Answer
$4.88$ ft/(mph),
$8.66$ ft/(mph), see explanations.
Work Step by Step
Step 1. Given the equation $s(v)=1.1v+0.054v^2$, we have $\frac{ds}{dv}=1.1+0.108v$
Step 2. At $v=35$mph, we have $\frac{ds}{dv}=1.1+0.108(35)=4.88$ ft/(mph); this means that for every 1mph increase of speed at 35mph, the increase in stopping distance is 4.88 ft.
Step 3. At $v=70$mph, we have $\frac{ds}{dv}=1.1+0.108(70)=8.66$ ft/(mph); this means that for every 1mph increase of speed at 70mph, the increase in stopping distance is 8.66 ft.
