A car moves along an `x` axis through a distance of 900 m, starting at rest ( at `x=0`) and ending at rest ( at `x=900m`). Through the first `1/4` of that distance, its acceleration is `+2.75m//s^(2)`. Through the rest of that distance, its acceleration is `-0.750m//s^(2)`. What are (a) its travel time through the 900 m and (b) its maximum speed ? (c) Graph position `x`, velocity v, and acceleration a versus time t for the trip.

Answer

To solve the problem step by step, we will break it down into parts (a), (b), and (c) as requested.

### Given Data:
- Total distance \( S = 900 \, \text{m} \)
- Initial velocity \( u = 0 \, \text{m/s} \) (at \( x = 0 \))
- Final velocity \( v = 0 \, \text{m/s} \) (at \( x = 900 \, \text{m} \))
- Acceleration in the first \( \frac{1}{4} \) distance \( a_1 = +2.75 \, \text{m/s}^2 \)
- Acceleration in the remaining distance \( a_2 = -0.75 \, \text{m/s}^2 \)

### Step 1: Calculate the distance for each segment
1. **First Segment (A to C)**:
- Distance \( S_1 = \frac{1}{4} \times 900 = 225 \, \text{m} \)

2. **Second Segment (C to B)**:
- Distance \( S_2 = 900 - 225 = 675 \, \text{m} \)

### Step 2: Calculate the maximum speed at point C
Using the third equation of motion:
\[ v^2 = u^2 + 2aS \]
For the first segment (A to C):
- \( u = 0 \)
- \( a = 2.75 \, \text{m/s}^2 \)
- \( S = 225 \, \text{m} \)

Substituting the values:
\[ v^2 = 0 + 2 \times 2.75 \times 225 \]
\[ v^2 = 1237.5 \implies v = \sqrt{1237.5} \approx 35.2 \, \text{m/s} \]

### Step 3: Calculate the time taken for each segment
1. **Time for the first segment (A to C)**:
Using the equation:
\[ S = ut + \frac{1}{2} a t^2 \]
Since \( u = 0 \):
\[ 225 = \frac{1}{2} \times 2.75 \times t_1^2 \]
\[ t_1^2 = \frac{225 \times 2}{2.75} \approx 163.64 \implies t_1 \approx 12.8 \, \text{s} \]

2. **Time for the second segment (C to B)**:
Using the first equation of motion:
\[ v = u + at \]
Where:
- \( v = 0 \)
- \( u = 35.2 \, \text{m/s} \)
- \( a = -0.75 \, \text{m/s}^2 \)

Rearranging gives:
\[ 0 = 35.2 - 0.75 t_2 \implies t_2 = \frac{35.2}{0.75} \approx 46.93 \, \text{s} \]

### Step 4: Total time for the journey
\[ T = t_1 + t_2 = 12.8 + 46.93 \approx 59.73 \, \text{s} \]

### Final Answers
(a) The total travel time through the 900 m is approximately **59.73 seconds**.

(b) The maximum speed attained by the car is approximately **35.2 m/s**.

### Step 5: Graphs
1. **Acceleration vs Time Graph**:
- From \( t = 0 \) to \( t_1 \) (12.8 s), \( a = 2.75 \, \text{m/s}^2 \) (constant).
- From \( t_1 \) to \( t = 59.73 \, \text{s} \), \( a = -0.75 \, \text{m/s}^2 \) (constant).

2. **Velocity vs Time Graph**:
- From \( t = 0 \) to \( t_1 \), the velocity increases linearly from 0 to 35.2 m/s.
- From \( t_1 \) to \( t = 59.73 \, \text{s} \), the velocity decreases linearly from 35.2 m/s to 0.

3. **Position vs Time Graph**:
- From \( t = 0 \) to \( t_1 \), the position follows a quadratic curve (parabola opening upwards).
- From \( t_1 \) to \( t = 59.73 \, \text{s} \), the position curve will be a downward-opening parabola.