Speed as a function of distance traveled

Question a)

Data

The equation of distance traveled as a function of time:

$S\left(t\right)=t^2+5·t+1$

The instants of time where to study the speed:

t₁=0 sg y t₂=3 sg

Resolution

To calculate the average speed, we must use the following equation:

We know the times (t₁ and t₂) between which we will calculate the speed, but we do not know directly the distance that the body has traveled in each of them. However, since we have the function that determines how much distance it has traveled at every moment of time, it will be enough to substitute both values of t in the function to find out.

For t₁ = 0 s

$$\begin{gathered} S\left(0\right)=s_1=0^2+5·0+1 m \Rightarrow \\ \boxed{s_1=1 m} \end{gathered}$$

For t₂=3 sg

$$\begin{gathered} S\left(3\right)=s_2=3^2+5·3+1 m \Rightarrow \\ \boxed{s_2=25 m} \end{gathered}$$

At this point, we have all the necessary information to calculate the average speed in the interval between 0 and 3 seconds, substituting in the first equation:

$$\begin{gathered} V_{avg}=\frac{s_2-s_1}{t_2-t_1}\Rightarrow \\ {V}_{avg}=\frac{25-1}{3-0}m/s\Rightarrow \\ \boxed{V_{avg}=8.33 m/s} \end{gathered}$$

Question b)

Instantaneous speed is calculated by solving the following equation:

To calculate Δs, we will take the time t₁ = t, and t₂ = t + ΔtPara calcular Δs, vamos a tomar como tiempo t₁=t y t₂=t+Δt:

 

$$\begin{gathered} s(t+\Delta t)=(t+\Delta t{)}^2+5·(t+\Delta t)+1 \\ s(t)=t^2+5t+1 \\ \Delta s=s(t+\Delta t)-s(t)=(t+\Delta t{)}^2+5·(t+\Delta t)+1 -(t^2+5t+1) \Rightarrow \\ \boxed{\Delta s=\Delta t^2+2·t·\Delta t+5·\Delta t} \end{gathered}$$

By applying the formula of instantaneous speed:

$$\begin{gathered} V\left(t\right)=\underset{\Delta t\to 0}{\lim }\frac{\Delta t^2+2·t·\Delta t+5·\Delta t}{\Delta t}\Rightarrow \\ \boxed{V\left(t\right)=2·t+5} \end{gathered}$$