cards","topicId":"838a8dd7-8e44-4e21-9a72-584d0314efac","topicSlug":"vectors-in-kinematics","cards":[{"id":"6fde1118-2e24-497e-9058-33ed4f8d7d49","cardType":"basic","front":"In two-dimensional kinematics, which direction does $\\mathbf{i}$ represent?","back":"The positive horizontal direction.","points":1,"mcqOptions":null},{"id":"7d2e484e-28ec-4d07-ac0d-40842c39d8ce","cardType":"basic","front":"In two-dimensional kinematics, which direction does $\\mathbf{j}$ represent?","back":"The positive vertical direction.","points":1,"mcqOptions":null},{"id":"3ba51515-aecb-45d3-ad7e-8c2a201a8e1f","cardType":"basic","front":"Why can velocity be negative in a chosen direction while speed cannot be negative?","back":"Velocity has direction; speed is a non-negative magnitude.","points":2,"mcqOptions":null},{"id":"d8e27ba9-44d0-4be4-adb0-570bc3d2bafe","cardType":"cloze","front":"For $\\mathbf{a}=a_x\\mathbf{i}+a_y\\mathbf{j}$, its magnitude is $\\text{[...]}$ and its direction satisfies $\\tan\\theta=\\frac{a_y}{a_x}$.","back":"For $\\mathbf{a}=a_x\\mathbf{i}+a_y\\mathbf{j}$, its magnitude is $\\boxed{|\\mathbf{a}|=\\sqrt{a_x^2+a_y^2}}$ and its direction satisfies $\\tan\\theta=\\frac{a_y}{a_x}$.","points":null,"mcqOptions":null},{"id":"35125106-c464-4d58-90c5-a55f706fcfdf","cardType":"cloze","front":"For $\\mathbf{a}=a_x\\mathbf{i}+a_y\\mathbf{j}$, its magnitude is $|\\mathbf{a}|=\\sqrt{a_x^2+a_y^2}$ and its direction satisfies $\\text{[...]}$.","back":"For $\\mathbf{a}=a_x\\mathbf{i}+a_y\\mathbf{j}$, its magnitude is $|\\mathbf{a}|=\\sqrt{a_x^2+a_y^2}$ and its direction satisfies $\\boxed{\\tan\\theta=\\frac{a_y}{a_x}}$.","points":null,"mcqOptions":null},{"id":"6708b8d3-eff4-4aac-adf4-8966b8893e0b","cardType":"basic","front":"What is the displacement from position $\\mathbf{r}_1$ to position $\\mathbf{r}_2$?","back":"$\\mathbf{r}_2-\\mathbf{r}_1$.","points":1,"mcqOptions":null},{"id":"a6f50b6c-0573-4d8b-a107-997bdefe76cd","cardType":"basic","front":"Why can the magnitude of displacement be less than the distance travelled?","back":"Displacement uses only the initial and final positions, not the path.","points":2,"mcqOptions":null},{"id":"27619e6a-dd88-45e8-bdb7-02b0812297c0","cardType":"basic","front":"How is the velocity vector oriented relative to a particle's path?","back":"It is tangent to the path in the instantaneous direction of motion.","points":1,"mcqOptions":null},{"id":"95d0ba5f-997a-4935-a958-b19ab1ea4e52","cardType":"cloze","front":"The differentiation relationships are $\\text{[...]}$ and $\\mathbf{a}=\\frac{d\\mathbf{v}}{dt}=\\frac{d^2\\mathbf{r}}{dt^2}$.","back":"The differentiation relationships are $\\boxed{\\mathbf{v}=\\frac{d\\mathbf{r}}{dt}}$ and $\\mathbf{a}=\\frac{d\\mathbf{v}}{dt}=\\frac{d^2\\mathbf{r}}{dt^2}$.","points":null,"mcqOptions":null},{"id":"1f6f0e85-68ae-4ea7-8fb7-d01f86b54a97","cardType":"cloze","front":"The differentiation relationships are $\\mathbf{v}=\\frac{d\\mathbf{r}}{dt}$ and $\\text{[...]}$.","back":"The differentiation relationships are $\\mathbf{v}=\\frac{d\\mathbf{r}}{dt}$ and $\\boxed{\\mathbf{a}=\\frac{d\\mathbf{v}}{dt}=\\frac{d^2\\mathbf{r}}{dt^2}}$.","points":null,"mcqOptions":null},{"id":"62c33a90-5ac6-4c81-a036-d2a2656e86d7","cardType":"basic","front":"How do you obtain acceleration directly from a position vector $\\mathbf{r}(t)$?","back":"Differentiate each component twice with respect to time.","points":2,"mcqOptions":null},{"id":"67b85ec0-7cff-4144-a1fd-bc39d825299f","cardType":"basic","front":"Why does vector integration require a constant in each component?","back":"Horizontal and vertical components can have different initial values.","points":2,"mcqOptions":null},{"id":"6e93e6c4-9c16-4c06-a31e-771547aa035a","cardType":"cloze","front":"For constant acceleration, final velocity is $\\text{[...]}$ and displacement is $\\mathbf{s}=\\mathbf{u}t+\\frac{1}{2}\\mathbf{a}t^2$.","back":"For constant acceleration, final velocity is $

1.10.8 Vectors in kinematics Flashcards

1.10.8 Vectors in kinematics Flashcards