Answer
a)
s(t) = -16t$^{2}$ + 1362
v(t) = -32t
b)
-48 feet per second
c)
v(1) = -32 feet per second
v(2) = -64 feet per second
d)
$\frac{\sqrt(1362)}{4}$ seconds
e)
v($\frac{\sqrt(1362)}{4}$) = -295.242 feet per second
Work Step by Step
a)
The given position function is s(t) = -16t$^{2}$ + v$_{0}$t + s$_{0}$, where v$_{0}$ is the initial velocity of the object, and s$_{0}$ is the initial position of the object.
Since the silver dollar is being dropped from the top of a building, it has no initial velocity right as it is released, so v$_{0}$ = 0. We are given that the building is 1362 feet tall, meaning that the coin is being dropped from that height as well. Since this is the height that the coin is being dropped from, this is the coin's initial height, so s$_{0}$ = 1362. Plugging in all of our variables into the position function, we get that the position function of the coin is s(t) = -16t$^{2}$ + 1362
We are then asked to find the velocity function of the coin. Velocity is the derivative of position, so we can use the power rule of the position function we just derived. Using the power rule, we can find that v(t) = -32t + 0, which simplifies to v(t) = -32t
b)
The average velocity of a function is determined by the slope of its secant line. We are given the interval as [1, 2]. Thus, to find the secant line, we can use the equation $\frac{s(2) - s(1)}{2-1}$. Plugging in 2 into our position equation, we get -16(2)$^{2}$ + 1362, so s(2) = 1298. Similarly, plugging 1 into our position equation, we get -16(1)$^{2}$ + 1362, meaning s(1) = 1364. Plugging all of this into our equation for the secant line, we get $\frac{1298 - 1364}{2-1}$ = -48. The units are feet per second, since it is velocity.
c)
The instantaneous velocity of the coin at a certain point in time can be found using the velocity function which we derived in part a (v(t) = -32t).
v(1) = -32(1) = -32 feet per second
v(2) = -32(2) = -64 feet per second
d)
The coin reaches ground level when position is equal to zero. As such, we can set our position function (s(t) = -16t$^{2}$ + 1362) equal to zero.
-16t$^{2}$ + 1362 = 0
-16t$^{2}$ = -1362
t$^{2}$ = $\frac{-1362}{-16}$ = $\frac{1362}{16}$
t = $\frac{\sqrt 1362}{\sqrt 16}$ = $\frac{\sqrt 1362}{4}$ seconds
e)
In order to find the velocity of the coin at impact, we have to plug in the time at which the coin hits the ground into our velocity function. As such:
v($\frac{\sqrt 1362}{4}$) = -32($\frac{\sqrt 1362}{4}$) = -295.242 feet per second
