6. Problem: 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 $100 \mathrm{ft} / \mathrm{s}$. At the same time, Edward started running away with an acceleration of $1 \mathrm{ft} / \mathrm{s}^{2}$. Assuming that air resistance causes the arrow to decelerate at $1 \mathrm{ft} / \mathrm{s}^{2}$ and that it does hit Edward, how fast was the arrow moving at the moment of impact (in $\mathrm{ft} / \mathrm{s}$ )?
Solution: We use the formula for distance, $d=\frac{1}{2} a t^{2}+v t+d_{0}$. Then after $t$ seconds, Edward is at location $1875+\frac{1}{2}(1)\left(t^{2}\right)$ from the prison. After $t$ seconds, the arrow is at location $\frac{1}{2}(-1)\left(t^{2}\right)+100 t$ from the prison. When the arrow hits Edward, both objects are at the same distance away from the tower. Hence $1875+\frac{1}{2}(1)\left(t^{2}\right)=\frac{1}{2}(-1)\left(t^{2}\right)+100 t$. Solving for $t$ yields $t^{2}-100 t+1875=0 \Rightarrow t=25$ or $t=75$. Then it must be $t=25$, because after the arrow hits Edward, he will stop running.
After 25 seconds, the arrow is moving at a velocity of $100-25(1)=75 \mathrm{ft} / \mathrm{s}$.