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

karush

Well-known member
Jan 31, 2012
2,724
cp_2_6.PNG

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
Mar 1, 2012
702
$x(t) = at^2-bt^3$

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

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

karush

Well-known member
Jan 31, 2012
2,724
$x(t) = at^2-bt^3$

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

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
Mar 1, 2012
702
$x(t)=1.50(t)^2-0.50(t)^3$

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

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

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