속도와 거리

(a) Setting $v(t) = 0$, we find $500 - 32t = 0$, and thus $t = \frac{500}{32} = \frac{125}{8} = 15.625$. This is the time that the rocket's velocity changes from positive to negative, and thus the point when the rocket's position changes from increasing to decreasing: it's when the rocket reaches its maximum height. (b) Since $v$ is a positive linear function that connects the points $(0,500)$ and $(15.625,0)$ on the interval $[0, 15.625]$, the region enclosed by the velocity function and the $t$-axis is a right triangle with vertical leg of length $500$ and horizontal leg of length $15.625$. The area of this triangle is $A = \frac{1}{2}bh = \frac{1}{2}(15.625)(500) = 3906.25$. Because this is the area under the velocity curve (where velocity is positive) this area represents the vertical distance traveled by the rocket on the time interval $[0, 15.625]$. (c) One function that has derivative $v(t) = s'(t) = 500-32t$ is $s(t) = 500t - 16t^2$. (d) Using our result in (c), $s(15.625) - s(0) = 500(15.625) - 16(15.625)^2 - (0 - 0) = 3906.25$. Thus, the change of the rocket's position on $[0,15.625]$ is $3906.25$ feet, which is also the distance the rocket traveled vertically on the time interval, since velocity is positive throughout. (e) We find that $s(5) - s(1) = 1616$, which tells us that the rocket rose $1616$ feet on the $4$-second interval from $t = 1$ to $t = 5$.