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 $(-9.8ah + 4.9h^2 – 4.9a^2) /h$
Best Answer
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 $\dfrac{f(t_2)−f(t_1)}{t_2−t_1}$.