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Math Help - Mechanics one

  1. #1
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    Post Mechanics one

    the question says:

    calculate the magnitude and direction of the acceleration of a particle that moves so that its position vector in metres is given by

    r=(8t-24^2)i + (6+4t-t^2)

    where t is time in seconds.

    I think i have worked out the acceleration to be -4i -2j

    and so as a result is the magnitude is 4.47

    but i am unsure how to work out direction. My teacher whipped through it and so i got lost... can you help please?
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  2. #2
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    Grandad's Avatar
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    Vector acceleration

    Hello BeckyC
    Quote Originally Posted by BeckyC View Post
    the question says:

    calculate the magnitude and direction of the acceleration of a particle that moves so that its position vector in metres is given by

    r=(8t-24^2)i + (6+4t-t^2)j

    where t is time in seconds.

    I think i have worked out the acceleration to be -4i -2j

    and so as a result is the magnitude is 4.47

    but i am unsure how to work out direction. My teacher whipped through it and so i got lost... can you help please?
    I'm guessing from your answer so far that you have made a couple of typos in your question, and it should be

    \vec{r} = (8t-2t^2)\vec{i} + (6 + 4t -t^2)\vec{j}

    In which case, you are right: acceleration = \vec{\ddot{r}} = -4\vec{i} - 2\vec{j}

    and its magnitude = |\vec{\ddot{r}}| = \sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47

    To find the direction, just think about the vector -4\vec{i} - 2\vec{j}: this vector is -4 units in the \vec{i} direction (i.e. 4 units to the left) and -2 units in the \vec{j} direction (i.e. 2 units downward.)

    So, draw a right-angled triangle having one side horizontal and 4 units long, and at its left-hand end a side 2 units long vertically downwards. Join the hypotenuse: this side represents the actual acceleration vector. See the attached diagram.

    So if the acceleration makes an angle \theta below the horizontal, \tan\theta = \frac{2}{4}= 0.5 \Rightarrow \theta \approx 26.6^o.

    Grandad
    Attached Thumbnails Attached Thumbnails Mechanics one-untitled.jpg  
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello BeckyCI'm guessing from your answer so far that you have made a couple of typos in your question, and it should be

    \vec{r} = (8t-2t^2)\vec{i} + (6 + 4t -t^2)\vec{j}

    In which case, you are right: acceleration = \vec{\ddot{r}} = -4\vec{i} - 2\vec{j}

    and its magnitude = |\vec{\ddot{r}}| = \sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47

    To find the direction, just think about the vector -4\vec{i} - 2\vec{j}: this vector is -4 units in the \vec{i} direction (i.e. 4 units to the left) and -2 units in the \vec{j} direction (i.e. 2 units downward.)

    So, draw a right-angled triangle having one side horizontal and 4 units long, and at its left-hand end a side 2 units long vertically downwards. Join the hypotenuse: this side represents the actual acceleration vector. See the attached diagram.

    So if the acceleration makes an angle \theta below the horizontal, \tan\theta = \frac{2}{4}= 0.5 \Rightarrow \theta \approx 26.6^o.

    Grandad

    thank you so so much!
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