Problem Analysis:

We are given:

• Distance $S = 87.5 \text{ m}$
• Initial velocity $v_1 = 5.0 \text{ m/s}$
• Final velocity $v_2 = 10 \text{ m/s}$
• Acceleration constraint $a(t) \le a_{max} = 1.0 \text{ m/s}^2$
• Acceleration is unidirectional (always positive).

We need to find the range of average acceleration $a_{av}$.

$$ a_{av} = \frac{v_2 – v_1}{\Delta t} = \frac{5}{\Delta t} $$

To find the range of $a_{av}$, we need the range of possible time intervals $\Delta t$.

Calculation Details:

Case 1: Minimum Time (Maximum Average Acceleration)

Strategy: Accelerate early to maximize velocity over the interval.

• Phase 1: Accel from 5 to 10 m/s at $a = 1 \text{ m/s}^2$.
  $\Delta t_1 = \frac{10-5}{1} = 5 \text{ s}$.
  Distance $s_1 = \frac{10^2 – 5^2}{2(1)} = \frac{75}{2} = 37.5 \text{ m}$.
• Phase 2: Constant velocity 10 m/s for remaining distance.
  Remaining distance $s_2 = 87.5 – 37.5 = 50 \text{ m}$.
  $\Delta t_2 = \frac{50}{10} = 5 \text{ s}$.
• Total Time $T_{min} = 10 \text{ s}$.
• $a_{av, max} = \frac{5 \text{ m/s}}{10 \text{ s}} = 0.5 \text{ m/s}^2$.

Case 2: Maximum Time (Minimum Average Acceleration)

Strategy: Accelerate as late as possible to keep velocity low.

• Phase 1: Constant velocity 5 m/s.
  Need to leave enough distance for the final acceleration. Acceleration distance is fixed at 37.5 m (at max acceleration for shortest distance during speed change).
  Distance $s_1 = 87.5 – 37.5 = 50 \text{ m}$.
  $\Delta t_1 = \frac{50}{5} = 10 \text{ s}$.
• Phase 2: Accel from 5 to 10 m/s at $a = 1 \text{ m/s}^2$.
  $\Delta t_2 = 5 \text{ s}$.
• Total Time $T_{max} = 15 \text{ s}$.
• $a_{av, min} = \frac{5 \text{ m/s}}{15 \text{ s}} = 0.33 \text{ m/s}^2$.

Result: $$ 0.33 \text{ m/s}^2 \le a_{av} \le 0.5 \text{ m/s}^2 $$