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#### karush

##### Well-known member

- Jan 31, 2012

- 3,004

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 3,004

- Mar 1, 2012

- 980

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

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

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- Jan 31, 2012

- 3,004

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

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:

- Mar 1, 2012

- 980

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

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

so if $t_2=2$ and $t_1=0$ then

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

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