1. ## Velocity vectors helpp

You program a dot on a computer screen to have a position given by $\displaystyle \vec{r}=(1.5cm + 2.0cm/s^2t^2)\hat{i}+((3.0m/s)t)\hat{j}$. At time $\displaystyle t_{10} > 0$ the dot's displacement from its t = 0 position has a magnitude 10.0cm a.) Find the magnitude and direction of the average velocity of the dot between t = 0 and $\displaystyle t_{10}$. b.) find the magnitude and direction of the instantaneous velocity at t = 0 and at $\displaystyle t_{10}$

a.) 5.0cm/s 37 degrees
b.) t = 0: 3.0cm/s, 90 degrees
t_{10} : 8.5cm/s,21 degrees

im stuck with the letter a what i did was substitute 10 as t and got 201.5 as V2 and i got 19.2cm/s as the average velocity
HELP

You program a dot on a computer screen to have a position given by $\displaystyle \vec{r}=(1.5cm + 2.0cm/s^2t^2)\hat{i}+((3.0m/s)t)\hat{j}$. At time $\displaystyle t_{10} > 0$ the dot's displacement from its t = 0 position has a magnitude 10.0cm a.) Find the magnitude and direction of the average velocity of the dot between t = 0 and $\displaystyle t_{10}$. b.) find the magnitude and direction of the instantaneous velocity at t = 0 and at $\displaystyle t_{10}$

a.) 5.0cm/s 37 degrees
b.) t = 0: 3.0cm/s, 90 degrees
t_{10} : 8.5cm/s,21 degrees

im stuck with the letter a what i did was substitute 10 as t and got 201.5 as V2 and i got 19.2cm/s as the average velocity
HELP
First write the expreasion for $\displaystyle \vec{r}$ more clearly (if necessary remove the units from inside the expression).

The average velocity is $\displaystyle \vec{v}=[\vec{r}(10) - \vec{r}(0)]/10$. Now you need to express this in amplitude angle form.

RonL

3. ok...

i dunno whats the meaning of 10.0cm at t = 0
is it the components of the x and y of the displacement?
(Vave)x = ((x-component) - 10cm) / 10
(Vave)y = ((y-component) - 10cm) / 10

idunno

ok...

i dunno whats the meaning of 10.0cm at t = 0
is it the components of the x and y of the displacement?
(Vave)x = ((x-component) - 10cm) / 10
(Vave)y = ((y-component) - 10cm) / 10

idunno
OK I think I see what is going on better now. $\displaystyle t_{10}$ is some unknown time $\displaystyle >0$ (not $\displaystyle t=10$ as I had assumed). We have to determine what $\displaystyle t_{10}$ is from the condition:

$\displaystyle |\vec{r}(t_{10}) - \vec{r}(0)|=10 \mbox{ cm}$

That is the magnitude of the displacement from $\displaystyle \vec{r}(0)$ at $\displaystyle t_{10}$ is $\displaystyle 10\mbox{ cm}$.

RonL