University Calculus: Early Transcendentals (3rd Edition)

- For $v=35$, $ds/dv=4.88$ feet/mph. - For $v=70$, $ds/dv=8.66$ feet/mph. The derivative represents the change in the stopping distance of the car at different speed.
$$s=1.1v+0.054v^2$$ Find $ds/dv$: $$\frac{ds}{dv}=(1.1v+0.054v^2)'=1.1+0.108v(ft/mph)$$ - For $v=35$: $$\frac{ds}{dv}=1.1+0.108\times35=4.88(ft/mph)$$ - For $v=70$: $$\frac{ds}{dv}=1.1+0.108\times70=8.66(ft/mph)$$ Since $s$ represents the total stopping distance of a moving car, its derivative will represent the change in the stopping distance of the car at different speed. To put simply, the faster the car is moving, the greater distance the car needs to stop when a barrier is perceived; the relationship is direct proportional, and that is represented in $ds/dv$.