \]

Calculating this:

\[ x(5) = 250 - 5(25) = 250 - 125 = 125 \text{ meters} \]

### Step 4: Calculate the total distance traveled by the particle.

The particle travels from the starting position (0 meters) to the maximum position (125 meters) and then back to the position at \( t = 6 \) seconds (120 meters).

The distance traveled is:

1. From 0 to 125 meters: \( 125 \) meters
2. From 125 meters back to 120 meters: \( 125 - 120 = 5 \) meters

Total distance traveled:

\[ \text{Total Distance} = 125 + 5 = 130 \text{ meters} \]

### Step 5: Analyze the options.

- The particle has indeed traveled a distance of 130 meters by \( t = 6 \) seconds.
- At \( t = 4 \) seconds and \( t = 12 \) seconds, the particle is at the same position of 120 meters.
- The particle can travel a distance of 130 meters, but it will never be at a position of 130 meters.
- The particle's velocity changes direction but not magnitude at different times.

### Conclusion

The correct alternatives are:
- A: The particle has traveled a distance of 130 meters by \( t = 6 \) seconds.
- B: At \( t = 4 \) and \( t = 12 \), the particle is at 120 meters.
- D: The particle does not have the same velocity at different times.