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

This problem seems incredibly simple to me, all I do is plug in $5$ for $t$ and I get $227.5$.
Simple, so I know that it went 227.5 meters in 5 seconds, simple enough. That gives me 45.5m/s except that is not the answer in the book. Where am I going wrong?
I am not sure what they mean by velocity, if they mean the average velocity or instaneous but since they didn't say I will use average since it is easier. 
$100+50(a+h)-4.9(a+h)^2 - (100+50-4.9a^2)) / h $   I get $(-9.8ah + 4.9h^2 - 4.9a^2) /h$
 A: The equation of the motion is
$$s=f(t) = 100 + 50t - 4.9t^2\text{ m}.$$
The velocity at the instant $t$ is by definition
$$\frac{ds}{dt}=f'(t) =50 - 4.9\times 2t\text{ m/s}.$$
At $t=5\text{ s}$
$$f'(5) =50 - 4.9\times 2\times 5=1\text{ m/s}.$$
Alternatively by the definition of the derivative, you get
$$\begin{aligned}f'(5)&=\lim_{h\rightarrow 0}\frac{f(5+h)-f(5)}{h}=\lim_{ h\rightarrow 0}\frac{
227.5+h-4.9h^{2}-227.5}{h} \\ &=\lim_{h\rightarrow 0}\;1-4.9h=1\text{ m/s}.\end{aligned}$$
Detailed computation
$$\begin{eqnarray*}
f(t) &=&100+50t-4.9t^{2} \\
f(5) &=&100+50\left( 5\right) -4.9\left( 5\right) ^{2}=227.\,5 \\
f(5+h) &=&100+50\left( 5+h\right) -4.9\left( 5+h\right) ^{2} \\
&=&227.5+h-4.9h^{2} \\
f(5+h)-f(5) &=&227.5+h-4.9h^{2}-227.\,5 \\
&=&h-4.9h^{2}
\end{eqnarray*}$$
and
$$\begin{equation*}
\frac{f(5+h)-f(5)}{h}=\frac{h-4.9h^{2}}{h}=\frac{h(1-4.9h)}{h}=1-4.9h.
\end{equation*}$$
Comment: The instantaneous velocity is the derivative of $s$, i.e. $f'(t)$. The average velocity between $t_1$ and $t_2$ is $\dfrac{f(t_2)−f(t_1)}{t_2−t_1}$.
A: When the problem says "find the velocity at $t=5$", they necessarily mean the instantaneous velocity (that is, the value of the derivative at $t=5$). This, because average velocities require two times: the initial time and the ending time.
If they were asking about the average velocity between $t=0$ and $t=5$, then the correct answer would be:
$$\begin{align*}
\text{avg velocity} &= \frac{\text{distance traveled}}{\text{time}} = \frac{\text{final position} - \text{initial position}}{5}\\
&= \frac{f(5)-f(0)}{5} = \frac{227.5 - 100}{5} = 25.5 \text{ m/s}.
\end{align*}$$
But they are not asking this. They are asking for the instantaneous velocity when $t=5$. That is, they are asking for $f'(5)$. That means computing
$$\begin{align*}
f'(t) &= \lim_{h\to 0}\frac{f(5+h)-f(5)}{h}\\
&\strut\\
&=\lim_{h\to 0}\frac{\bigl( 100 + 50(5+h) - 4.9(5+h)^2\bigr) - \bigl(100 + 50(5) - 4.9(5^2)\bigr)}{h}\\
&\strut\\
&=\lim_{h\to 0}\frac{\bigl( 100 + 250 + 50h - 4.9(25 + 10h + h^2)\bigr) - \bigl(100 + 250 - 4.9(25)\bigr)}{h}\\
&\strut\\
&=\lim_{h\to 0}\frac{\bigl( 350 + 50h - 122.5 - 49h - 4.9h^2\bigr) - \bigl( 350 - 122.5\bigr)}{h}\\
&\strut\\
&=\lim_{h\to 0}\frac{227.5 + h - 4.9h^2 - 227.5}{h}\\
&\strut\\
&=\lim_{h\to 0}\frac{h - 4.9h^2}{h}.
\end{align*}$$
Again, nothing but algebra so far. If we try evaluating at $h=0$, we get (as you might expect by now) $\frac{0}{0}$. But there is a very obvious factor of $h$ that can be cancelled, so we have:
$$f'(5) = \lim_{h\to 0}\frac{h-4.9h^2}{h} = \lim_{h\to 0}\frac{1-4.9h}{1} = 1-4.9(0) = 1.$$
So the instantaneous velocity at $t=5$ is $1$. What are the units? The units of $f$ are meters, the units of $t$ are seconds. Since $f'(t)$ is computed as a fraction of (values of $f$)/(values of $t$), it is measured in (units of $f$)/(units of $t$) = meters/sec. 
That is, the instantaneous velocity at $t=5$ is $1$ meter/sec.
What about the instantaneous speed? Velocity includes direction: positive velocity means movement left-to-right; negative velocity means movement right-to-left. Speed is just the size of the velocity. That is, the speed is the absolute value of the velocity (this is true for both average speed and for instantaneous speed). So the instantaneous speed at $t=5$ is $|f'(5)|$ meters per second, which in this case is $1$ meter per second, same as the velocity. 
A: To obtain the instantaneous velocity, compute the derivative $f'(t) = d f(t) / dt$. The velocity at $t=5$ is $f'(5)$.
