The motion of the bus can be divided into three phases:

1. Acceleration from rest for 5 s at 2.1 \(\text{m/s}^2\):

   The final velocity \(v = u + at = 0 + 2.1 \times 5 = 10.5 \text{ m/s}\).

2. Constant speed for 24 s:

   The velocity remains 10.5 \(\text{m/s}\).

3. Deceleration to stop in 6 s:

   The initial velocity for this phase is 10.5 \(\text{m/s}\) and the final velocity is 0 \(\text{m/s}\).

The velocity-time graph is a trapezium with vertices at \((0, 0)\), \((5, 10.5)\), \((29, 10.5)\), and \((35, 0)\).

The area under the velocity-time graph gives the total distance traveled:

Area = \(\frac{1}{2} \times (24 + 35) \times 10.5\) or \(\frac{1}{2} \times 5 \times 10.5 + 24 \times 10.5 + \frac{1}{2} \times 6 \times 10.5\).

Calculating the area:

Area = \(\frac{1}{2} \times 59 \times 10.5 = 309.75 \text{ m}\).

Thus, the distance between the two bus stops is 309.75 m or 310 m.