Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 14

Suppose that a rock is thrown upward on the planet Mars with a velocity of 10$m/s$, its height (in meters) after $t$ seconds is given
by $H = 10 t - 1.86 t ^2$
a.) Find the velocity of the rock after one second.
b.) Find the velocity of the rock when $t=a$.
c.) At what time will the rock hit the surface?
d.) At what velocity will the rock hit the surface?

a.) From the definition of instantaneous velocity,

$
\displaystyle
\nu ( a ) = \lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}\\
$



$
\begin{equation}
\begin{aligned}
\qquad
f(t)
& = 10t-1.86t^2\\
\nu(a)
& = \lim \limits_{h \to 0} \frac{10(a+h) - 1.86(a+h)^2 - [10(a) - 1.86(a)^2]}{h}\\
\nu(a)
& = \lim \limits_{h \to 0} \frac{\cancel{10a} + 10h - \cancel{1.86a^2} - 3.72ah - 1.86h^2 - \cancel{10a} + \cancel{1.86a^2}}{h}\\
\nu(a)
& = \lim \limits_{h \to 0} \frac{\cancel{h}(10-3.72a-1.86h)}{\cancel{h}}\\
\nu(a)
& = \lim \limits_{h \to 0} (10 - 3.72a - 1.86h)\\
\nu(a)
& = 10 - 3.72a - 1.86(0)\\
\nu(a)
& = 10-3.72a
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{The velocity after } 1s \text{ is } v(1) & = 10 - 3.72(1) = 6.28 m/s\\
\end{aligned}
\end{equation}
$


b.) The velocity of the rock when $t=a$ is $\nu(a) = 10 - 3.72a$
c.) The rock will hit the surface at $H(t) = 0$ so,

$
\begin{equation}
\begin{aligned}
0 = & 10 t - 1.86t^2\\
\frac{1.86 \cancel{t^2}}{\cancel{t}} = & \frac{10 \cancel{t}}{\cancel{t}}\\
t = & \frac{10}{1.86} s
\end{aligned}
\end{equation}
$


d.) The velocity when the rock hits the surface...

$
\begin{equation}
\begin{aligned}
\nu (t) & = 10 - 3.72t && ;\text{ where } t = \frac{10}{1.86}\\
\nu & = 10 - 3.72 \left( \frac{10}{1.86} \right)\\
\nu & = -10 \frac{m}{s} &&; \text{ velocity is negative since the rock is falling.}
\end{aligned}
\end{equation}
$