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Math Help - Velocity vectors helpp

  1. #1
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    Velocity vectors helpp

    You program a dot on a computer screen to have a position given by \vec{r}=(1.5cm + 2.0cm/s^2t^2)\hat{i}+((3.0m/s)t)\hat{j}. At time 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 t_{10}. b.) find the magnitude and direction of the instantaneous velocity at t = 0 and at t_{10}

    the answer is
    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
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    You program a dot on a computer screen to have a position given by \vec{r}=(1.5cm + 2.0cm/s^2t^2)\hat{i}+((3.0m/s)t)\hat{j}. At time 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 t_{10}. b.) find the magnitude and direction of the instantaneous velocity at t = 0 and at t_{10}

    the answer is
    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 \vec{r} more clearly (if necessary remove the units from inside the expression).

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

    RonL
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  3. #3
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    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
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    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. t_{10} is some unknown time >0 (not t=10 as I had assumed). We have to determine what t_{10} is from the condition:

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

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


    RonL
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