# Derivation of Distance to Get Velocity and Acceleration

Derivatives reveal a new connection between distance $s\left(t\right)$, velocity $v\left(t\right)$ and acceleration $a\left(t\right)$. By differentiating, you can find an expression for velocity $v\left(t\right)$ and acceleration $a\left(t\right)$ using the provided distance $s\left(t\right)$. This is how they are connected:

Rule

### Distance,VelocityandAcceleration

If you have the position (distance) $s\left(t\right)$, then it follows that $\begin{array}{llll}\hfill v\left(t\right)& ={s}^{\prime }\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill a\left(t\right)& ={v}^{\prime }\left(t\right)={s}^{″}\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 1

A plane flies from London to New York. The aircraft’s position above the Atlantic is given by

 $s\left(t\right)=\text{}-8.27\text{}{t}^{3}+\text{}111.81\text{}{t}^{2}+\text{}302.82\text{}t+\text{}400.9\text{},$

 $s\left(t\right)=\text{}-8.27\text{}{t}^{3}+\text{}111.81\text{}{t}^{2}+\text{}302.82\text{}t+\text{}400.9\text{},$

where $t$ is the number of hours after departure, and the unit of $s\left(t\right)$ is kilometers.
1.
How far has the plane flown after four hours?
2.
What is the velocity of the plane after four hours?
3.
Find the acceleration of the plane after four hours.
1.
You insert $t=4$ into $s\left(t\right)$:
$\begin{array}{llll}\hfill s\left(4\right)& =-8.27\cdot {4}^{3}+111.81\cdot {4}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}+302.82\cdot 4+400.9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2871.86\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill s\left(4\right)& =-8.27\cdot {4}^{3}+111.81\cdot {4}^{2}+302.82\cdot 4+400.9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2871.86\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

After four hours, the plane has flown $2871.86$ km.
2.
To find the velocity of the plane after four hours, you have to find the velocity $v\left(t\right)$ of the plane. You find that by differentiating $s\left(t\right)$:
$\begin{array}{llll}\hfill v\left(t\right)& ={s}^{\prime }\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-24.81{t}^{2}+223.62t\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}+302.82\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill v\left(4\right)& ={s}^{\prime }\left(4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-24.81\cdot {4}^{2}+223.62\cdot 4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}+302.82\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =800.34\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 800\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill v\left(t\right)& ={s}^{\prime }\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-24.81{t}^{2}+223.62t+302.82\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill v\left(4\right)& ={s}^{\prime }\left(4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-24.81\cdot {4}^{2}+223.62\cdot 4+302.82\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =800.34\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 800\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

After four hours, the velocity of the plane is about $800$ km/h.
3.
To find out what the acceleration is after four hours, you have to differentiate the velocity function $v\left(t\right)$. That gives you $\begin{array}{llll}\hfill a\left(t\right)& ={v}^{\prime }\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={s}^{″}\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-49.62t+223.62\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill a\left(4\right)& ={v}^{\prime }\left(4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={s}^{″}\left(4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-49.62\cdot 4+223.62\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =25.14\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

After four hours, the acceleration of the plane is $25.14$ km/h2. This tells you that the plane is increasing its speed four hours into the flight.