속도와 거리

$18{s}^{2}=6{\int }_{s}^{2s}2xdx$

The net increase is 1 unit.

In dry conditions, with initial velocity ${v}_{0}=30$ m/s, $D=64.3$ and, if ${v}_{0}=25,D=44.64.$ In wet conditions, if ${v}_{0}=30,$ and $D=180$ and if ${v}_{0}=25,D=125.$

(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 $t = 5$.

(a) The plot below shows that the rate of pollution appears to be increasing at an increasing rate as time goes on. (b) If we think about an interval on which the rate of pollution is relatively constant and suppose the interval is $\triangle t$ wide, then the area under the curve on that interval is approximately the area of the rectangle whose width is $\triangle t$ and whose height is $p(t)$, the polution rate on that interval. When we compute the product $p(t) \cdot \triangle t$ to find the area, the units involved are “units of pollution per day” times “days”, which results in an overall unit of “units of pollution”. Thus, the area under the curve tells us the total pollution emitted over the time interval $[0,30]$. (c) One way to estimate the total amount of pollution is to suppose the rate of pollution is constant on each of the intervals $[0,6]$, $[6,12]$, $\ldots$. For instance, we could assume the rate on $[0,6]$ is always $7$ units per day, and then the rate jumps to $8$ units per day on $[6,12]$, and so on. Doing so, and using the ideas we discussed in (b), one estimate for the total pollution, $P$, is $$P = 7 \cdot 6 + 8 \cdot 6 + 10 \cdot 6 + 13 \cdot 6 + 18 \cdot 6 = 336$$ total units of pollution. Note, for example, that when we compute $13 \cdot 6$ this represents an estimate of the total amount of pollution emitted on the interval $[18,24]$ since we are assuming the plant emits $13$ units of pollution per day for 6 days. It is apparent that this particular estimate is an underestimate of the total amount of pollution since the rate is always increasing but we are choosing the lowest value of that rate on a given interval.