Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 37

Suppose that a particle moves along a straight line with equation of motion
$s = f(t) = 100 + 50 t - 4.9 t^2$ , where $s$ is measured in meters and $t$
in seconds. Determine the velocity and the speed when $t=5$.



Based from the definition of instantenous velocity,



$
\quad
\begin{equation}
\begin{aligned}
\nu (a) &= \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}\\
f(t) &= 100 + 50t - 4.9t^2
\end{aligned}
\end{equation}
$



$
\quad
\begin{equation}
\begin{aligned}
\nu(t) &= \lim\limits_{h \to 0} \frac{100+50(t+h)-4.9(t+h)^2 - [100+50(t)-4.9(t)^2]}{h}\\
\nu(t) &= \lim\limits_{h \to 0} \frac{ \cancel{100} + \cancel{50t} + 50h - \cancel{4.9t^2} - 9.8th - 4.9h^2 - \cancel{100} - \cancel{50t} + \cancel{4.9t^2} }{h}\\
\nu(t) &= \lim\limits_{h \to 0} \frac{50h-9.8th-4.9h^2}{h}\\
\nu(t) &= \lim\limits_{h \to 0} \frac{\cancel{h} (50-9.8t - 4.9h)}{\cancel{h}}\\
\nu(t) &= \lim\limits_{h \to 0} (50- 9.8t - 4.9h)\\
\nu(t) &= 50-9.8t-4.9(0)\\
\nu(t) &= 50-9.8t
\end{aligned}
\end{equation}
$



The velocity after $5s$ is $\displaystyle \nu(5) = 50 - 9.8(5) = 1 \frac{m}{s}$



The velocity and speed have equal values except that the velocity is referred as a vector quantity which has
magnitude and direction.


Therefore, the speed and velocity of the particle are $\displaystyle 1 \frac{m}{s}$ and $\displaystyle 1 \frac{m}{s}$
east respectively. Assuming that the particle is moving to the east.