# Physics2.6 Calculate the average velocity of the car in different time interval

#### karush

##### Well-known member

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 hw before classes start on Aug 26

Mahalo

#### skeeter

##### Well-known member
MHB Math Helper
$x(t) = at^2-bt^3$

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

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

#### karush

##### Well-known member
$x(t) = at^2-bt^3$

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

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

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

Last edited:

#### skeeter

##### Well-known member
MHB Math Helper
$x(t)=1.50(t)^2-0.50(t)^3$

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

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

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