# What is the average velocity of a particle, whose position can be determined by x=10t^2, from t_"initial" = 2.0 " s" to t_"final"=3.0 " s" and again, from t_"initial"=2.0 " s" to t_"final"=2.1 " s"?

Mar 27, 2016

#### Answer:

From 2 to 3 seconds, the average velocity is $\overline{v} = 50 \text{ m/s}$ and from 2 to 2.1 seconds, the average velocity is $\overline{v} = 41 \text{ m/s}$.

#### Explanation:

The average velocity is defined as

$\overline{v} = \frac{\Delta x}{\Delta t} = \left({x}_{\text{final"-x_"initial")/(t_"final"-t_"initial}}\right)$

So for the time interval from ${t}_{\text{initial"=2.0 " s}}$ to ${t}_{\text{final"=3.0 " s}}$, we have ${x}_{\text{initial"=10(2.0)^2=40 " m}}$ and ${x}_{\text{final"=10(3.0)^2=90 " m}}$. Thus, the average velocity for this time interval is

bar(v)=(90 " m"- 40 " m")/(3.0 " s" - 2.0 " s")=(50 " m")/(1.0 " s")=50 " m/s"

For the time interval from ${t}_{\text{initial"=2.0 " s}}$ to ${t}_{\text{final"=2.1 " s}}$, we have ${x}_{\text{initial"=10(2.0)^2=40 " m}}$ and ${x}_{\text{final"=10(2.1)^2=44.1 " m}}$. Thus, the average velocity for this time interval is

bar(v)=(44.1 " m"- 40 " m")/(2.1 " s" - 2.0 " s")=(4.1 " m")/(0.1 " s")=41 " m/s"