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"#?

1 Answer
Mar 27, 2016

Answer:

From 2 to 3 seconds, the average velocity is #bar(v)=50 " m/s"# and from 2 to 2.1 seconds, the average velocity is #bar(v)=41 " m/s"#.

Explanation:

The average velocity is defined as

#bar(v)=(Deltax)/(Deltat)=(x_"final"-x_"initial")/(t_"final"-t_"initial")#

So for the time interval from #t_"initial"=2.0 " s"# to #t_"final"=3.0 " s"#, we have #x_"initial"=10(2.0)^2=40 " m"# and #x_"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_"initial"=2.0 " s"# to #t_"final"=2.1 " s"#, we have #x_"initial"=10(2.0)^2=40 " m"# and #x_"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"#