MHB 2.6 Calculate the average velocity of the car in different time interval

[karush]

View attachment 9199

OK I just had time to post and hopefully ok but still typos maybe
the graph was done in Deimos wanted to try tikx but not sure about the polynomial

trying to as many physics homework before classes start on Aug 26

Mahalo

 

[skeeter]

$x(t) = at^2-bt^3$

$\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x(t_2)-x(t_1)}{t_2-t_1}$

units will be in $m/s$, not $m/s^2$ ... try again

 

[karush]

$x(t) = at^2-bt^3$

$\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x(t_2)-x(t_1)}{t_2-t_1}$

units will be in $m/s$, not $m/s^2$ ... try again

$x(t)=1.50(t)^2-0.50(t)^3$

$\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x(t_2)-x(t_1)}{t_2-t_1}$
so if $t_2=2$ and $t_1=0$ then

$\bar{v}=\dfrac{x(2)-x(0)}{2-0}=\dfrac{1.50(2)^2-0.50(2)^3-0}{2}=\dfrac{6-4}{2}=2\, m/s$

 

[skeeter]

$x(t)=1.50(t)^2-0.50(t)^3$

$\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x(t_2)-x(t_1)}{t_2-t_1}$
so if $t_2=2$ and $t_1=0$ then

$\bar{v}=\dfrac{x(2)-x(0)}{2-0}=\dfrac{1.50(2)^2-0.50(2)^3-0}{2}=\dfrac{6-4}{2}=2\, m/s$

$\dfrac{6-4}{2} = \dfrac{2}{2} = 1 \, m/s$