Edward, the author of this test, had to escape from prison to work in the grading room today. He stopped to rest at a place $1,875$ feet from the prison and was spotted by a guard with a crossbow. 

The guard fired an arrow with an initial velocity of $100 \dfrac{\text{ft}}{\text{s}}$.  At the same time, Edward started running away with an acceleration of $1 \dfrac{\text{ft}}{\text{s}^2}$.  Assuming that air resistance causes the arrow to decelerate at $1 \dfrac{\text{ft}}{\text{s}^2}$, and that it does hit Edward, how fast was the arrow moving at the moment of impact (in $\dfrac{\text{ft}}{\text{s}}$)?
1. **Define the position functions for the arrow and Edward:**
   - Let \( a(t) \) be the position of the arrow at time \( t \).
   - Let \( e(t) \) be the position of Edward at time \( t \).

2. **Determine the velocity and acceleration functions:**
   - The acceleration of the arrow is given as \( a''(t) = -1 \, \text{ft/s}^2 \).
   - Integrate the acceleration to find the velocity:
     \[
     a'(t) = \int a''(t) \, dt = \int -1 \, dt = -t + C
     \]
     Given the initial velocity of the arrow is \( 100 \, \text{ft/s} \) at \( t = 0 \):
     \[
     a'(0) = 100 \implies -0 + C = 100 \implies C = 100
     \]
     Therefore, the velocity of the arrow is:
     \[
     a'(t) = -t + 100
     \]

3. **Integrate the velocity to find the position of the arrow:**
   \[
   a(t) = \int a'(t) \, dt = \int (-t + 100) \, dt = -\frac{1}{2}t^2 + 100t + C
   \]
   Assuming the arrow starts from the origin (0 feet):
   \[
   a(0) = 0 \implies -\frac{1}{2}(0)^2 + 100(0) + C = 0 \implies C = 0
   \]
   Therefore, the position of the arrow is:
   \[
   a(t) = -\frac{1}{2}t^2 + 100t
   \]

4. **Determine the velocity and position functions for Edward:**
   - The acceleration of Edward is given as \( e''(t) = 1 \, \text{ft/s}^2 \).
   - Integrate the acceleration to find the velocity:
     \[
     e'(t) = \int e''(t) \, dt = \int 1 \, dt = t + C
     \]
     Given Edward starts from rest:
     \[
     e'(0) = 0 \implies 0 + C = 0 \implies C = 0
     \]
     Therefore, the velocity of Edward is:
     \[
     e'(t) = t
     \]

5. **Integrate the velocity to find the position of Edward:**
   \[
   e(t) = \int e'(t) \, dt = \int t \, dt = \frac{1}{2}t^2 + C
   \]
   Given Edward starts 1875 feet from the prison:
   \[
   e(0) = 1875 \implies \frac{1}{2}(0)^2 + C = 1875 \implies C = 1875
   \]
   Therefore, the position of Edward is:
   \[
   e(t) = \frac{1}{2}t^2 + 1875
   \]

6. **Find the time \( t \) when the arrow hits Edward:**
   - Set the positions equal to each other:
     \[
     a(t) = e(t) \implies -\frac{1}{2}t^2 + 100t = \frac{1}{2}t^2 + 1875
     \]
   - Combine and simplify:
     \[
     -\frac{1}{2}t^2 + 100t = \frac{1}{2}t^2 + 1875 \implies -t^2 + 100t = 1875 \implies t^2 - 100t + 1875 = 0
     \]
   - Solve the quadratic equation:
     \[
     t = \frac{100 \pm \sqrt{100^2 - 4 \cdot 1 \cdot 1875}}{2 \cdot 1} = \frac{100 \pm \sqrt{10000 - 7500}}{2} = \frac{100 \pm \sqrt{2500}}{2} = \frac{100 \pm 50}{2}
     \]
     Therefore, the solutions are:
     \[
     t = \frac{150}{2} = 75 \quad \text{and} \quad t = \frac{50}{2} = 25
     \]
   - We take the first time \( t = 25 \) seconds when the arrow first hits Edward.

7. **Calculate the velocity of the arrow at \( t = 25 \):**
   \[
   a'(25) = -25 + 100 = 75 \, \text{ft/s}
   \]

The final answer is \(\boxed{75 \, \text{ft/s}}\)